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Elementary Linear Algebra

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Page 1: Elementary Linear Algebra

ELEMENTARY

LINEAR ALGEBRA

K� R� MATTHEWS

DEPARTMENT OF MATHEMATICS

UNIVERSITY OF QUEENSLAND

Second Online Version� December ����

Comments to the author at krm�maths�uq�edu�au

All contents copyright c����� Keith R� Matthews

Department of Mathematics

University of Queensland

All rights reserved

Page 2: Elementary Linear Algebra

Contents

� LINEAR EQUATIONS �

��� Introduction to linear equations � � � � � � � � � � � � � � � � � �

��� Solving linear equations � � � � � � � � � � � � � � � � � � � � � �

��� The Gauss�Jordan algorithm � � � � � � � � � � � � � � � � � � �

�� Systematic solution of linear systems� � � � � � � � � � � � � � �

�� Homogeneous systems � � � � � � � � � � � � � � � � � � � � � � ��

��� PROBLEMS � � � � � � � � � � � � � � � � � � � � � � � � � � � ��

� MATRICES ��

��� Matrix arithmetic � � � � � � � � � � � � � � � � � � � � � � � � � ��

��� Linear transformations � � � � � � � � � � � � � � � � � � � � � � ��

��� Recurrence relations � � � � � � � � � � � � � � � � � � � � � � � ��

�� PROBLEMS � � � � � � � � � � � � � � � � � � � � � � � � � � � ��

�� Non�singular matrices � � � � � � � � � � � � � � � � � � � � � � ��

��� Least squares solution of equations � � � � � � � � � � � � � � � �

��� PROBLEMS � � � � � � � � � � � � � � � � � � � � � � � � � � � �

� SUBSPACES ��

��� Introduction � � � � � � � � � � � � � � � � � � � � � � � � � � � �

��� Subspaces of Fn � � � � � � � � � � � � � � � � � � � � � � � � �

��� Linear dependence � � � � � � � � � � � � � � � � � � � � � � � � �

�� Basis of a subspace � � � � � � � � � � � � � � � � � � � � � � � � ��

�� Rank and nullity of a matrix � � � � � � � � � � � � � � � � � � �

��� PROBLEMS � � � � � � � � � � � � � � � � � � � � � � � � � � � ��

� DETERMINANTS ��

�� PROBLEMS � � � � � � � � � � � � � � � � � � � � � � � � � � � �

i

Page 3: Elementary Linear Algebra

� COMPLEX NUMBERS ��

�� Constructing the complex numbers � � � � � � � � � � � � � � � ���� Calculating with complex numbers � � � � � � � � � � � � � � � ���� Geometric representation of C � � � � � � � � � � � � � � � � � � �� Complex conjugate � � � � � � � � � � � � � � � � � � � � � � � � ��� Modulus of a complex number � � � � � � � � � � � � � � � � � ���� Argument of a complex number � � � � � � � � � � � � � � � � � ����� De Moivre s theorem � � � � � � � � � � � � � � � � � � � � � � � ����� PROBLEMS � � � � � � � � � � � � � � � � � � � � � � � � � � � ���

� EIGENVALUES AND EIGENVECTORS ���

��� Motivation � � � � � � � � � � � � � � � � � � � � � � � � � � � � ����� De�nitions and examples � � � � � � � � � � � � � � � � � � � � � ������ PROBLEMS � � � � � � � � � � � � � � � � � � � � � � � � � � � ��

� Identifying second degree equations ���

��� The eigenvalue method � � � � � � � � � � � � � � � � � � � � � � ������ A classi�cation algorithm � � � � � � � � � � � � � � � � � � � � ����� PROBLEMS � � � � � � � � � � � � � � � � � � � � � � � � � � � ��

� THREEDIMENSIONAL GEOMETRY ���

��� Introduction � � � � � � � � � � � � � � � � � � � � � � � � � � � � ����� Three�dimensional space � � � � � � � � � � � � � � � � � � � � � ���� Dot product � � � � � � � � � � � � � � � � � � � � � � � � � � � � ���� Lines � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � ����� The angle between two vectors � � � � � � � � � � � � � � � � � ������ The cross�product of two vectors � � � � � � � � � � � � � � � � ������ Planes � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � ������ PROBLEMS � � � � � � � � � � � � � � � � � � � � � � � � � � � ��

� FURTHER READING ���

� BIBLIOGRAPHY ���

�� INDEX ���

ii

Page 4: Elementary Linear Algebra

List of Figures

��� Gauss�Jordan algorithm � � � � � � � � � � � � � � � � � � � � � ��

��� Re�ection in a line � � � � � � � � � � � � � � � � � � � � � � � � ��

��� Projection on a line � � � � � � � � � � � � � � � � � � � � � � � ��

�� Area of triangle OPQ� � � � � � � � � � � � � � � � � � � � � � � ��

�� Complex addition and subtraction � � � � � � � � � � � � � � � ��

�� Complex conjugate � � � � � � � � � � � � � � � � � � � � � � � � ���� Modulus of a complex number � � � � � � � � � � � � � � � � � ��

� Apollonius circles � � � � � � � � � � � � � � � � � � � � � � � � � ���

� Argument of a complex number � � � � � � � � � � � � � � � � � ���� Argument examples � � � � � � � � � � � � � � � � � � � � � � � ��

�� The nth roots of unity� � � � � � � � � � � � � � � � � � � � � � � ����� The roots of zn � a� � � � � � � � � � � � � � � � � � � � � � � � ���

��� Rotating the axes � � � � � � � � � � � � � � � � � � � � � � � � � ���

��� An ellipse example � � � � � � � � � � � � � � � � � � � � � � � � ����� ellipse� standard form � � � � � � � � � � � � � � � � � � � � � � ���

��� hyperbola� standard forms � � � � � � � � � � � � � � � � � � � � ����� parabola� standard forms �i� and �ii� � � � � � � � � � � � � � � ���

�� parabola� standard forms �iii� and �iv� � � � � � � � � � � � � � ���

��� �st parabola example � � � � � � � � � � � � � � � � � � � � � � � ����� �nd parabola example � � � � � � � � � � � � � � � � � � � � � � ��

��� Equality and addition of vectors � � � � � � � � � � � � � � � � ��

��� Scalar multiplication of vectors� � � � � � � � � � � � � � � � � � ����� Representation of three�dimensional space � � � � � � � � � � � �

�� The vector�

AB� � � � � � � � � � � � � � � � � � � � � � � � � � � ��� The negative of a vector� � � � � � � � � � � � � � � � � � � � � � ��

iii

Page 5: Elementary Linear Algebra

iv

��� �a� Equality of vectors� �b� Addition and subtraction of vectors������ Position vector as a linear combination of i� j and k� � � � � � ����� Representation of a line� � � � � � � � � � � � � � � � � � � � � � ������ The line AB� � � � � � � � � � � � � � � � � � � � � � � � � � � � ������� The cosine rule for a triangle� � � � � � � � � � � � � � � � � � � ������� Pythagoras theorem for a right�angled triangle� � � � � � � � ������� Distance from a point to a line� � � � � � � � � � � � � � � � � � ������� Projecting a segment onto a line� � � � � � � � � � � � � � � � � ������ The vector cross�product� � � � � � � � � � � � � � � � � � � � � ����� Vector equation for the plane ABC� � � � � � � � � � � � � � � ������� Normal equation of the plane ABC� � � � � � � � � � � � � � � ������� The plane ax� by � cz � d� � � � � � � � � � � � � � � � � � � � ������� Line of intersection of two planes� � � � � � � � � � � � � � � � � ������� Distance from a point to the plane ax� by � cz � d� � � � � � ��

Page 6: Elementary Linear Algebra

Chapter �

LINEAR EQUATIONS

��� Introduction to linear equations

A linear equation in n unknowns x�� x�� � � � � xn is an equation of the form

a�x� � a�x� � � � �� anxn � b�

where a�� a�� � � � � an� b are given real numbers�

For example� with x and y instead of x� and x�� the linear equation�x� �y � � describes the line passing through the points ��� � and ��� ��

Similarly� with x� y and z instead of x�� x� and x�� the linear equation �x � �y � �z � �� describes the plane passing through the points��� �� �� ��� �� �� ��� �� ��

A system of m linear equations in n unknowns x�� x�� � � � � xn is a familyof linear equations

a��x� � a��x� � � � �� a�nxn � b�

a��x� � a��x� � � � �� a�nxn � b����

am�x� � am�x� � � � �� amnxn � bm�

We wish to determine if such a system has a solution� that is to ndout if there exist numbers x�� x�� � � � � xn which satisfy each of the equationssimultaneously� We say that the system is consistent if it has a solution�Otherwise the system is called inconsistent�

Page 7: Elementary Linear Algebra

� CHAPTER �� LINEAR EQUATIONS

Note that the above system can be written concisely as

nXj��

aijxj � bi� i � �� �� � � � � m�

The matrix �����

a�� a�� � � � a�na�� a�� � � � a�n���

���am� am� � � � amn

�����

is called the coe�cient matrix of the system� while the matrix

�����

a�� a�� � � � a�n b�a�� a�� � � � a�n b����

������

am� am� � � � amn bm

�����

is called the augmented matrix of the system�Geometrically� solving a system of linear equations in two �or three

unknowns is equivalent to determining whether or not a family of lines �orplanes has a common point of intersection�

EXAMPLE ����� Solve the equation

�x� �y � ��

Solution� The equation �x � �y � � is equivalent to �x � � � �y orx � �� �

�y� where y is arbitrary� So there are in nitely many solutions�

EXAMPLE ����� Solve the system

x� y � z � �

x� y � z � ��

Solution� We subtract the second equation from the rst� to get �y � �and y � �

�� Then x � y � z � �

�� z� where z is arbitrary� Again there are

in nitely many solutions�

EXAMPLE ����� Find a polynomial of the form y � a��a�x�a�x��a�x

which passes through the points ���� ��� ���� �� ��� �� ��� ��

Page 8: Elementary Linear Algebra

���� INTRODUCTION TO LINEAR EQUATIONS �

Solution� When x has the values ��� ��� �� �� then y takes correspondingvalues ��� �� �� � and we get four equations in the unknowns a�� a�� a�� a��

a� � �a� � �a� � ��a� � ��

a� � a� � a� � a� � �

a� � a� � a� � a� � �

a� � �a� � �a� � �a� � ��

This system has the unique solution a� � ������ a� � �������� a� ��������a� � �������� So the required polynomial is

y ���

���

���

���x�

��

��x� �

��

���x��

In ���� pages ������ there are examples of systems of linear equationswhich arise from simple electrical networks using Kirchho��s laws for electrical circuits�

Solving a system consisting of a single linear equation is easy� However ifwe are dealing with two or more equations� it is desirable to have a systematicmethod of determining if the system is consistent and to nd all solutions�

Instead of restricting ourselves to linear equations with rational or realcoe�cients� our theory goes over to the more general case where the coef cients belong to an arbitrary �eld� A �eld F is a set F which possessesoperations of addition and multiplication which satisfy the familiar rules ofrational arithmetic� There are ten basic properties that a eld must have�

THE FIELD AXIOMS�

�� �a� b � c � a� �b� c for all a� b� c in F �

�� �abc � a�bc for all a� b� c in F �

�� a� b � b� a for all a� b in F �

�� ab � ba for all a� b in F �

�� there exists an element � in F such that � � a � a for all a in F �

�� there exists an element � in F such that �a � a for all a in F �

Page 9: Elementary Linear Algebra

� CHAPTER �� LINEAR EQUATIONS

�� to every a in F � there corresponds an additive inverse �a in F � satisfying

a� ��a � ��

�� to every non�zero a in F � there corresponds a multiplicative inverse

a�� in F � satisfying

aa�� � ��

�� a�b� c � ab� ac for all a� b� c in F �

��� � �� ��

With standard de nitions such as a � b � a � ��b anda

b� ab�� for

b �� �� we have the following familiar rules�

��a� b � ��a � ��b� �ab�� � a��b���

���a � a� �a���� � a�

��a� b � b� a� �a

b�� �

b

a�

a

b�

c

d�

ad� bc

bd�

a

b

c

d�

ac

bd�

ab

ac�

b

c�

a�bc

� � ac

b�

��ab � ��ab � a��b�

��ab

�a

b�

a

�b�

�a � ��

��a�� � ��a���

Fields which have only nitely many elements are of great interest inmany parts of mathematics and its applications� for example to coding theory� It is easy to construct elds containing exactly p elements� where p isa prime number� First we must explain the idea of modular addition andmodular multiplication� If a is an integer� we de ne a �mod p to be theleast remainder on dividing a by p� That is� if a � bp� r� where b and r areintegers and � � r � p� then a �mod p � r�

For example� �� �mod � � �� � �mod � � �� � �mod � � ��

Page 10: Elementary Linear Algebra

���� INTRODUCTION TO LINEAR EQUATIONS �

Then addition and multiplication mod p are de ned by

a� b � �a� b �mod p

a� b � �ab �mod p�

For example� with p � �� we have � � � � � �mod� � � and � � � ��� �mod� � �� Here are the complete addition and multiplication tablesmod ��

� � � � � � � �� � � � � � � �� � � � � � � �� � � � � � � �� � � � � � � �� � � � � � � �� � � � � � � �� � � � � � � �

� � � � � � � �� � � � � � � �� � � � � � � �� � � � � � � �� � � � � � � �� � � � � � � �� � � � � � � �� � � � � � � �

If we now let Zp � f�� �� � � � � p� �g� then it can be proved thatZp formsa eld under the operations of modular addition and multiplication mod p�For example� the additive inverse of � in Z� is �� so we write �� � � whencalculating in Z�� Also the multiplicative inverse of � in Z� is � � so we write��� � � when calculating in Z��

In practice� we write a� b and a� b as a�b and ab or a�b when dealingwith linear equations over Zp�

The simplest eld isZ�� which consists of two elements �� � with additionsatisfying ��� � �� So in Z�� �� � � and the arithmetic involved in solvingequations over Z� is very simple�

EXAMPLE ����� Solve the following system over Z��

x� y � z � �

x� z � ��

Solution� We add the rst equation to the second to get y � �� Then x ��� z � �� z� with z arbitrary� Hence the solutions are �x� y� z � ��� �� �and ��� �� ��

We use Q and R to denote the elds of rational and real numbers� respectively� Unless otherwise stated� the eld used will be Q�

Page 11: Elementary Linear Algebra

� CHAPTER �� LINEAR EQUATIONS

��� Solving linear equations

We show how to solve any system of linear equations over an arbitrary eld�using the GAUSS�JORDAN algorithm� We rst need to de ne some terms�

DEFINITION ����� �Row�echelon form� A matrix is in row�echelon

form if

�i all zero rows �if any are at the bottom of the matrix and

�ii if two successive rows are non�zero� the second row starts with morezeros than the rst �moving from left to right�

For example� the matrix ����

� � � �� � � �� � � �� � � �

����

is in row�echelon form� whereas the matrix����

� � � �� � � �� � � �� � � �

����

is not in row�echelon form�

The zero matrix of any size is always in row�echelon form�

DEFINITION ����� �Reduced row�echelon form� A matrix is in re�duced row�echelon form if

�� it is in row�echelon form�

�� the leading �leftmost non�zero entry in each non�zero row is ��

�� all other elements of the column in which the leading entry � occursare zeros�

For example the matrices

� �� �

�and

����

� � � � � �� � � � � �� � � � � �� � � � � �

����

Page 12: Elementary Linear Algebra

���� SOLVING LINEAR EQUATIONS �

are in reduced row�echelon form� whereas the matrices

�� � � �

� � �� � �

�� and

�� � � �

� � �� � �

��

are not in reduced row�echelon form� but are in row�echelon form�

The zero matrix of any size is always in reduced row�echelon form�

Notation� If a matrix is in reduced row�echelon form� it is useful to denotethe column numbers in which the leading entries � occur� by c�� c�� � � � � cr�with the remaining column numbers being denoted by cr��� � � � � cn� wherer is the number of non�zero rows� For example� in the �� � matrix above�we have r � �� c� � �� c� � �� c� � �� c� � �� c� � �� c � ��

The following operations are the ones used on systems of linear equationsand do not change the solutions�

DEFINITION ����� �Elementary row operations� There are threetypes of elementary row operations that can be performed on matrices�

�� Interchanging two rows�

Ri � Rj interchanges rows i and j�

�� Multiplying a row by a non�zero scalar�

Ri � tRi multiplies row i by the non�zero scalar t�

�� Adding a multiple of one row to another row�

Rj � Rj � tRi adds t times row i to row j�

DEFINITION ����� �Row equivalence�Matrix A is row�equivalent to matrix B if B is obtained from A by a sequence of elementary row operations�

EXAMPLE ����� Working from left to right�

A �

�� � � �

� � �� �� �

�� R� � R� � �R�

�� � � �

� �� �� �� �

��

R� � R�

�� � � �

� �� �� �� �

�� R� � �R�

�� � � �

� �� �� �� �

�� � B�

Page 13: Elementary Linear Algebra

� CHAPTER �� LINEAR EQUATIONS

Thus A is row�equivalent to B� Clearly B is also row�equivalent to A� byperforming the inverse row�operationsR� �

�R�� R�� R�� R�� R���R�

on B�

It is not di�cult to prove that if A and B are row�equivalent augmentedmatrices of two systems of linear equations� then the two systems have thesame solution sets � a solution of the one system is a solution of the other�For example the systems whose augmented matrices are A and B in theabove example are respectively

� �

x� �y � ��x� y � �x� y � �

and

� �

�x� �y � �x� y � ��x� y � �

and these systems have precisely the same solutions�

��� The Gauss�Jordan algorithm

We now describe the GAUSS�JORDAN ALGORITHM� This is a processwhich starts with a given matrix A and produces a matrix B in reduced row�echelon form� which is row�equivalent to A� If A is the augmented matrixof a system of linear equations� then B will be a much simpler matrix thanA from which the consistency or inconsistency of the corresponding systemis immediately apparent and in fact the complete solution of the system canbe read o��

STEP ��

Find the rst non�zero column moving from left to right� �column c�and select a non�zero entry from this column� By interchanging rows� ifnecessary� ensure that the rst entry in this column is non�zero� Multiplyrow � by the multiplicative inverse of a�c� thereby converting a�c� to �� Foreach non�zero element aic� � i � �� �if any in column c�� add �aic� timesrow � to row i� thereby ensuring that all elements in column c�� apart fromthe rst� are zero�

STEP �� If the matrix obtained at Step � has its �nd� � � � � mth rows allzero� the matrix is in reduced row�echelon form� Otherwise suppose thatthe rst column which has a non�zero element in the rows below the rst iscolumn c�� Then c� � c�� By interchanging rows below the rst� if necessary�ensure that a�c� is non�zero� Then convert a�c� to � and by adding suitablemultiples of row � to the remaing rows� where necessary� ensure that allremaining elements in column c� are zero�

Page 14: Elementary Linear Algebra

���� SYSTEMATIC SOLUTION OF LINEAR SYSTEMS� �

The process is repeated and will eventually stop after r steps� eitherbecause we run out of rows� or because we run out of non�zero columns� Ingeneral� the nal matrix will be in reduced row�echelon form and will haver non�zero rows� with leading entries � in columns c�� � � � � cr� respectively�

EXAMPLE �����

�� � � � �

� � �� �� � �� �

�� R� � R�

�� � � �� �

� � � �� � �� �

��

R� ��

�R�

�� � � �� �

� � � �� � �� �

�� R� � R� � �R�

�� � � �� �

� � � �� � � ���

��

R� ��

�R�

�� � � �� �

� � � �� � � ���

�� �

R� � R� � R�

R� � R� � �R�

�� � � � �

� � � �� � � ���

��

R� ���

��R�

�� � � � �

� � � �� � � �

�� R� � R� �

�R�

�� � � � �

� � � �� � � �

��

The last matrix is in reduced row�echelon form�

REMARK ����� It is possible to show that a given matrix over an arbitrary eld is row�equivalent to precisely one matrix which is in reducedrow�echelon form�

A �ow�chart for the Gauss�Jordan algorithm� based on ��� page ��� is presented in gure ��� below�

��� Systematic solution of linear systems�

Suppose a system of m linear equations in n unknowns x�� � � � � xn has augmented matrix A and that A is row�equivalent to a matrix B which is inreduced row�echelon form� via the Gauss�Jordan algorithm� Then A and Bare m� �n� �� Suppose that B has r non�zero rows and that the leadingentry � in row i occurs in column number ci� for � � i � r� Then

� � c� � c� � � � � � � cr � n� ��

Page 15: Elementary Linear Algebra

�� CHAPTER �� LINEAR EQUATIONS

START

�InputA� m� n

�i � �� j � �

�� �

�Are the elements in thejth column on and belowthe ith row all zero�

j � j � ������

R YesNo�

Is j � n�

YesNo

Let apj be the rst non�zeroelement in column j on or

below the ith row

�Is p � i�

Yes

PPPPPq No

Interchange thepth and ith rows

�������

Divide the ith row by aij

�Subtract aqj times the ithrow from the qth row forfor q � �� � � � � m �q �� i

�Set ci � j

�Is i � m���

��Is j � n��

i � i� �j � j � �

No

No

Yes

Yes �

��

Print A�c�� � � � � ci

STOP

Figure ���� Gauss�Jordan algorithm�

Page 16: Elementary Linear Algebra

���� SYSTEMATIC SOLUTION OF LINEAR SYSTEMS� ��

Also assume that the remaining column numbers are cr��� � � � � cn��� where

� � cr�� � cr�� � � � � � cn � n� ��

Case �� cr � n � �� The system is inconsistent� For the last non�zerorow of B is ��� �� � � � � �� and the corresponding equation is

�x� � �x� � � � �� �xn � ��

which has no solutions� Consequently the original system has no solutions�

Case �� cr � n� The system of equations corresponding to the non�zerorows of B is consistent� First notice that r � n here�

If r � n� then c� � �� c� � �� � � � � cn � n and

B �

������������

� � � � � � d�� � � � � � d����

���� � � � � � dn� � � � � � ����

���� � � � � � �

�������������

There is a unique solution x� � d�� x� � d�� � � � � xn � dn�

If r � n� there will be more than one solution �in nitely many if the eld is in nite� For all solutions are obtained by taking the unknownsxc� � � � � � xcr as dependent unknowns and using the r equations corresponding to the non�zero rows of B to express these unknowns in terms of theremaining independent unknowns xcr�� � � � � � xcn � which can take on arbitrary values�

xc� � b�n�� � b�cr��xcr�� � � � � � b�cnxcn���

xcr � br n�� � brcr��xcr�� � � � � � brcnxcn �

In particular� taking xcr�� � �� � � � � xcn�� � � and xcn � �� � respectively�produces at least two solutions�

EXAMPLE ����� Solve the system

x� y � �

x� y � �

�x� �y � ��

Page 17: Elementary Linear Algebra

�� CHAPTER �� LINEAR EQUATIONS

Solution� The augmented matrix of the system is

A �

�� � � �

� �� �� � �

��

which is row equivalent to

B �

�� � � �

� � ��

� � �

�� �

We read o� the unique solution x � �

�� y � ��

��

�Here n � �� r � �� c� � �� c� � �� Also cr � c� � � � � � n � � andr � n�

EXAMPLE ����� Solve the system

�x� � �x� � �x� � ��x� � �x� � x� � ���x� � �x� � x� � ��

Solution� The augmented matrix is

A �

�� � � �� �

� � � ��� � �� �

��

which is row equivalent to

B �

�� � � � �

� � � �� � � �

�� �

We read o� inconsistency for the original system��Here n � �� r � �� c� � �� c� � �� Also cr � c� � � � n � ��

EXAMPLE ����� Solve the system

x� � x� � x� � �

x� � x� � x� � ��

Page 18: Elementary Linear Algebra

���� SYSTEMATIC SOLUTION OF LINEAR SYSTEMS� ��

Solution� The augmented matrix is

A �

� �� � �� � �� �

which is row equivalent to

B �

� � � �

� � �� �

��

The complete solution is x� ��

�� x� �

�� x�� with x� arbitrary�

�Here n � �� r � �� c� � �� c� � �� Also cr � c� � � � � � n � � andr � n�

EXAMPLE ����� Solve the system

�x� � �x� � �x� � �x � �

�x� � x� � �x� � �x � �

�x� � �x� � x� � �x� � �x� � x � �

�x� � �x� � ��x� � ��x� � �x � ��

Solution� The augmented matrix is

A �

����

� � � � �� �� �� � � � �� �� �� �� � � �� � �� �� � �� ��� � �

����

which is row equivalent to

B �

����

� ��

�� ��

��

� �

��

� � � �

���

�� �

� � � � � � �

� � � � � � �

���� �

The complete solution is

x� ��

��� �

�x� �

��

x� �

x��

x� ��

�� �

�x� �

�x��

x ��

��

with x�� x�� x� arbitrary��Here n � �� r � �� c� � �� c� � �� c� � �� cr � c� � � � � � n � �� r � n�

Page 19: Elementary Linear Algebra

�� CHAPTER �� LINEAR EQUATIONS

EXAMPLE ����� Find the rational number t for which the following system is consistent and solve the system for this value of t�

x� y � �

x� y � �

�x� y � t�

Solution� The augmented matrix of the system is

A �

�� � � �

� �� �� �� t

��

which is row�equivalent to the simpler matrix

B �

�� � � �

� � �� � t � �

�� �

Hence if t �� � the system is inconsistent� If t � � the system is consistentand

B �

�� � � �

� � �� � �

���

�� � � �

� � �� � �

�� �

We read o� the solution x � �� y � ��

EXAMPLE ���� For which rationals a and b does the following systemhave �i no solution� �ii a unique solution� �iii in nitely many solutions�

x� �y � �z � �

�x� �y � az � �

�x� �y � �z � b�

Solution� The augmented matrix of the system is

A �

�� � �� � �

� �� a �� �� � b

��

Page 20: Elementary Linear Algebra

���� SYSTEMATIC SOLUTION OF LINEAR SYSTEMS� ��

�R� � R� � �R�

R� � R� � �R�

�� � �� � �

� � a� � ��� � �� b� ��

��

R� � R� � �R�

�� � �� � �

� � a� � ��� � ��a� � b� �

�� � B�

Case �� a �� �� Then ��a� � �� � and we see that B can be reduced toa matrix of the form �

� � � � u� � � v

� � � b���a��

��

and we have the unique solution x � u� y � v� z � �b� �����a� ��

Case �� a � �� Then

B �

�� � �� � �

� � �� ��� � � b� �

�� �

If b �� � we get no solution� whereas if b � � then

B �

�� � �� � �

� � �� ��� � � �

�� R� � R� � �R�

�� � � �� ��

� � �� ��� � � �

��� We

read o� the complete solution x � �� � z� y � �� � �z� with z arbitrary�

EXAMPLE ���� Find the reduced row�echelon form of the following matrix over Z��

� � � �� � � �

��

Hence solve the system

�x� y � �z � �

�x� �y � z � ��

over Z��

Solution�

Page 21: Elementary Linear Algebra

�� CHAPTER �� LINEAR EQUATIONS

� � � �� � � �

�R� � R� �R�

� � � �� � �� ��

��

� � � �� � � �

R� � �R�

� � � �� � � �

�R�� R� �R�

� � � �� � � �

��

The last matrix is in reduced row�echelon form�To solve the system of equations whose augmented matrix is the given

matrix over Z�� we see from the reduced row�echelon form that x � � andy � � � �z � � � z� where z � �� �� �� Hence there are three solutionsto the given system of linear equations� �x� y� z � ��� �� �� ��� �� � and��� �� ��

��� Homogeneous systems

A system of homogeneous linear equations is a system of the form

a��x� � a��x� � � � �� a�nxn � �

a��x� � a��x� � � � �� a�nxn � ����

am�x� � am�x� � � � �� amnxn � ��

Such a system is always consistent as x� � �� � � � � xn � � is a solution�This solution is called the trivial solution� Any other solution is called anon�trivial solution�

For example the homogeneous system

x� y � �

x� y � �

has only the trivial solution� whereas the homogeneous system

x� y � z � �

x� y � z � �

has the complete solution x � �z� y � �� z arbitrary� In particular� takingz � � gives the non�trivial solution x � ��� y � �� z � ��

There is simple but fundamental theorem concerning homogeneous systems�

THEOREM ����� A homogeneous system of m linear equations in n un�knowns always has a non�trivial solution if m � n�

Page 22: Elementary Linear Algebra

���� PROBLEMS ��

Proof� Suppose that m � n and that the coe�cient matrix of the systemis row�equivalent to B� a matrix in reduced row�echelon form� Let r be thenumber of non�zero rows in B� Then r � m � n and hence n � r � � andso the number n � r of arbitrary unknowns is in fact positive� Taking oneof these unknowns to be � gives a non�trivial solution�

REMARK ����� Let two systems of homogeneous equations in n unknowns have coe�cient matrices A and B� respectively� If each row of B isa linear combination of the rows of A �i�e� a sum of multiples of the rowsof A and each row of A is a linear combination of the rows of B� then it iseasy to prove that the two systems have identical solutions� The converse istrue� but is not easy to prove� Similarly if A and B have the same reducedrow�echelon form� apart from possibly zero rows� then the two systems haveidentical solutions and conversely�

There is a similar situation in the case of two systems of linear equations�not necessarily homogeneous� with the proviso that in the statement ofthe converse� the extra condition that both the systems are consistent� isneeded�

��� PROBLEMS

�� Which of the following matrices of rationals is in reduced row�echelonform�

�a

�� � � � � ��

� � � � �� � � � �

�� �b

�� � � � � �

� � � � ��� � � �� �

�� �c

�� � � � �

� � � �� � � ��

��

�d

����

� � � � �� � � � ��� � � � �� � � � �

���� �e

����

� � � � �� � � � �� � � � �� � � � �

���� �f

����

� � � �� � � �� � � �� � � �

����

�g

����

� � � � �� � � � �� � � � ��� � � � �

����� �Answers� �a� �e� �g�

�� Find reduced row�echelon forms which are row�equivalent to the followingmatrices�

�a

� � �� � �

��b

� � �� � �

��c

�� � � �

� � �� � �

�� �d

�� � � �

� � ��� � �

�� �

Page 23: Elementary Linear Algebra

�� CHAPTER �� LINEAR EQUATIONS

�Answers�

�a

� � �� � �

��b

� � ��� � �

��c

�� � � �

� � �� � �

�� �d

�� � � �

� � �� � �

����

�� Solve the following systems of linear equations by reducing the augmentedmatrix to reduced row�echelon form�

�a x� y � z � � �b x� � x� � x� � �x� � ���x� �y � z � � �x� � x� � �x� � �x� � �x� y � z � �� ��x� � �x� � ��x� � �x� � �

�c �x� y � �z � � �d �x� � �x� � �x� � ��x� y � �z � �

��x� � �x� � �

x� y � z � � �x� � �x� � �x� � �x� � ��x� �y � ��z � � �x� � �x� � �x� � �

�Answers� �a x � ��� y � �

�� z � �

�� �b inconsistent�

�c x � ��

�� �z� y � ��

�� �z� with z arbitrary�

�d x� ��

�� �x�� x� � ��

�� ��

�x�� x� � �� �

�x�� with x� arbitrary��

�� Show that the following system is consistent if and only if c � �a � �band solve the system in this case�

�x� y � �z � a

�x� y � �z � b

��x� �y � ��z � c�

�Answer� x � a�b�

� �

�z� y � ��a��b

�� �

�z� with z arbitrary��

�� Find the value of t for which the following system is consistent and solvethe system for this value of t�

x� y � �

tx � y � t

�� � tx� �y � ��

�Answer� t � �� x � �� y � ���

Page 24: Elementary Linear Algebra

���� PROBLEMS ��

�� Solve the homogeneous system

��x� � x� � x� � x� � �

x� � �x� � x� � x� � �

x� � x� � �x� � x� � �

x� � x� � x� � �x� � ��

�Answer� x� � x� � x� � x�� with x� arbitrary��

�� For which rational numbers � does the homogeneous system

x � ��� �y � �

��� �x� y � �

have a non�trivial solution�

�Answer� � � �� ���

�� Solve the homogeneous system

�x� � x� � x� � x� � �

�x� � x� � x� � x� � ��

�Answer� x� � ��

�x�� x� � ��

�x� � x�� with x� and x� arbitrary��

�� Let A be the coe�cient matrix of the following homogeneous system ofn equations in n unknowns�

��� nx� � x� � � � �� xn � �

x� � ��� nx� � � � �� xn � �

� � � � �

x� � x� � � � �� ��� nxn � ��

Find the reduced row�echelon form of A and hence� or otherwise� prove thatthe solution of the above system is x� � x� � � � �� xn� with xn arbitrary�

��� Let A �

a b

c d

�be a matrix over a eld F � Prove that A is row�

equivalent to

� �� �

�if ad � bc �� �� but is row�equivalent to a matrix

whose second row is zero� if ad� bc � ��

Page 25: Elementary Linear Algebra

�� CHAPTER �� LINEAR EQUATIONS

��� For which rational numbers a does the following system have �i nosolutions �ii exactly one solution �iii in nitely many solutions�

x� �y � �z � �

�x� y � �z � �

�x� y � �a� � ��z � a� ��

�Answer� a � ��� no solution� a � �� in nitely many solutions� a �� ��exactly one solution��

��� Solve the following system of homogeneous equations over Z��

x� � x� � x� � �

x� � x� � x� � �

x� � x� � x� � x� � �

x� � x� � ��

�Answer� x� � x� � x� � x�� x� � x�� with x� and x� arbitrary elements ofZ���

��� Solve the following systems of linear equations over Z��

�a �x� y � �z � � �b �x� y � �z � ��x� y � �z � � �x� y � �z � ��x� y � �z � � x� y � ��

�Answer� �a x � �� y � �� z � �� �b x � � � �z� y � � � �z� with z anarbitrary element of Z���

��� If ���� � � � � �n and ���� � � � � �n are solutions of a system of linear equations� prove that

���� t�� � t��� � � � � ��� t�n � t�n

is also a solution�

��� If ���� � � � � �n is a solution of a system of linear equations� prove thatthe complete solution is given by x� � �� � y�� � � � � xn � �n � yn� where�y�� � � � � yn is the general solution of the associated homogeneous system�

Page 26: Elementary Linear Algebra

���� PROBLEMS ��

��� Find the values of a and b for which the following system is consistent�Also nd the complete solution when a � b � ��

x� y � z � w � �

ax� y � z � w � b

�x� �y � aw � � � a�

�Answer� a �� � or a � � � b� x � �� �z� y � �z � w� with z� w arbitrary��

��� Let F � f�� �� a� bg be a eld consisting of � elements�

�a Determine the addition and multiplication tables of F � �Hint� Provethat the elements ���� ���� ��a� ��b are distinct and deduce that� � �� � � � � �� then deduce that � � � � ��

�b A matrix A� whose elements belong to F � is de ned by

A �

�� � a b a

a b b �� � � a

�� �

prove that the reduced row�echelon form of A is given by the matrix

B �

�� � � � �

� � � b

� � � �

�� �

Page 27: Elementary Linear Algebra

��

Page 28: Elementary Linear Algebra

Chapter �

MATRICES

��� Matrix arithmetic

A matrix over a �eld F is a rectangular array of elements from F � The sym�bol Mm�n�F � denotes the collection of all m�n matrices over F � Matriceswill usually be denoted by capital letters and the equation A � �aij � meansthat the element in the ith row and jth column of the matrix A equalsaij � It is also occasionally convenient to write aij � �A�ij� For the presentall matrices will have rational entries unless otherwise stated�

EXAMPLE ����� The formula aij � ���i � j� for � � i � � � � j � �de�nes a � � matrix A � �aij � namely

A �

������

������ �

DEFINITION ����� �Equality of matrices� MatricesA andB are saidto be equal if A and B have the same size and corresponding elements areequal� that is A and B � Mm�n�F � and A � �aij �� B � �bij� with aij � bijfor � � i � m� � � j � n�

DEFINITION ����� �Addition of matrices� Let A � �aij � and B ��bij� be of the same size� Then A � B is the matrix obtained by addingcorresponding elements of A and B� that is

A� B � �aij � � �bij � � �aij � bij ��

Page 29: Elementary Linear Algebra

�� CHAPTER �� MATRICES

DEFINITION ����� �Scalar multiple of a matrix� Let A � �aij� andt � F �that is t is a scalar�� Then tA is the matrix obtained by multiplyingall elements of A by t� that is

tA � t�aij � � �taij ��

DEFINITION ����� �Additive inverse of a matrix� Let A � �aij � �Then �A is the matrix obtained by replacing the elements of A by theiradditive inverses� that is

�A � ��aij � � ��aij ��

DEFINITION ����� �Subtraction of matrices� Matrix subtraction isde�ned for two matrices A � �aij � and B � �bij� of the same size in theusual way� that is

A� B � �aij �� �bij � � �aij � bij ��

DEFINITION ����� �The zero matrix� For each m� n the matrix inMm�n�F � all of whose elements are zero is called the zero matrix �of sizem� n� and is denoted by the symbol ��

The matrix operations of addition scalar multiplication additive inverseand subtraction satisfy the usual laws of arithmetic� �In what follows s andt will be arbitrary scalars and A� B� C are matrices of the same size��

�� �A� B� � C � A� �B � C��

�� A� B � B � A�

� � � A � A�

�� A� ��A� � ��

�� �s� t�A � sA� tA �s� t�A � sA� tA�

�� t�A �B� � tA � tB t�A �B� � tA � tB�

�� s�tA� � �st�A�

�� �A � A �A � � ����A � �A��� tA � �� t � � or A � ��

Other similar properties will be used when needed�

Page 30: Elementary Linear Algebra

���� MATRIX ARITHMETIC ��

DEFINITION ���� �Matrix product� Let A � �aij � be a matrix ofsize m � n and B � �bjk� be a matrix of size n � p� �that is the numberof columns of A equals the number of rows of B�� Then AB is the m � pmatrix C � �cik� whose �i� k�th element is de�ned by the formula

cik �nX

j��

aijbjk � ai�b�k � � � �� ainbnk�

EXAMPLE �����

��

�� � �

��� �� �

��

��� � � �� � �� � � �� � � � � �� � � � � �� �

��

��� ��� ��

��

��

�� �� �

��� � �

��

�� � � ��

����� � �

� �� �� �

��

���

� � �

� �� �

��

��� �

� ��

�����

��

�� ��� ��

� �� ��� ��

��

�� �� �

��

Matrix multiplication obeys many of the familiar laws of arithmetic apartfrom the commutative law�

�� �AB�C � A�BC� if A� B� C are m� n� n � p� p� q respectively�

�� t�AB� � �tA�B � A�tB� A��B� � ��A�B � ��AB�� � �A� B�C � AC � BC if A and B are m� n and C is n � p�

�� D�A� B� � DA�DB if A and B are m� n and D is p�m�

We prove the associative law only�First observe that �AB�C and A�BC� are both of size m� q�Let A � �aij �� B � �bjk�� C � �ckl�� Then

��AB�C�il �

pXk��

�AB�ikckl �

pXk��

� nX

j��

aijbjk

�A ckl

pXk��

nXj��

aijbjkckl�

Page 31: Elementary Linear Algebra

�� CHAPTER �� MATRICES

Similarly

�A�BC��il �nX

j��

pXk��

aijbjkckl�

However the double summations are equal� For sums of the form

nXj��

pXk��

djk and

pXk��

nXj��

djk

represent the sum of the np elements of the rectangular array �djk� by rowsand by columns respectively� Consequently

��AB�C�il � �A�BC��il

for � � i � m� � � l � q� Hence �AB�C � A�BC��

The system of m linear equations in n unknowns

a��x� � a��x� � � � �� a�nxn � b�

a��x� � a��x� � � � �� a�nxn � b����

am�x� � am�x� � � � �� amnxn � bm

is equivalent to a single matrix equation�����

a�� a�� � � � a�na�� a�� � � � a�n���

���am� am� � � � amn

����������

x�x����

xn

����� �

�����

b�b����

bm

����� �

that is AX � B where A � �aij � is the coe�cient matrix of the system

X �

�����

x�x����xn

����� is the vector of unknowns and B �

�����

b�b����bm

����� is the vector of

constants�Another useful matrix equation equivalent to the above system of linear

equations is

x�

�����

a��a�����

am�

������ x�

�����

a��a�����

am�

������ � � �� xn

�����

a�na�n���

amn

����� �

�����

b�b����bm

����� �

Page 32: Elementary Linear Algebra

���� LINEAR TRANSFORMATIONS ��

EXAMPLE ����� The system

x� y � z � �

x� y � z � ��

is equivalent to the matrix equation

�� � �� �� �

��� x

yz

�� �

���

and to the equation

x

���

�� y

��

���� z

���

��

���

��

��� Linear transformations

An ndimensional column vector is an n � � matrix over F � The collectionof all ndimensional column vectors is denoted by Fn�

Every matrix is associated with an important type of function called alinear transformation�

DEFINITION ����� �Linear transformation� With A �Mm�n�F � weassociate the function TA � Fn � Fm de�ned by TA�X� � AX for allX � Fn� More explicitly using components the above function takes theform

y� � a��x� � a��x� � � � �� a�nxn

y� � a��x� � a��x� � � � �� a�nxn���

ym � am�x� � am�x� � � � �� amnxn�

where y�� y�� � � � � ym are the components of the column vector TA�X��

The function just de�ned has the property that

TA�sX � tY � � sTA�X� � tTA�Y � �����

for all s� t � F and all ndimensional column vectors X� Y � For

TA�sX � tY � � A�sX � tY � � s�AX� � t�AY � � sTA�X� � tTA�Y ��

Page 33: Elementary Linear Algebra

�� CHAPTER �� MATRICES

REMARK ����� It is easy to prove that if T � Fn � Fm is a functionsatisfying equation ��� then T � TA where A is the m � n matrix whosecolumns are T �E��� � � � � T �En� respectively where E�� � � � � En are the ndimensional unit vectors de�ned by

E� �

�����

������

����� � � � � � En �

�����

������

����� �

One wellknown example of a linear transformation arises from rotatingthe �x� y�plane in ��dimensional Euclidean space anticlockwise through �radians� Here a point �x� y� will be transformed into the point �x�� y��where

x� � x cos � � y sin �

y� � x sin � � y cos ��

In dimensional Euclidean space the equations

x� � x cos � � y sin �� y� � x sin � � y cos �� z� � z�x� � x� y� � y cos�� z sin �� z� � y sin� � z cos��

x� � x cos� � z sin�� y� � y� z� � x sin� � z cos��

correspond to rotations about the positive z� x� yaxes anticlockwise through�� �� � radians respectively�

The product of two matrices is related to the product of the correspond�ing linear transformations�

If A ism�n and B is n�p then the function TATB � F p � Fm obtainedby �rst performing TB then TA is in fact equal to the linear transformationTAB� For if X � F p we have

TATB�X� � A�BX� � �AB�X � TAB�X��

The following example is useful for producing rotations in dimensionalanimated design� �See ��� pages ��������

EXAMPLE ����� The linear transformation resulting from successivelyrotating dimensional space about the positive z� x� yaxes anticlockwisethrough �� �� � radians respectively is equal to TABC where

Page 34: Elementary Linear Algebra

���� LINEAR TRANSFORMATIONS ��

l

�x� y�

�x�� y��

��������

����

����

���

Figure ���� Re�ection in a line�

C �

�� cos � � sin � �

sin � cos � �� � �

�� B �

�� � � �

� cos� � sin�� sin � cos�

���

A �

�� cos� � � sin�

� � �sin� � cos�

���

The matrix ABC is quite complicated�

A�BC� �

�� cos� � � sin �

� � �sin� � cos�

���� cos � � sin � �

cos� sin � cos� cos � � sin�sin � sin � sin� cos � cos�

��

�� cos� cos � � sin� sin� sin � � cos� sin � � sin� sin� sin � � sin� cos�

cos� sin � cos� cos � � sin �sin� cos � � cos� sin� sin � � sin� sin � � cos� sin � cos � cos� cos�

���

EXAMPLE ����� Another example of a linear transformation arising fromgeometry is re�ection of the plane in a line l inclined at an angle � to thepositive xaxis�

We reduce the problem to the simpler case � � � where the equationsof transformation are x� � x� y� � �y� First rotate the plane clockwisethrough � radians thereby taking l into the xaxis� next re�ect the plane inthe xaxis� then rotate the plane anticlockwise through � radians therebyrestoring l to its original position�

Page 35: Elementary Linear Algebra

� CHAPTER �� MATRICES

l

�x� y�

�x�� y��

��������

����

���

Figure ���� Projection on a line�

In terms of matrices we get transformation equations�x�y�

��

�cos � � sin �sin � cos �

� �� �� ��

��cos ���� � sin ����sin ���� cos ����

� �xy

�cos � sin �sin � � cos �

� �cos � sin �

� sin � cos �

��x

y

�cos �� sin ��sin �� � cos ��

��xy

��

The more general transformation�x�y�

�� a

�cos � � sin �sin � cos �

� �x

y

��

�u

v

�� a � ��

represents a rotation followed by a scaling and then by a translation� Suchtransformations are important in computer graphics� See �� ����

EXAMPLE ����� Our last example of a geometrical linear transformationarises from projecting the plane onto a line l through the origin inclinedat angle � to the positive xaxis� Again we reduce that problem to thesimpler case where l is the xaxis and the equations of transformation arex� � x� y� � ��

In terms of matrices we get transformation equations�x�y�

��

�cos � � sin �sin � cos �

��� �� �

��cos ���� � sin ����sin ���� cos ����

��xy

Page 36: Elementary Linear Algebra

���� RECURRENCE RELATIONS �

�cos � �sin � �

��cos � sin �

� sin � cos �

� �x

y

�cos� � cos � sin �

sin � cos � sin� �

� �xy

��

��� Recurrence relations

DEFINITION ����� �The identity matrix� The n � n matrix In ���ij � de�ned by �ij � � if i � j� �ij � � if i �� j is called the n� n identity

matrix of order n� In other words the columns of the identity matrix oforder n are the unit vectors E�� � � � � En respectively�

For example I� �

�� �� �

��

THEOREM ����� If A is m� n then ImA � A � AIn�

DEFINITION ����� �kth power of a matrix� If A is an n�n matrixwe de�ne Ak recursively as follows� A� � In and Ak� � AkA for k � ��

For example A� � A�A � InA � A and hence A� � A�A � AA�

The usual index laws hold provided AB � BA�

�� AmAn � Amn �Am�n � Amn�

�� �AB�n � AnBn�

� AmBn � BnAm�

�� �A� B�� � A� � �AB � B��

�� �A� B�n �nXi��

ni

�AiBn�i �

�� �A� B��A�B� � A� � B��

We now state a basic property of the natural numbers�

AXIOM ����� �PRINCIPLE OF MATHEMATICAL INDUCTION�

If for each n � �� Pn denotes a mathematical statement and

�i� P� is true�

Page 37: Elementary Linear Algebra

� CHAPTER �� MATRICES

�ii� the truth of Pn implies that of Pn� for each n � ��

then Pn is true for all n � ��

EXAMPLE ����� Let A �

�� �

�� ���� Prove that

An �

�� � �n �n��n �� �n

�if n � ��

Solution� We use the principle of mathematical induction�

Take Pn to be the statement

An �

�� � �n �n��n �� �n

��

Then P� asserts that

A� �

�� � �� � �� ��� � � �� �� �

��

�� �

�� ����

which is true� Now let n � � and assume that Pn is true� We have to deducethat

An� �

�� � ��n� �� ��n� �����n� �� �� ��n� ��

��

�� � �n �n� ���n� � ��� �n

��

Now

An� � AnA

�� � �n �n��n �� �n

��� �

�� ���

��� � �n�� � ��n����� �� � �n�� � ��n��������n�� � ��� �n����� ���n�� � ��� �n�����

�� � �n �n� ���n� � ��� �n

��

and �the induction goes through��

The last example has an application to the solution of a system of re�currence relations�

Page 38: Elementary Linear Algebra

���� PROBLEMS

EXAMPLE ����� The following system of recurrence relations holds forall n � ��

xn� � �xn � �yn

yn� � ��xn � �yn�

Solve the system for xn and yn in terms of x� and y��

Solution� Combine the above equations into a single matrix equation�xn�yn�

��

�� �

�� ��� �

xnyn

��

or Xn� � AXn where A �

�� �

�� ���and Xn �

�xnyn

��

We see that

X� � AX�

X� � AX� � A�AX�� � A�X�

���

Xn � AnX��

�The truth of the equation Xn � AnX� for n � � strictly speakingfollows by mathematical induction� however for simple cases such as theabove it is customary to omit the strict proof and supply instead a fewlines of motivation for the inductive statement��

Hence the previous example gives�xnyn

�� Xn �

�� � �n �n��n �� �n

� �x�y�

��� � �n�x� � ��n�y����n�x� � ��� �n�y�

��

and hence xn � ����n�x���ny� and yn � ���n�x������n�y� for n � ��

��� PROBLEMS

�� Let A� B� C� D be matrices de�ned by

A �

�� ��� �� �

�� � B �

�� � � ��� � ��� �

�� �

Page 39: Elementary Linear Algebra

� CHAPTER �� MATRICES

C �

�� � ��

� ��

�� � D �

�� ��� �

��

Which of the following matrices are de�ned� Compute those matriceswhich are de�ned�

A� B� A� C� AB� BA� CD� DC� D��

�Answers� A� C� BA� CD� D���� � ��

� � �

�� �

�� � ��

�� ���� �

��

�� ���

�� ���� ��

�� �

�� ��� ��

���

�� Let A �

� �� � �� � �

�� Show that if B is a � � such that AB � I�

then

B �

�� a b�a� � �� ba� � b

��

for suitable numbers a and b� Use the associative law to show that�BA��B � B�

� If A �

�a b

c d

� prove that A� � �a� d�A� �ad� bc�I� � ��

�� If A �

�� � � �

� use the fact A� � �A � I� and mathematical

induction to prove that

An �� n � ��

�A�

� n

�I� if n � ��

�� A sequence of numbers x�� x�� � � � � xn� � � � satis�es the recurrence rela�tion xn� � axn�bxn�� for n � � where a and b are constants� Provethat �

xn�xn

�� A

�xnxn��

��

Page 40: Elementary Linear Algebra

���� PROBLEMS �

where A �

�a b

� �

�and hence express

�xn�xn

�in terms of

�x�x�

��

If a � � and b � � use the previous question to �nd a formula forxn in terms of x� and x��

�Answer�

xn � n � �

�x� �

� n

�x���

�� Let A �

��a �a�� �

��

�a� Prove that

An �

��n� ��an �nan�nan�� ��� n�an

�if n � ��

�b� A sequence x�� x�� � � � � xn� � � � satis�es the recurrence relation xn� ��axn � a�xn�� for n � �� Use part �a� and the previous questionto prove that xn � nan��x� � ��� n�anx� for n � ��

�� Let A �

�a bc d

�and suppose that �� and �� are the roots of the

quadratic polynomial x���a�d�x�ad�bc� ��� and �� may be equal��

Let kn be de�ned by k� � �� k� � � and for n � �

kn �nXi��

�n�i� �i��� �

Prove thatkn� � ��� � ���kn � ����kn���

if n � �� Also prove that

kn �

���n� � �n������� ��� if �� �� ��

n�n��� if �� � ���

Use mathematical induction to prove that if n � �

An � knA� ����kn��I��

�Hint� Use the equation A� � �a� d�A� �ad� bc�I���

Page 41: Elementary Linear Algebra

� CHAPTER �� MATRICES

�� Use Question � to prove that if A �

�� �� �

� then

An � n

�� �� �

������n��

� �� �� ��

if n � ��

�� The Fibonacci numbers are de�ned by the equations F� � �� F� � �and Fn� � Fn � Fn�� if n � �� Prove that

Fn ��p�

��� �

p�

�n

����p�

�n�

if n � ��

��� Let r � � be an integer� Let a and b be arbitrary positive integers�Sequences xn and yn of positive integers are de�ned in terms of a andb by the recurrence relations

xn� � xn � ryn

yn� � xn � yn�

for n � � where x� � a and y� � b�

Use Question � to prove that

xnyn� p

r as n��

��� Non�singular matrices

DEFINITION ����� �Nonsingular matrix�

A square matrix A � Mn�n�F � is called non�singular or invertible ifthere exists a matrix B �Mn�n�F � such that

AB � In � BA�

Any matrix B with the above property is called an inverse of A� If A doesnot have an inverse A is called singular�

Page 42: Elementary Linear Algebra

���� NON�SINGULAR MATRICES �

THEOREM ����� �Inverses are unique�

If A has inverses B and C then B � C�

Proof� Let B and C be inverses of A� Then AB � In � BA and AC �In � CA� Then B�AC� � BIn � B and �BA�C � InC � C� Hence becauseB�AC� � �BA�C we deduce that B � C�

REMARK ����� If A has an inverse it is denoted by A��� So

AA�� � In � A��A�

Also if A is nonsingular it follows that A�� is also nonsingular and

�A����� � A�

THEOREM ����� If A and B are nonsingular matrices of the same sizethen so is AB� Moreover

�AB��� � B��A���

Proof�

�AB��B��A��� � A�BB���A�� � AInA�� � AA�� � In�

Similarly�B��A����AB� � In�

REMARK ����� The above result generalizes to a product of m nonsingular matrices� If A�� � � � � Am are nonsingular n � n matrices then theproduct A� � � �Am is also nonsingular� Moreover

�A� � � �Am��� � A��m � � �A��� �

�Thus the inverse of the product equals the product of the inverses in the

reverse order��

EXAMPLE ����� If A and B are n � n matrices satisfying A� � B� ��AB�� � In prove that AB � BA�

Solution� Assume A� � B� � �AB�� � In� Then A� B� AB are nonsingular and A�� � A� B�� � B� �AB��� � AB�

But �AB��� � B��A�� and hence AB � BA�

Page 43: Elementary Linear Algebra

� CHAPTER �� MATRICES

EXAMPLE ����� A �

�� �� �

�is singular� For suppose B �

�a b

c d

�is an inverse of A� Then the equation AB � I� gives�

� �� �

� �a bc d

��

�� �� �

and equating the corresponding elements of column � of both sides gives thesystem

a� �c � �

�a� �c � �

which is clearly inconsistent�

THEOREM ����� Let A �

�a b

c d

�and � � ad � bc �� �� Then A is

nonsingular� Also

A�� � ���

�d �b�c a

��

REMARK ����� The expression ad � bc is called the determinant of A

and is denoted by the symbols detA or

���� a b

c d

�����Proof� Verify that the matrix B � ���

�d �b�c a

�satis�es the equation

AB � I� � BA�

EXAMPLE ����� Let

A �

�� � � �

� � �� � �

�� �

Verify that A� � �I� deduce that A is nonsingular and �nd A���

Solution� After verifying that A� � �I� we notice that

A

��

�A�

�� I� �

��

�A�

�A�

Hence A is nonsingular and A�� � �

�A��

Page 44: Elementary Linear Algebra

���� NON�SINGULAR MATRICES �

THEOREM ����� If the coe�cient matrix A of a system of n equationsin n unknowns is nonsingular then the system AX � B has the uniquesolution X � A��B�

Proof� Assume that A�� exists�

�� �Uniqueness�� Assume that AX � B� Then

�A��A�X � A��B�

InX � A��B�

X � A��B�

�� �Existence�� Let X � A��B� Then

AX � A�A��B� � �AA���B � InB � B�

THEOREM ����� �Cramer�s rule for � equations in � unknowns�

The system

ax� by � e

cx� dy � f

has a unique solution if � �

���� a bc d

���� �� � namely

x ���

�� y �

��

��

where

�� �

���� e b

f d

���� and �� �

���� a e

c f

���� �Proof� Suppose � �� �� Then A �

�a b

c d

�has inverse

A�� � ���

�d �b�c a

and we know that the system

A

�xy

��

�ef

Page 45: Elementary Linear Algebra

�� CHAPTER �� MATRICES

has the unique solution�xy

�� A��

�ef

��

�d �b�c a

� �ef

��

�de� bf

�ce � af

��

���

��

��

���������

��

Hence x � ����� y � �����

COROLLARY ����� The homogeneous system

ax� by � �

cx� dy � �

has only the trivial solution if � �

���� a bc d

���� �� ��

EXAMPLE ����� The system

�x� �y � ���

�x� �y � ��

has the unique solution x � ����� y � ���� where

� �

���� � �� ��

���� � ���� �� �

���� ��� ��� ��

���� � ����� �� �

���� � ���� ��

���� � �� ��So x � ��

�and y � ���

��

THEOREM ����� Let A be a square matrix� If A is nonsingular thehomogeneous system AX � � has only the trivial solution� Equivalentlyif the homogenous system AX � � has a nontrivial solution then A issingular�

Proof� If A is nonsingular and AX � � then X � A��� � ��

REMARK ����� If A��� � � � � A�n denote the columns of A then the equa�tion

AX � x�A�� � � � �� xnA�n

holds� Consequently theorem ����� tells us that if there exist scalars x�� � � � � xnnot all zero such that

x�A�� � � � �� xnA�n � ��

Page 46: Elementary Linear Algebra

���� NON�SINGULAR MATRICES ��

that is if the columns of A are linearly dependent then A is singular� Anequivalent way of saying that the columns of A are linearly dependent is thatone of the columns of A is expressible as a sum of certain scalar multiplesof the remaining columns of A� that is one column is a linear combination

of the remaining columns�

EXAMPLE �����

A �

�� � �

� � � � �

��

is singular� For it can be veri�ed that A has reduced rowechelon form�� � � �

� � �� � �

��

and consequently AX � � has a nontrivial solution x � ��� y � ��� z � ��

REMARK ����� More generally if A is rowequivalent to a matrix con�taining a zero row then A is singular� For then the homogeneous systemAX � � has a nontrivial solution�

An important class of nonsingular matrices is that of the elementary

row matrices�

DEFINITION ����� �Elementary row matrices� There are three typesEij� Ei�t�� Eij�t� corresponding to the three kinds of elementary row oper�ation�

�� Eij � �i �� j� is obtained from the identity matrix In by interchangingrows i and j�

�� Ei�t�� �t �� �� is obtained by multiplying the ith row of In by t�

� Eij�t�� �i �� j� is obtained from In by adding t times the jth row ofIn to the ith row�

EXAMPLE ����� �n � ��

E�� �

�� � � �

� � �� � �

�� � E����� �

�� � � �

� �� �� � �

�� � E������ �

�� � � �

� � ��� � �

�� �

Page 47: Elementary Linear Algebra

�� CHAPTER �� MATRICES

The elementary row matrices have the following distinguishing property�

THEOREM ���� If a matrix A is premultiplied by an elementary rowmatrix the resulting matrix is the one obtained by performing the corre�sponding elementary rowoperation on A�

EXAMPLE ����

E��

�� a b

c d

e f

�� �

�� � � �

� � �� � �

���� a b

c d

e f

�� �

�� a b

e f

c d

�� �

COROLLARY ����� The three types of elementary rowmatrices are nonsingular� Indeed

�� E��ij � Eij �

�� E��i �t� � Ei�t����

� �Eij�t���� � Eij��t��

Proof� Taking A � In in the above theorem we deduce the followingequations�

EijEij � In

Ei�t�Ei�t��� � In � Ei�t

���Ei�t� if t �� �

Eij�t�Eij��t� � In � Eij��t�Eij�t��

EXAMPLE ����� Find the � matrix A � E����E�����E�� explicitly�Also �nd A���

Solution�

A � E����E�����

�� � � �� � �� � �

�� � E����

�� � � �� � �� � �

�� �

�� � � �

� � �� � �

�� �

To �nd A�� we have

A�� � �E����E�����E�����

� E���� �E�������� �E�����

��

� E��E������E������

Page 48: Elementary Linear Algebra

���� NON�SINGULAR MATRICES �

� E��E�������� � � �� � �� � �

��

� E��

�� � � �

� � ��

� � �

�� �

�� � � ��

� � �� � �

�� �

REMARK ����� Recall that A and B are rowequivalent if B is obtainedfrom A by a sequence of elementary row operations� If E�� � � � � Er are therespective corresponding elementary row matrices then

B � Er �� � ��E��E�A�� � � �� � �Er � � �E��A � PA�

where P � Er � � �E� is nonsingular� Conversely if B � PA where P isnonsingular then A is rowequivalent to B� For as we shall now see P isin fact a product of elementary row matrices�

THEOREM ����� Let A be nonsingular n� n matrix� Then

�i� A is rowequivalent to In

�ii� A is a product of elementary row matrices�

Proof� Assume thatA is nonsingular and let B be the reduced rowechelonform of A� Then B has no zero rows for otherwise the equation AX � �would have a nontrivial solution� Consequently B � In�

It follows that there exist elementary row matrices E�� � � � � Er such thatEr �� � � �E�A� � � �� � B � In and hence A � E��� � � �E��r a product ofelementary row matrices�

THEOREM ���� Let A be n � n and suppose that A is rowequivalentto In� Then A is nonsingular and A�� can be found by performing thesame sequence of elementary row operations on In as were used to convertA to In�

Proof� Suppose that Er � � �E�A � In� In other words BA � In whereB � Er � � �E� is nonsingular� Then B���BA� � B��In and so A � B��which is nonsingular�

Also A�� � B��

���� B � Er ��� � � �E�In� � � �� which shows that A��

is obtained from In by performing the same sequence of elementary rowoperations as were used to convert A to In�

Page 49: Elementary Linear Algebra

�� CHAPTER �� MATRICES

REMARK ���� It follows from theorem ����� that if A is singular thenA is rowequivalent to a matrix whose last row is zero�

EXAMPLE ���� Show that A �

�� �� �

�is nonsingular �nd A�� and

express A as a product of elementary row matrices�

Solution� We form the partitionedmatrix �AjI�� which consists ofA followedby I�� Then any sequence of elementary row operations which reduces A toI� will reduce I� to A

��� Here

�AjI�� ��� � � �� � � �

R� � R� �R�

�� � � �� �� �� �

R� � ����R�

�� � � �� � � ��

R� � R� � �R�

�� � �� �� � � ��

��

Hence A is rowequivalent to I� and A is nonsingular� Also

A�� �

� �� �� ��

��

We also observe that

E������E�����E������A � I��

Hence

A�� � E������E�����E������A � E�����E�����E������

The next result is the converse of Theorem ����� and is useful for provingthe nonsingularity of certain types of matrices�

THEOREM ������ Let A be an n � n matrix with the property thatthe homogeneous system AX � � has only the trivial solution� Then A isnonsingular� Equivalently if A is singular then the homogeneous systemAX � � has a nontrivial solution�

Page 50: Elementary Linear Algebra

���� NON�SINGULAR MATRICES ��

Proof� If A is n � n and the homogeneous system AX � � has only thetrivial solution then it follows that the reduced rowechelon form B of Acannot have zero rows and must therefore be In� Hence A is nonsingular�

COROLLARY ����� Suppose that A and B are n � n and AB � In�Then BA � In�

Proof� Let AB � In where A and B are n � n� We �rst show that Bis nonsingular� Assume BX � �� Then A�BX� � A� � � so �AB�X ��� InX � � and hence X � ��

Then from AB � In we deduce �AB�B�� � InB�� and hence A � B���

The equation BB�� � In then gives BA � In�

Before we give the next example of the above criterion for non�singularitywe introduce an important matrix operation�

DEFINITION ����� �The transpose of a matrix� Let A be an m�nmatrix� Then At the transpose of A is the matrix obtained by interchangingthe rows and columns of A� In other words if A � �aij � then

At�ji� aij �

Consequently At is n �m�

The transpose operation has the following properties�

�� At�t� A�

�� �A B�t � At Bt if A and B are m� n�

� �sA�t � sAt if s is a scalar�

�� �AB�t � BtAt if A is m� n and B is n� p�

�� If A is nonsingular then At is also nonsingular and At���

� A��

�t�

�� X tX � x�� � � � �� x�n if X � �x�� � � � � xn�t is a column vector�

We prove only the fourth property� First check that both �AB�t and BtAt

have the same size �p � m�� Moreover corresponding elements of bothmatrices are equal� For if A � �aij � and B � �bjk� we have

�AB�t�ki

� �AB�ik

�nX

j��

aijbjk

Page 51: Elementary Linear Algebra

�� CHAPTER �� MATRICES

�nX

j��

Bt�kj

At�ji

� BtAt

�ki�

There are two important classes of matrices that can be de�ned conciselyin terms of the transpose operation�

DEFINITION ����� �Symmetric matrix� A real matrixA is called sym�

metric if At � A� In other words A is square �n � n say� and aji � aij forall � � i � n� � � j � n� Hence

A �

�a bb c

�is a general �� � symmetric matrix�

DEFINITION ����� �Skewsymmetric matrix� A real matrixA is calledskew�symmetric if At � �A� In other words A is square �n � n say� andaji � �aij for all � � i � n� � � j � n�

REMARK ����� Taking i � j in the de�nition of skewsymmetric matrixgives aii � �aii and so aii � �� Hence

A �

�� b�b �

�is a general �� � skewsymmetric matrix�

We can now state a second application of the above criterion for nonsingularity�

COROLLARY ����� Let B be an n � n skewsymmetric matrix� ThenA � In �B is nonsingular�

Proof� Let A � In � B where Bt � �B� By Theorem ������ it su�ces toshow that AX � � implies X � ��

We have �In �B�X � � so X � BX � Hence X tX � X tBX �Taking transposes of both sides gives

�X tBX�t � �X tX�t

X tBt�X t�t � X t�X t�t

X t��B�X � X tX

�X tBX � X tX � X tBX�

Hence X tX � �X tX and X tX � �� But if X � �x�� � � � � xn�t then X tX �

x�� � � � �� x�n � � and hence x� � �� � � � � xn � ��

Page 52: Elementary Linear Algebra

���� LEAST SQUARES SOLUTION OF EQUATIONS ��

��� Least squares solution of equations

Suppose AX � B represents a system of linear equations with real coe��cients which may be inconsistent because of the possibility of experimentalerrors in determining A or B� For example the system

x � �

y � �

x� y � ����

is inconsistent�It can be proved that the associated system AtAX � AtB is always

consistent and that any solution of this system minimizes the sum r��� � � ��r�m where r�� � � � � rm �the residuals� are de�ned by

ri � ai�x� � � � �� ainxn � bi�

for i � �� � � � � m� The equations represented by AtAX � AtB are called thenormal equations corresponding to the system AX � B and any solutionof the system of normal equations is called a least squares solution of theoriginal system�

EXAMPLE ����� Find a least squares solution of the above inconsistentsystem�

Solution� Here A �

�� � �

� �� �

�� � X �

�x

y

�� B �

�� �

� ����

���

Then AtA �

�� � �� � �

��� � �� �� �

�� �

�� �� �

��

Also AtB �

�� � �� � �

��� ��

����

�� �

�����������

��

So the normal equations are

�x� y � �����

x� �y � �����

which have the unique solution

x � ����

� y �

�����

Page 53: Elementary Linear Algebra

�� CHAPTER �� MATRICES

EXAMPLE ����� Points �x�� y��� � � � � �xn� yn� are experimentally deter�mined and should lie on a line y � mx� c� Find a least squares solution tothe problem�

Solution� The points have to satisfy

mx� � c � y����

mxn � c � yn�

or Ax � B where

A �

���

x� ����

���xn �

��� � X �

�m

c

�� B �

���

y����yn

��� �

The normal equations are given by �AtA�X � AtB� Here

AtA �

�x� � � � xn� � � � �

����x� ����

���xn �

��� �

�x�� � � � �� x�n x� � � � �� xnx� � � � �� xn n

Also

AtB �

�x� � � � xn� � � � �

����y����yn

��� �

�x�y� � � � �� xnyny� � � � �� yn

��

It is not di�cult to prove that

� � det �AtA� �X

��i�j�n

�xi � xj���

which is positive unless x� � � � � � xn� Hence if not all of x�� � � � � xn areequal AtA is nonsingular and the normal equations have a unique solution�This can be shown to be

m ��

X��i�j�n

�xi � xj��yi � yj�� c ��

X��i�j�n

�xiyj � xjyi��xi � xj��

REMARK ����� The matrix AtA is symmetric�

Page 54: Elementary Linear Algebra

���� PROBLEMS ��

��� PROBLEMS

�� Let A �

�� �

� �

�� Prove that A is nonsingular �nd A�� and

express A as a product of elementary row matrices�

�Answer� A�� �

��

��� �

���

��

��

A � E���� �E��� �E����� is one such decomposition��

�� A square matrix D � �dij� is called diagonal if dij � � for i �� j� �Thatis the o��diagonal elements are zero�� Prove that premultiplicationof a matrix A by a diagonal matrix D results in matrix DA whoserows are the rows of A multiplied by the respective diagonal elementsof D� State and prove a similar result for postmultiplication by adiagonal matrix�

Let diag �a�� � � � � an� denote the diagonal matrix whose diagonal ele�ments dii are a�� � � � � an respectively� Show that

diag �a�� � � � � an�diag �b�� � � � � bn� � diag �a�b�� � � � � anbn�

and deduce that if a� � � �an �� � then diag �a�� � � � � an� is nonsingularand

�diag �a�� � � � � an���� � diag �a��� � � � � � a��n ��

Also prove that diag �a�� � � � � an� is singular if ai � � for some i�

� Let A �

�� � � �

� � � � �

��� Prove that A is nonsingular �nd A�� and

express A as a product of elementary row matrices�

�Answers� A�� �

�� ��� � ��

�� �

�� �

��

A � E��E��� �E��E����E�����E������E������ is one such decompo�sition��

Page 55: Elementary Linear Algebra

�� CHAPTER �� MATRICES

�� Find the rational number k for which the matrix A �

�� � � k

�� �� ��

��

is singular� �Answer� k � � ��

�� Prove thatA �

�� �

�� ���is singular and �nd a nonsingular matrix

P such that PA has last row zero�

�� If A �

�� �

� �

� verify that A� � �A � � I� � � and deduce that

A�� � � �

���A� �I���

�� Let A �

�� � � ��

� � �� � �

���

�i� Verify that A� � A� � A� I��

�ii� Express A� in terms of A�� A and I� and hence calculate A�

explicitly�

�iii� Use �i� to prove that A is nonsingular and �nd A�� explicitly�

�Answers� �ii� A� � �A� � �A� I� �

�� ��� �� ��

�� � ��� �� �

���

�iii� A�� � A� � A� I� �

�� �� � �

� � ��� � �

����

�� �i� Let B be an n�n matrix such that B� � �� If A � In�B provethat A is nonsingular and A�� � In �B �B��

Show that the system of linear equations AX � b has the solution

X � b� Bb� B�b�

�ii� If B �

�� � r s

� � t� � �

�� verify that B� � � and use �i� to determine

�I� �B��� explicitly�

Page 56: Elementary Linear Algebra

���� PROBLEMS ��

�Answer�

�� � r s � rt

� � t� � �

����

�� Let A be n� n�

�i� If A� � � prove that A is singular�

�ii� If A� � A and A �� In prove that A is singular�

��� Use Question � to solve the system of equations

x� y � z � a

z � b

�x� y � �z � c

where a� b� c are given rationals� Check your answer using the GaussJordan algorithm�

�Answer� x � �a � b� c� y � �a� �b� c� z � b��

��� Determine explicitly the following products of � elementary rowmatrices�

�i� E��E�� �ii� E����E�� �iii� E��� �E���� � �iv� �E���������

�v� E���� �vi� �E�������� �vii� �E�����E���������

�Answers� �i�

�� � � �

� � �� � �

�� �ii�

�� � � �

� � �� � �

�� �iii�

�� �� �� � �� � �

��

�iv�

�� �

���� �

� � �� � �

�� �v�

�� � � �

� � �� � �

�� �vi�

�� � �� �

� � �� � �

�� �vii�

�� � �� �

� � ��� � �

����

��� Let A be the following product of �� � elementary row matrices�

A � E����E��E��� ��

Find A and A�� explicitly�

�Answers� A �

����

� � �� � � �� � � �� � � �

���� � A�� �

����

� � � �� � � �� � �

��

� � � �

������

Page 57: Elementary Linear Algebra

�� CHAPTER �� MATRICES

� � Determine which of the following matrices over Z� are nonsingularand �nd the inverse where possible�

�a�

����

� � � �� � � �� � � �� � � �

���� �b�

����

� � � �� � � �� � � �� � � �

�����

�Answer� �a�

����

� � � �� � � �� � � �� � � �

������

��� Determine which of the following matrices are nonsingular and �ndthe inverse where possible�

�a�

�� � � ��� � �� � �

�� �b�

�� � � �� � �� � �

�� �c�

�� � � �

� � �� � �

��

�d�

�� � � �

� �� �� � �

�� �e�

����

� � � �� � � �� � � �� � � �

���� �f�

�� � �

� � �� � �

���

�Answers� �a�

�� � � �

� � �

� �� ��

�� �b�

�� ��

�� �

� � ��

��� ��

�� �d�

�� �

�� �

� ��

��

� � �

��

�e�

����

� �� � � � � �� �� � � ��� � � �

������

��� Let A be a nonsingular n � n matrix� Prove that At is nonsingularand that �At��� � �A���t�

��� Prove that A �

�a bc d

�has no inverse if ad� bc � ��

�Hint� Use the equation A� � �a� d�A� �ad� bc�I� � ���

Page 58: Elementary Linear Algebra

���� PROBLEMS �

��� Prove that the real matrix A �

�� � a b

�a � c�b �c �

�� is nonsingular by

proving that A is rowequivalent to I��

��� If P��AP � B prove that P��AnP � Bn for n � ��

��� Let A �

��

��

�� P �

��

�� �

�� Verify that P��AP �

��

���

� �

�and deduce that

An ��

� � �

���

��

��

�n �� �

��

��

��� Let A �

�a b

c d

�be aMarkovmatrix� that is a matrix whose elements

are nonnegative and satisfy a�c � � � b�d� Also let P �

�b �c ��

��

Prove that if A �� I� then

�i� P is nonsingular and P��AP �

�� �� a� d� �

�ii� An � �

b� c

�b b

c c

�as n� if A ��

�� �� �

��

��� If X �

�� � �

�� �

�� and Y �

�� ��

�� �nd XX t� X tX� Y Y t� Y tY �

�Answers�

�� � �� ��

�� �� ��� � ��

�� � � � ��

�� ��

��

�� � � ��� � ���� �� ��

�� � ����

��� Prove that the system of linear equations

x� �y � �x� y � �

x� �y � ��

is inconsistent and �nd a least squares solution of the system�

�Answer� x � �� y � ������

Page 59: Elementary Linear Algebra

�� CHAPTER �� MATRICES

� � The points ��� ��� ��� ��� ��� ���� � � ��� ��� �� are required to lie on aparabola y � a � bx � cx�� Find a least squares solution for a� b� c�Also prove that no parabola passes through these points�

�Answer� a � �

�� b � ��� c � ���

��� If A is a symmetric n�n real matrix and B is n�m prove that BtAB

is a symmetric m�m matrix�

��� If A is m� n and B is n�m prove that AB is singular if m � n�

��� Let A and B be n � n� If A or B is singular prove that AB is alsosingular�

Page 60: Elementary Linear Algebra

Chapter �

SUBSPACES

��� Introduction

Throughout this chapter� we will be studying Fn� the set of all n�dimensionalcolumn vectors with components from a �eld F � We continue our study ofmatrices by considering an important class of subsets of Fn called subspaces�These arise naturally for example� when we solve a system of m linear ho�mogeneous equations in n unknowns�

We also study the concept of linear dependence of a family of vectors�This was introduced brie�y in Chapter �� Remark ������ Other topics dis�cussed are the row space� column space and null space of a matrix over F �the dimension of a subspace� particular examples of the latter being the rankand nullity of a matrix�

��� Subspaces of F n

DEFINITION ����� A subset S of Fn is called a subspace of Fn if

� The zero vector belongs to S �that is� � � S

�� If u � S and v � S� then u � v � S �S is said to be closed undervector addition

�� If u � S and t � F � then tu � S �S is said to be closed under scalarmultiplication �

EXAMPLE ����� Let A � Mm�n�F � Then the set of vectors X � Fn

satisfying AX � � is a subspace of Fn called the null space of A and isdenoted here by N�A � �It is sometimes called the solution space of A�

��

Page 61: Elementary Linear Algebra

�� CHAPTER �� SUBSPACES

Proof� � A� � �� so � � N�A �� If X� Y � N�A � then AX � � andAY � �� so A�X � Y � AX �AY � � � � � � and so X � Y � N�A �� If X � N�A and t � F � then A�tX � t�AX � t� � �� so tX � N�A �

For example� if A �

� ��

�� then N�A � f�g� the set consisting of

just the zero vector� If A �

� �� �

�� then N�A is the set of all scalar

multiples of ���� �t�

EXAMPLE ����� Let X�� � � � � Xm � Fn� Then the set consisting of alllinear combinations x�X� � � � �� xmXm� where x�� � � � � xm � F � is a sub�space of Fn� This subspace is called the subspace spanned or generated byX�� � � � � Xm and is denoted here by hX�� � � � � Xmi� We also call X�� � � � � Xm

a spanning family for S � hX�� � � � � Xmi�

Proof� � � � �X� � � � � � �Xm� so � � hX�� � � � � Xmi �� If X� Y �hX�� � � � � Xmi� then X � x�X� � � � �� xmXm and Y � y�X� � � � �� ymXm�so

X � Y � �x�X� � � � �� xmXm � �y�X� � � � �� ymXm

� �x� � y� X� � � � �� �xm � ym Xm � hX�� � � � � Xmi�

�� If X � hX�� � � � � Xmi and t � F � then

X � x�X� � � � �� xmXm

tX � t�x�X� � � � �� xmXm

� �tx� X� � � � �� �txm Xm � hX�� � � � � Xmi�

For example� if A �Mm�n�F � the subspace generated by the columns of Ais an important subspace of Fm and is called the column space of A� Thecolumn space of A is denoted here by C�A � Also the subspace generatedby the rows of A is a subspace of Fn and is called the row space of A and isdenoted by R�A �

EXAMPLE ����� For example Fn � hE�� � � � � Emi� where E�� � � � � En arethe n�dimensional unit vectors� For if X � �x�� � � � � xn�t � Fn� then X �x�E� � � � �� xnEn�

EXAMPLE ����� Find a spanning family for the subspace S ofR� de�nedby the equation �x� �y � �z � ��

Page 62: Elementary Linear Algebra

���� SUBSPACES OF FN ��

Solution� �S is in fact the null space of ��� ��� ��� so S is indeed a subspaceof R��

If �x� y� z�t � S� then x � �

�y � �

�z� Then�

� x

y

z

�� �

��

�y � �

�z

y

z

�� � y

��

�� � z

�� ��

��

and conversely� Hence ���� � ��t and ���

�� �� �t form a spanning family for

S�The following result is easy to prove�

LEMMA ����� Suppose each of X�� � � � � Xr is a linear combination ofY�� � � � � Ys� Then any linear combination of X�� � � � � Xr is a linear combi�nation of Y�� � � � � Ys�

As a corollary we have

THEOREM ����� Subspaces hX�� � � � � Xri and hY�� � � � � Ysi are equal ifeach ofX�� � � � � Xr is a linear combination of Y�� � � � � Ys and each of Y�� � � � � Ysis a linear combination of X�� � � � � Xr�

COROLLARY ����� Subspaces hX�� � � � � Xr� Z�� � � � � Zti and hX�� � � � � Xriare equal if each of Z�� � � � � Zt is a linear combination of X�� � � � � Xr�

EXAMPLE ����� If X and Y are vectors in Rn� then

hX� Y i � hX � Y� X � Y i�

Solution� Each of X � Y and X � Y is a linear combination of X and Y �Also

X �

��X � Y �

��X � Y and Y �

��X � Y �

��X � Y �

so each of X and Y is a linear combination of X � Y and X � Y �

There is an important application of Theorem ���� to row equivalentmatrices �see De�nition ���� �

THEOREM ����� If A is row equivalent to B� then R�A � R�B �

Proof� Suppose that B is obtained from A by a sequence of elementary rowoperations� Then it is easy to see that each row of B is a linear combinationof the rows of A� But A can be obtained from B by a sequence of elementaryoperations� so each row of A is a linear combination of the rows of B� Henceby Theorem ����� R�A � R�B �

Page 63: Elementary Linear Algebra

�� CHAPTER �� SUBSPACES

REMARK ����� If A is row equivalent to B� it is not always true thatC�A � C�B �

For example� if A �

�and B �

� � �

�� then B is in fact the

reduced row�echelon form of A� However we see that

C�A �

��

��

���

��

��

and similarly C�B �

���

���

Consequently C�A �� C�B � as

�� C�A but

��� C�B �

��� Linear dependence

We now recall the de�nition of linear dependence and independence of afamily of vectors in Fn given in Chapter ��

DEFINITION ����� Vectors X�� � � � � Xm in Fn are said to be linearly

dependent if there exist scalars x�� � � � � xm� not all zero� such that

x�X� � � � �� xmXm � ��

In other words� X�� � � � � Xm are linearly dependent if some Xi is expressibleas a linear combination of the remaining vectors�

X�� � � � � Xm are called linearly independent if they are not linearly depen�dent� Hence X�� � � � � Xm are linearly independent if and only if the equation

x�X� � � � �� xmXm � �

has only the trivial solution x� � �� � � � � xm � ��

EXAMPLE ����� The following three vectors in R�

X� �

��

��

�� � X� �

�� �

�� � X� �

�� �

��

��

are linearly dependent� as �X� � �X� � �� X� � ��

Page 64: Elementary Linear Algebra

���� LINEAR DEPENDENCE ��

REMARK ����� If X�� � � � � Xm are linearly independent and

x�X� � � � �� xmXm � y�X� � � � �� ymXm�

then x� � y�� � � � � xm � ym� For the equation can be rewritten as

�x� � y� X� � � � �� �xm � ym Xm � �

and so x� � y� � �� � � � � xm � ym � ��

THEOREM ����� A family ofm vectors in Fn will be linearly dependentif m � n� Equivalently� any linearly independent family of m vectors in Fn

must satisfy m � n�

Proof� The equation

x�X� � � � �� xmXm � �

is equivalent to n homogeneous equations inm unknowns� By Theorem ����such a system has a non�trivial solution if m � n�

The following theorem is an important generalization of the last resultand is left as an exercise for the interested student�

THEOREM ����� A family of s vectors in hX�� � � � � Xri will be linearlydependent if s � r� Equivalently� a linearly independent family of s vectorsin hX�� � � � � Xri must have s � r�

Here is a useful criterion for linear independence which is sometimescalled the left�to�right test�

THEOREM ����� Vectors X�� � � � � Xm in Fn are linearly independent if

�a X� �� �

�b For each k with � k � m� Xk is not a linear combination ofX�� � � � � Xk���

One application of this criterion is the following result�

THEOREM ����� Every subspace S of Fn can be represented in the formS � hX�� � � � � Xmi� where m � n�

Page 65: Elementary Linear Algebra

�� CHAPTER �� SUBSPACES

Proof� If S � f�g� there is nothing to prove � we take X� � � and m � �

So we assume S contains a non�zero vector X� then hX�i � S as S is asubspace� If S � hX�i� we are �nished� If not� S will contain a vector X��not a linear combination of X� then hX�� X�i � S as S is a subspace� IfS � hX�� X�i� we are �nished� If not� S will contain a vector X� which isnot a linear combination of X� and X�� This process must eventually stop�for at stage k we have constructed a family of k linearly independent vectorsX�� � � � � Xk� all lying in Fn and hence k � n�

There is an important relationship between the columns of A and B� ifA is row�equivalent to B�

THEOREM ����� Suppose that A is row equivalent toB and let c�� � � � � crbe distinct integers satisfying � ci � n� Then

�a Columns A�c� � � � � � A�cr of A are linearly dependent if and only if thecorresponding columns of B are linearly dependent indeed more istrue�

x�A�c� � � � �� xrA�cr � �� x�B�c� � � � �� xrB�cr � ��

�b Columns A�c� � � � � � A�cr of A are linearly independent if and only if thecorresponding columns of B are linearly independent�

�c If � cr�� � n and cr�� is distinct from c�� � � � � cr� then

A�cr�� � z�A�c� � � � �� zrA�cr � B�cr�� � z�B�c� � � � �� zrB�cr �

Proof� First observe that if Y � �y�� � � � � yn�t is an n�dimensional column

vector and A is m� n� then

AY � y�A�� � � � �� ynA�n�

Also AY � � � BY � �� if B is row equivalent to A� Then �a follows bytaking yi � xcj if i � cj and yi � � otherwise�

�b is logically equivalent to �a � while �c follows from �a as

A�cr�� � z�A�c� � � � �� zrA�cr � z�A�c� � � � �� zrA�cr � �� A�cr�� � �

� z�B�c� � � � �� zrB�cr � �� B�cr�� � �

� B�cr�� � z�B�c� � � � �� zrB�cr �

Page 66: Elementary Linear Algebra

���� BASIS OF A SUBSPACE �

EXAMPLE ����� The matrix

A �

�� � �

� � � �� � � � ��

��

has reduced row�echelon form equal to

B �

�� � � � �

� � � �� � � �

�� �

We notice that B��� B�� and B�� are linearly independent and hence so areA��� A�� and A��� Also

B�� � �B�� � �B��

B�� � �� B�� � �B�� � �B���

so consequently

A�� � �A�� � �A��

A�� � �� A�� � �A�� � �A���

��� Basis of a subspace

We now come to the important concept of basis of a vector subspace�

DEFINITION ����� Vectors X�� � � � � Xm belonging to a subspace S aresaid to form a basis of S if

�a Every vector in S is a linear combination of X�� � � � � Xm

�b X�� � � � � Xm are linearly independent�

Note that �a is equivalent to the statement that S � hX�� � � � � Xmi as weautomatically have hX�� � � � � Xmi � S� Also� in view of Remark ���� above��a and �b are equivalent to the statement that every vector in S is uniquelyexpressible as a linear combination of X�� � � � � Xm�

EXAMPLE ����� The unit vectors E�� � � � � En form a basis for Fn�

Page 67: Elementary Linear Algebra

�� CHAPTER �� SUBSPACES

REMARK ����� The subspace f�g� consisting of the zero vector alone�does not have a basis� For every vector in a linearly independent familymust necessarily be non�zero� �For example� if X� � �� then we have thenon�trivial linear relation

X� � �X� � � � �� �Xm � �

and X�� � � � � Xm would be linearly independent�

However if we exclude this case� every other subspace of Fn has a basis�

THEOREM ����� A subspace of the form hX�� � � � � Xmi� where at leastone of X�� � � � � Xm is non�zero� has a basis Xc� � � � � � Xcr � where � c� �

� � �� cr � m�

Proof� �The left�to�right algorithm � Let c� be the least index k for whichXk is non�zero� If c� � m or if all the vectors Xk with k � c� are linearcombinations of Xc� � terminate the algorithm and let r � � Otherwise letc� be the least integer k � c� such that Xk is not a linear combination ofXc� �

If c� � m or if all the vectors Xk with k � c� are linear combinationsof Xc� and Xc� � terminate the algorithm and let r � �� Eventually thealgorithm will terminate at the r�th stage� either because cr � m� or becauseall vectors Xk with k � cr are linear combinations of Xc� � � � � � Xcr �

Then it is clear by the construction ofXc� � � � � � Xcr � using Corollary ����that

�a hXc� � � � � � Xcri � hX�� � � � � Xmi

�b the vectors Xc� � � � � � Xcr are linearly independent by the left�to�righttest�

Consequently Xc� � � � � � Xcr form a basis �called the left�to�right basis forthe subspace hX�� � � � � Xmi�

EXAMPLE ����� Let X and Y be linearly independent vectors in Rn�Then the subspace h�� �X� X� �Y� X�Y i has left�to�right basis consistingof �X� �Y �

A subspace S will in general have more than one basis� For example� anypermutation of the vectors in a basis will yield another basis� Given oneparticular basis� one can determine all bases for S using a simple formula�This is left as one of the problems at the end of this chapter�

We settle for the following important fact about bases�

Page 68: Elementary Linear Algebra

���� BASIS OF A SUBSPACE ��

THEOREM ����� Any two bases for a subspace S must contain the samenumber of elements�

Proof� For if X�� � � � � Xr and Y�� � � � � Ys are bases for S� then Y�� � � � � Ysform a linearly independent family in S � hX�� � � � � Xri and hence s � r byTheorem ������ Similarly r � s and hence r � s�

DEFINITION ����� This number is called the dimension of S and iswritten dimS� Naturally we de�ne dim f�g � ��

It follows from Theorem ���� that for any subspace S of Fn� we must havedimS � n�

EXAMPLE ����� If E�� � � � � En denote the n�dimensional unit vectors inFn� then dim hE�� � � � � Eii � i for � i � n�

The following result gives a useful way of exhibiting a basis�

THEOREM ����� A linearly independent family of m vectors in a sub�space S� with dimS � m� must be a basis for S�

Proof� Let X�� � � � � Xm be a linearly independent family of vectors in asubspace S� where dim S � m� We have to show that every vector X � S isexpressible as a linear combination ofX�� � � � � Xm� We consider the followingfamily of vectors in S� X�� � � � � Xm� X � This family contains m� elementsand is consequently linearly dependent by Theorem ������ Hence we have

x�X� � � � �� xmXm � xm��X � �� ���

where not all of x�� � � � � xm�� are zero� Now if xm�� � �� we would have

x�X� � � � �� xmXm � ��

with not all of x�� � � � � xm zero� contradictiong the assumption thatX� � � � � Xm

are linearly independent� Hence xm�� �� � and we can use equation �� toexpress X as a linear combination of X�� � � � � Xm�

X ��x�xm��

X� � � � ���xmxm��

Xm�

Page 69: Elementary Linear Algebra

�� CHAPTER �� SUBSPACES

��� Rank and nullity of a matrix

We can now de�ne three important integers associated with a matrix�

DEFINITION ����� Let A �Mm�n�F � Then

�a column rankA �dimC�A

�b row rankA �dimR�A

�c nullityA �dimN�A �

We will now see that the reduced row�echelon form B of a matrix A allowsus to exhibit bases for the row space� column space and null space of A�Moreover� an examination of the number of elements in each of these baseswill immediately result in the following theorem�

THEOREM ����� Let A �Mm�n�F � Then

�a column rankA � row rankA

�b column rankA�nullityA � n�

Finding a basis for R�A � The r non�zero rows of B form a basis for R�A and hence row rankA � r�

For we have seen earlier that R�A � R�B � Also

R�B � hB��� � � � � Bm�i

� hB��� � � � � Br�� � � � � � �i

� hB��� � � � � Br�i�

The linear independence of the non�zero rows of B is proved as follows� Letthe leading entries of rows � � � � � r of B occur in columns c�� � � � � cr� Supposethat

x�B�� � � � �� xrBr� � ��

Then equating components c�� � � � � cr of both sides of the last equation� givesx� � �� � � � � xr � �� in view of the fact that B is in reduced row� echelonform�

Finding a basis for C�A � The r columns A�c� � � � � � A�cr form a basis forC�A and hence column rank A � r� For it is clear that columns c�� � � � � crof B form the left�to�right basis for C�B and consequently from parts �b and �c of Theorem ������ it follows that columns c�� � � � � cr of A form theleft�to�right basis for C�A �

Page 70: Elementary Linear Algebra

���� RANK AND NULLITY OF A MATRIX ��

Finding a basis for N�A � For notational simplicity� let us suppose that c� �� � � � � cr � r� Then B has the form

B �

������������

� � � � � b�r�� � � � b�n� � � � � b�r�� � � � b�n���

��� � � ����

��� � � ����

� � � � � brr�� � � � brn� � � � � � � � � � ����

��� � � ����

��� � � ����

� � � � � � � � � � �

���

Then N�B and hence N�A are determined by the equations

x� � ��b�r�� xr�� � � � �� ��b�n xn���

xr � ��brr�� xr�� � � � �� ��brn xn�

where xr��� � � � � xn are arbitrary elements of F � Hence the general vector Xin N�A is given by�

���������

x����xrxr�����xn

��

� xr��

����������

�b�r�����

�brr������

��� � � �� xn

����������

�bn���

�brn����

��

����

� xr��X� � � � �� xnXn�r�

Hence N�A is spanned by X�� � � � � Xn�r� as xr��� � � � � xn are arbitrary� AlsoX�� � � � � Xn�r are linearly independent� For equating the right hand side ofequation ��� to � and then equating components r � � � � � � n of both sidesof the resulting equation� gives xr�� � �� � � � � xn � ��

Consequently X�� � � � � Xn�r form a basis for N�A �

Theorem ���� now follows� For we have

row rankA � dimR�A � r

column rankA � dimC�A � r�

Hencerow rankA � column rankA�

Page 71: Elementary Linear Algebra

�� CHAPTER �� SUBSPACES

Also

column rankA � nullityA � r � dimN�A � r � �n� r � n�

DEFINITION ����� The common value of column rankA and row rankAis called the rank of A and is denoted by rankA�

EXAMPLE ����� Given that the reduced row�echelon form of

A �

�� � �

� � � �� � � � ��

��

equal to

B �

�� � � � �

� � � �� � � �

�� �

�nd bases for R�A � C�A and N�A �

Solution� �� �� �� �� ��� ��� � �� �� �� and ��� �� �� � �� form a basis forR�A � Also

A�� �

��

��

�� � A�� �

�� ��

�� � A�� �

��

��

��

form a basis for C�A �

Finally N�A is given by

������

x�x�x�x�x�

�� �

������

��x� � x���x� � �x�

x���x�x�

�� � x�

������

������

��� x�

������

���

��

�� � x�X� � x�X��

where x� and x� are arbitrary� Hence X� and X� form a basis for N�A �

Here rankA � � and nullityA � ��

EXAMPLE ����� Let A �

� �� �

�� Then B �

� �� �

�is the reduced

row�echelon form of A�

Page 72: Elementary Linear Algebra

���� PROBLEMS ��

Hence �� �� is a basis for R�A and

��

�is a basis for C�A � Also N�A

is given by the equation x� � ��x�� where x� is arbitrary� Then�x�x�

��

���x�

x�

�� x�

���

and hence

���

�is a basis for N�A �

Here rankA � and nullityA � �

EXAMPLE ����� Let A �

� �� �

�� Then B �

� ��

�is the reduced

row�echelon form of A�Hence �� ��� ��� � form a basis for R�A while �� ��� ��� �� form a basis

for C�A � Also N�A � f�g�Here rankA � � and nullityA � ��

We conclude this introduction to vector spaces with a result of greattheoretical importance�

THEOREM ����� Every linearly independent family of vectors in a sub�space S can be extended to a basis of S�

Proof� Suppose S has basis X�� � � � � Xm and that Y�� � � � � Yr is a linearlyindependent family of vectors in S� Then

S � hX�� � � � � Xmi � hY�� � � � � Yr� X�� � � � � Xmi�

as each of Y�� � � � � Yr is a linear combination of X�� � � � � Xm�Then applying the left�to�right algorithm to the second spanning family

for S will yield a basis for S which includes Y�� � � � � Yr�

��� PROBLEMS

� Which of the following subsets of R� are subspaces�

�a �x� y� satisfying x � �y

�b �x� y� satisfying x � �y and �x � y

�c �x� y� satisfying x � �y �

�d �x� y� satisfying xy � �

Page 73: Elementary Linear Algebra

�� CHAPTER �� SUBSPACES

�e �x� y� satisfying x � � and y � ��

�Answer� �a and �b ��

�� If X� Y� Z are vectors in Rn� prove that

hX� Y� Zi � hX � Y� X � Z� Y � Zi�

�� Determine if X� �

����

��

�� � X� �

����

��

�� and X� �

����

�� are linearly

independent in R��

�� For which real numbers � are the following vectors linearly independentin R��

X� �

�� �

��

�� � X� �

�� �

�� � X� �

�� ���

�� �

�� Find bases for the row� column and null spaces of the following matrixover Q�

A �

����

� � � � � � �� � � �� � �

�� �

�� Find bases for the row� column and null spaces of the following matrixover Z��

A �

����

� � � � �� � �

�� �

�� Find bases for the row� column and null spaces of the following matrixover Z��

A �

����

� � �� � � � �� � � � �� � � � � �

�� �

Page 74: Elementary Linear Algebra

���� PROBLEMS ��

�� Find bases for the row� column and null spaces of the matrix A de�nedin section ��� Problem �� �Note� In this question� F is a �eld of fourelements�

�� If X�� � � � � Xm form a basis for a subspace S� prove that

X�� X� �X�� � � � � X� � � � ��Xm

also form a basis for S�

�� Let A �

�a b c

�� Find conditions on a� b� c such that �a rankA �

�b rankA � ��

�Answer� �a a � b � c �b at least two of a� b� c are distinct��

� Let S be a subspace of Fn with dimS � m� If X�� � � � � Xm are vectorsin S with the property that S � hX�� � � � � Xmi� prove that X� � � � � Xm

form a basis for S�

�� Find a basis for the subspace S of R� de�ned by the equation

x� �y � �z � ��

Verify that Y� � ��� �� �t � S and �nd a basis for S which includesY��

�� Let X�� � � � � Xm be vectors in Fn� If Xi � Xj � where i � j� prove thatX�� � � �Xm are linearly dependent�

�� Let X�� � � � � Xm�� be vectors in Fn� Prove that

dim hX�� � � � � Xm��i � dim hX�� � � � � Xmi

if Xm�� is a linear combination of X�� � � � � Xm� but

dim hX�� � � � � Xm��i � dim hX�� � � � � Xmi�

if Xm�� is not a linear combination of X�� � � � � Xm�

Deduce that the system of linear equations AX � B is consistent� ifand only if

rank �AjB� � rankA�

Page 75: Elementary Linear Algebra

�� CHAPTER �� SUBSPACES

�� Let a�� � � � � an be elements of F � not all zero� Prove that the set ofvectors �x�� � � � � xn�

t where x�� � � � � xn satisfy

a�x� � � � �� anxn � �

is a subspace of Fn with dimension equal to n� �

�� Prove Lemma ����� Theorem ����� Corollary ���� and Theorem ������

�� Let R and S be subspaces of Fn� with R � S� Prove that

dimR � dimS

and that equality implies R � S� �This is a very useful way of provingequality of subspaces�

�� Let R and S be subspaces of Fn � If R S is a subspace of Fn� provethat R � S or S � R�

�� Let X�� � � � � Xr be a basis for a subspace S� Prove that all bases for Sare given by the family Y�� � � � � Yr� where

Yi �rX

j��

aijXj �

and where A � �aij � �Mr�r�F is a non�singular matrix�

Page 76: Elementary Linear Algebra

Chapter �

DETERMINANTS

DEFINITION ����� If A �

�a�� a��a�� a��

�� we de�ne the determinant of

A� �also denoted by detA�� to be the scalar

detA � a��a�� � a��a���

The notation

���� a�� a��a�� a��

���� is also used for the determinant of A�

If A is a real matrix� there is a geometrical interpretation of detA� IfP � �x�� y�� and Q � �x�� y�� are points in the plane� forming a triangle

with the origin O � ��� ��� then apart from sign� �

���� x� y�x� y�

���� is the area

of the triangle OPQ� For� using polar coordinates� let x� � r� cos �� and

y� � r� sin ��� where r� � OP and �� is the angle made by the ray�OP with

the positive x�axis� Then triangle OPQ has area �

�OP � OQ sin�� where

� � �POQ� If triangle OPQ has anti�clockwise orientation� then the ray�OQ makes angle �� � �� � � with the positive x�axis� �See Figure ���

Also x� � r� cos �� and y� � r� sin ��� Hence

AreaOPQ �

�OP �OQ sin�

�OP �OQ sin ��� � ���

�OP �OQ�sin �� cos �� � cos �� sin ���

��OQ sin �� �OP cos �� �OQ cos �� �OP sin ���

Page 77: Elementary Linear Algebra

�� CHAPTER �� DETERMINANTS

x

y

����

����

Q

P

O

���

������������

Figure � Area of triangle OPQ�

��y�x� � x�y��

���� x� y�x� y�

���� �Similarly� if triangle OPQ has clockwise orientation� then its area equals

���

���� x� y�x� y�

�����For a general triangle P�P�P�� with Pi � �xi� yi�� i � � �� �� we can

take P� as the origin� Then the above formula gives

���� x� � x� y� � y�x� � x� y� � y�

���� or �

���� x� � x� y� � y�x� � x� y� � y�

���� �according as vertices P�P�P� are anti�clockwise or clockwise oriented�

We now give a recursive de�nition of the determinant of an n�n matrixA � �aij �� n � ��

DEFINITION ����� �Minor� Let Mij�A� �or simply Mij if there is noambiguity� denote the determinant of the �n� �� �n� � submatrix of Aformed by deleting the i�th row and j�th column of A� �Mij�A� is calledthe �i� j� minor of A��

Assume that the determinant function has been de�ned for matrices ofsize �n����n��� Then detA is de�ned by the so�called �rst�row Laplace

Page 78: Elementary Linear Algebra

��

expansion

detA � a��M���A�� a��M���A� � � � �� �����nM�n�A�

�nX

j��

�����ja�jM�j�A��

For example� if A � �aij � is a �� � matrix� the Laplace expansion gives

detA � a��M���A�� a��M���A� � a��M���A�

� a��

���� a�� a��a�� a��

����� a��

���� a�� a��a�� a��

����� a��

���� a�� a��a�� a��

����� a���a��a�� � a��a���� a���a��a�� � a��a��� � a���a��a�� � a��a���

� a��a��a�� � a��a��a�� � a��a��a�� � a��a��a�� � a��a��a�� � a��a��a���

�The recursive de�nition also works for �� � determinants� if we de�ne thedeterminant of a � matrix �t� to be the scalar t

detA � a��M���A�� a��M���A� � a��a�� � a��a����

EXAMPLE ����� If P�P�P� is a triangle with Pi � �xi� yi�� i � � �� ��then the area of triangle P�P�P� is

������x� y� x� y� x� y�

������ or �

������x� y� x� y� x� y�

������ �

according as the orientation of P�P�P� is anti�clockwise or clockwise�

For from the de�nition of �� � determinants� we have

������x� y� x� y� x� y�

������ �

�x�

���� y� y�

����� y�

���� x� x�

��������� x� y�x� y�

�����

���� x� � x� y� � y�x� � x� y� � y�

���� �

One property of determinants that follows immediately from the de�ni�tion is the following

THEOREM ����� If a row of a matrix is zero� then the value of the de�terminant is zero�

Page 79: Elementary Linear Algebra

� CHAPTER �� DETERMINANTS

�The corresponding result for columns also holds� but here a simple proofby induction is needed��

One of the simplest determinants to evaluate is that of a lower triangularmatrix�

THEOREM ����� Let A � �aij �� where aij � � if i � j� Then

detA � a��a�� � � �ann� ���

An important special case is when A is a diagonal matrix�If A �diag �a�� � � � � an� then detA � a� � � � an� In particular� for a scalar

matrix tIn� we have det �tIn� � tn�

Proof� Use induction on the size n of the matrix�The result is true for n � �� Now let n � � and assume the result true

for matrices of size n � � If A is n � n� then expanding detA along row gives

detA � a��

���������

a�� � � � � �a�� a�� � � � ����an� an� � � � ann

���������� a���a�� � � �ann�

by the induction hypothesis�

If A is upper triangular� equation � remains true and the proof is againan exercise in induction� with the slight di�erence that the column versionof theorem ��� is needed�

REMARK ����� It can be shown that the expanded form of the determi�nant of an n � n matrix A consists of n� signed products �a�i�a�i� � � � anin �where �i�� i�� � � � � in� is a permutation of �� �� � � � � n�� the sign being or�� according as the number of inversions of �i�� i�� � � � � in� is even or odd�An inversion occurs when ir � is but r � s� �The proof is not easy and isomitted��

The de�nition of the determinant of an n � n matrix was given in termsof the �rst�row expansion� The next theorem says that we can expandthe determinant along any row or column� �The proof is not easy and isomitted��

Page 80: Elementary Linear Algebra

��

THEOREM �����

detA �nX

j��

���i�jaijMij�A�

for i � � � � � � n �the so�called i�th row expansion� and

detA �nX

i��

���i�jaijMij�A�

for j � � � � � � n �the so�called j�th column expansion��

REMARK ����� The expression ���i�j obeys the chess�board patternof signs �

����� � � � � �� � � � � �

� � � � � ����

� �

The following theorems can be proved by straightforward inductions onthe size of the matrix

THEOREM ����� A matrix and its transpose have equal determinants�that is

detAt � detA�

THEOREM ����� If two rows of a matrix are equal� the determinant iszero� Similarly for columns�

THEOREM ���� If two rows of a matrix are interchanged� the determi�nant changes sign�

EXAMPLE ����� If P� � �x�� y�� and P� � �x�� y�� are distinct points�then the line through P� and P� has equation

������x y x� y� x� y�

������ � ��

Page 81: Elementary Linear Algebra

�� CHAPTER �� DETERMINANTS

For� expanding the determinant along row � the equation becomes

ax� by � c � ��

where

a �

���� y� y�

���� � y� � y� and b � �

���� x� x�

���� � x� � x��

This represents a line� as not both a and b can be zero� Also this line passesthrough Pi� i � � �� For the determinant has its �rst and i�th rows equalif x � xi and y � yi and is consequently zero�

There is a corresponding formula in three�dimensional geometry� IfP�� P�� P� are non�collinear points in three�dimensional space� with Pi ��xi� yi� zi�� i � � �� �� then the equation

��������

x y z x� y� z� x� y� z� x� y� z�

��������� �

represents the plane through P�� P�� P�� For� expanding the determinantalong row � the equation becomes ax� by � cz � d � �� where

a �

������y� z� y� z� y� z�

������ � b � �������x� z� x� z� x� z�

������ � c �������x� y� x� y� x� y�

������ �

As we shall see in chapter �� this represents a plane if at least one of a� b� cis non�zero� However� apart from sign and a factor �

�� the determinant

expressions for a� b� c give the values of the areas of projections of triangleP�P�P� on the �y� z�� �x� z� and �x� y� planes� respectively� Geometrically�it is then clear that at least one of a� b� c is non�zero� It is also possible togive an algebraic proof of this fact�

Finally� the plane passes through Pi� i � � �� � as the determinant hasits �rst and i�th rows equal if x � xi� y � yi� z � zi and is consequentlyzero� We now work towards proving that a matrix is non�singular if itsdeterminant is non�zero�

DEFINITION ����� �Cofactor� The �i� j� cofactor of A� denoted byCij�A� �or Cij if there is no ambiguity� is de�ned by

Cij�A� � ���i�jMij�A��

Page 82: Elementary Linear Algebra

��

REMARK ����� It is important to notice that Cij�A�� like Mij�A�� doesnot depend on aij � Use will be made of this observation presently�

In terms of the cofactor notation� Theorem ����� takes the form

THEOREM ����

detA �nX

j��

aijCij�A�

for i � � � � � � n and

detA �nX

i��

aijCij�A�

for j � � � � � � n�

Another result involving cofactors is

THEOREM ����� Let A be an n� n matrix� Then

�a�nX

j��

aijCkj�A� � � if i �� k�

Also

�b�nX

i��

aijCik�A� � � if j �� k�

Proof�If A is n�n and i �� k� let B be the matrix obtained from A by replacing

row k by row i� Then detB � � as B has two identical rows�Now expand detB along row k� We get

� � detB �nX

j��

bkjCkj�B�

�nX

j��

aijCkj�A��

in view of Remark �����

Page 83: Elementary Linear Algebra

�� CHAPTER �� DETERMINANTS

DEFINITION ����� �Adjoint� If A � �aij � is an n � n matrix� the ad�

joint of A� denoted by adjA� is the transpose of the matrix of cofactors�Hence

adjA �

�����

C�� C�� � � � Cn�

C�� C�� � � � Cn�

������

C�n C�n � � � Cnn

� �

Theorems ���� and ���� may be combined to give

THEOREM ����� Let A be an n� n matrix� Then

A�adjA� � �detA�In � �adjA�A�

Proof�

�A adjA�ik �nX

j��

aij�adjA�jk

�nX

j��

aijCkj�A�

� �ikdetA

� ��detA�In�ik�

Hence A�adjA� � �detA�In� The other equation is proved similarly�

COROLLARY ����� �Formula for the inverse� If detA �� �� then Ais non�singular and

A�� �

detAadjA�

EXAMPLE ����� The matrix

A �

�� � �

� �� � �

is non�singular� For

detA �

���� � �� �

����� �

���� �� �

����� �

���� �� �

����� �� � �� �

� �� �� ��

Page 84: Elementary Linear Algebra

��

Also

A�� �

��

�� C�� C�� C��

C�� C�� C��

C�� C�� C��

� �

������������

���� � �� �

���� �

���� � �� �

�������� � �� �

����

���� �� �

�������� �� �

���� �

���� � �

�������� �� �

���� �

���� �� �

�������� � �

����

� �

�� �� � ��

� �� ��� � ��

� �

The following theorem is useful for simplifying and numerically evaluatinga determinant� Proofs are obtained by expanding along the correspondingrow or column�

THEOREM ������ The determinant is a linear function of each row andcolumn�For example

�a�

������a�� � a��� a�� � a��� a�� � a���

a�� a�� a��a�� a�� a��

������ �������a�� a�� a��a�� a�� a��a�� a�� a��

�������������a��� a��� a���a�� a�� a��a�� a�� a��

������

�b�

������ta�� ta�� ta��a�� a�� a��a�� a�� a��

������ � t

������a�� a�� a��a�� a�� a��a�� a�� a��

������ �

COROLLARY ����� If a multiple of a row is added to another row� thevalue of the determinant is unchanged� Similarly for columns�

Proof� We illustrate with a � � � example� but the proof is really quitegeneral�

������a�� � ta�� a�� � ta�� a�� � ta��

a�� a�� a��a�� a�� a��

������ �������a�� a�� a��a�� a�� a��a�� a�� a��

�������������ta�� ta�� ta��a�� a�� a��a�� a�� a��

������

Page 85: Elementary Linear Algebra

�� CHAPTER �� DETERMINANTS

������a�� a�� a��a�� a�� a��a�� a�� a��

������� t

������a�� a�� a��a�� a�� a��a�� a�� a��

������

������a�� a�� a��a�� a�� a��a�� a�� a��

������� t� �

������a�� a�� a��a�� a�� a��a�� a�� a��

������ �

To evaluate a determinant numerically� it is advisable to reduce the matrixto row�echelon form� recording any sign changes caused by row interchanges�together with any factors taken out of a row� as in the following examples�

EXAMPLE ����� Evaluate the determinant������ � � � �� � �

������ �

Solution� Using row operations R� � R� � R� and R� � R� � �R� andthen expanding along the �rst column� gives

������ � � � �� � �

������ �

������ � �� �� ��� �� ��

������ ����� �� ���� ��

����

� ��

���� ��� ��

���� � ������ ��

���� � ���EXAMPLE ����� Evaluate the determinant��������

� � �� � � �

���������

Solution���������

� � �� � � �

���������

��������

� � �� �� �� � �� ��� � �

��������

Page 86: Elementary Linear Algebra

� ��

��������

� � �� � �� ��� � �

��������

� ��

��������

� � �� � �� ��� � �

��������

� �

��������

� � �� � �� � �� ��

��������

� �

��������

� � �� � �� � � ��

��������� ���

EXAMPLE ���� �Vandermonde determinant� Prove that

������ a b ca� b� c�

������ � �b� a��c� a��c� b��

Solution� Subtracting column from columns � and � � then expandingalong row � gives

������ a b ca� b� c�

������ �

������ � �a b� a c� aa� b� � a� c� � a�

�������

���� b� a c� a

b� � a� c� � a�

����� �b� a��c� a�

���� b� a c� a

���� � �b� a��c� a��c� b��

REMARK ����� From theorems ����� ���� and corollary ����� we de�duce

�a� det �EijA� � �detA�

�b� det �Ei�t�A� � t detA� if t �� ��

Page 87: Elementary Linear Algebra

�� CHAPTER �� DETERMINANTS

�c� det �Eij�t�A� �detA�

It follows that if A is row�equivalent toB� then detB � cdetA� where c �� ��Hence detB �� � � detA �� � and detB � � � detA � �� Consequentlyfrom theorem ����� and remark ������ we have the following important result

THEOREM ������ Let A be an n � n matrix� Then

�i� A is non�singular if and only if detA �� ��

�ii� A is singular if and only if detA � ��

�iii� the homogeneous system AX � � has a non�trivial solution if andonly if detA � ��

EXAMPLE ���� Find the rational numbers a for which the followinghomogeneous system has a non�trivial solution and solve the system forthese values of a

x� �y � �z � �

ax� �y � �z � �

�x� y � az � ��

Solution� The coe�cient determinant of the system is

� �

������ �� �a � �� a

������ �

������ �� �� � � �a �� �a� � a� �

�������

���� � � �a �� �a� a� �

����� �� � �a��a� ��� ���� �a�

� �a� � �a� �� � ��a� ���a� ���

So � � � � a � �� or a � � and these values of a are the only values forwhich the given homogeneous system has a non�trivial solution�

If a � ��� the coe�cient matrix has reduced row�echelon form equal to

�� � �

� ��� � �

Page 88: Elementary Linear Algebra

��

and so the complete solution is x � z� y � �z� with z arbitrary� If a � ��the coe�cient matrix has reduced row�echelon form equal to�

� � � �� � �

and so the complete solution is x � �z� y � z� with z arbitrary�

EXAMPLE ����� Find the values of t for which the following system isconsistent and solve the system in each case

x� y �

tx � y � t

� � t�x� �y � ��

Solution� Suppose that the given system has a solution �x�� y��� Then thefollowing homogeneous system

x� y � z � �

tx � y � tz � �

� � t�x � �y � �z � �

will have a non�trivial solution

x � x�� y � y�� z � ��

Hence the coe�cient determinant � is zero� However

� �

������ t t

� t � �

������ �������

� �t � t �

� t � t �� t

������ ����� � t �� t �� t

���� � ��t����t��

Hence t � or t � �� If t � � the given system becomes

x� y �

x� y �

�x� �y � �

which is clearly inconsistent� If t � �� the given system becomes

x� y �

�x� y � �

�x� �y � �

Page 89: Elementary Linear Algebra

� CHAPTER �� DETERMINANTS

which has the unique solution x � � y � ��

To �nish this section� we present an old ����� method of solving asystem of n equations in n unknowns called Cramer�s rule � The method isnot used in practice� However it has a theoretical use as it reveals explicitlyhow the solution depends on the coe�cients of the augmented matrix�

THEOREM ������ �Cramer s rule� The system of n linear equationsin n unknowns x�� � � � � xn

a��x� � a��x� � � � �� a�nxn � b�

a��x� � a��x� � � � �� a�nxn � b����

an�x� � an�x� � � � �� annxn � bn

has a unique solution if � � det �aij � �� �� namely

x� ���

�� x� �

��

�� � � � � xn �

�n

��

where �i is the determinant of the matrix formed by replacing the i�thcolumn of the coe�cient matrix A by the entries b�� b�� � � � � bn�

Proof� Suppose the coe�cient determinant � �� �� Then by corollary ����A�� exists and is given by A�� � �

�adjA and the system has the unique

solution�����

x�x����xn

� � A��

�����

b�b����bn

� �

�����

C�� C�� � � � Cn�

C�� C�� � � � Cn�

������

C�n C�n � � � Cnn

�����

b�b����bn

�����

b�C�� � b�C�� � � � �� bnCn�

b�C�� � b�C�� � � � �� bnCn�

���bnC�n � b�C�n � � � �� bnCnn

� �

However the i�th component of the last vector is the expansion of �i alongcolumn i� Hence �

����x�x����xn

� �

�����

��

��

����n

� �

�����

��������

����n��

� �

Page 90: Elementary Linear Algebra

���� PROBLEMS ��

��� PROBLEMS

� If the points Pi � �xi� yi�� i � � �� �� form a quadrilateral with ver�tices in anti�clockwise orientation� prove that the area of the quadri�lateral equals

����� x� x�y� y�

���� ����� x� x�y� y�

��������� x� x�y� y�

��������� x� x�y� y�

������

�This formula generalizes to a simple polygon and is known as theSurveyor�s formula��

�� Prove that the following identity holds by expressing the left�handside as the sum of � determinants ������

a� x b� y c� z

x� u y � v z � wu� a v � b w � c

������ � �

������a b c

x y zu v w

������ �

�� Prove that ������n� �n� �� �n� ���

�n� �� �n� ��� �n� ���

�n� ��� �n� ��� �n� ��

������ � ���

� Evaluate the following determinants

�a�

�������� �� ���� �� ���� �� ��

������ �b�

��������

� � �� � �� � � � � �� �

���������

�Answers �a� ��������� �b� �����

�� Compute the inverse of the matrix

A �

�� � ��

� � � ��

by �rst computing the adjoint matrix�

�Answer A�� � ��

��

�� � � �

�� � �� ��

���

Page 91: Elementary Linear Algebra

�� CHAPTER �� DETERMINANTS

�� Prove that the following identities hold

�i�

�������a �b b� c

�b �a a� ca � b a� b b

������ � ���a� b���a� b��

�ii�

������b� c b cc c� a a

b a a� b

������ � �a�b� � c���

�� Let Pi � �xi� yi�� i � � �� �� If x�� x�� x� are distinct� prove that thereis precisely one curve of the form y � ax� � bx � c passing throughP�� P� and P��

�� Let

A �

�� �

� � k k �

� �

Find the values of k for which detA � � and hence� or otherwise�determine the value of k for which the following system has more thanone solution

x� y � z �

�x� �y � kz � �

x� ky � �z � ��

Solve the system for this value of k and determine the solution forwhich x� � y� � z� has least value�

�Answer k � �� x � ���� y � ���� z � �����

�� By considering the coe�cient determinant� �nd all rational numbers aand b for which the following system has �i� no solutions� �ii� exactlyone solution� �iii� in�nitely many solutions

x� �y � bz � �

ax� �z � �

�x� �y � �

Solve the system in case �iii��

�Answer �i� ab � � and a �� �� no solution� ab �� �� unique solution�a � �� b � � in�nitely many solutions� x � ��

�z� �

�� y � �

�z�

� with

z arbitrary��

Page 92: Elementary Linear Algebra

���� PROBLEMS ��

�� Express the determinant of the matrix

B �

����

� � � � � �t� �� � �� t t

as as polynomial in t and hence determine the rational values of t forwhich B�� exists�

�Answer detB � �t� ����t� �� t �� � and t �� ����

� If A is a �� � matrix over a �eld and detA �� �� prove that

�i� det �adjA� � �detA���

�ii� �adjA��� �

detAA � adj �A����

�� Suppose that A is a real �� � matrix such that AtA � I��

�i� Prove that At�A� I�� � ��A� I��t�

�ii� Prove that detA � ��

�iii� Use �i� to prove that if detA � � then det �A� I�� � ��

�� If A is a square matrix such that one column is a linear combination ofthe remaining columns� prove that detA � �� Prove that the conversealso holds�

� Use Cramer�s rule to solve the system

��x� �y � z � x� �y � z �

��x� y � z � ���

�Answer x � �� y � �� z � ��

�� Use remark ��� to deduce that

detEij � �� detEi�t� � t� detEij�t� �

and use theorem ����� and induction� to prove that

det �BA� � detB detA�

if B is non�singular� Also prove that the formula holds when B issingular�

Page 93: Elementary Linear Algebra

�� CHAPTER �� DETERMINANTS

�� Prove that��������

a� b� c a� b a aa� b a� b� c a a

a a a� b� c a� ba a a � b a� b� c

��������� c���b�c��a��b�c��

�� Prove that��������

� u� u� u� u�u� � u� u� u�u� u� � u� u�u� u� u� � u�

��������� � u� � u� � u� � u��

�� Let A � Mn�n�F �� If At � �A� prove that detA � � if n is odd and

� �� � in F �

�� Prove that ��������

r r r r r r

��������� �� r���

��� Express the determinant

������ a� � bc a�

b� � ca b�

c� � ab c�

������as the product of one quadratic and four linear factors�

�Answer �b� a��c� a��c� b��a� b� c��b�� bc� c� � ac� ab� a����

Page 94: Elementary Linear Algebra

Chapter �

COMPLEX NUMBERS

��� Constructing the complex numbers

One way of introducing the �eld C of complex numbers is via the arithmeticof �� � matrices�

DEFINITION ����� A complex number is a matrix of the form�x �yy x

��

where x and y are real numbers�

Complex numbers of the form

�x �� x

�are scalar matrices and are called

real complex numbers and are denoted by the symbol fxg�The real complex numbers fxg and fyg are respectively called the real

part and imaginary part of the complex number

�x �yy x

��

The complex number

�� ��� �

�is denoted by the symbol i�

We have the identities�x �yy x

��

�x �� x

��

�� �yy �

��

�x �� x

��

�� ��� �

� �y �� y

� fxg� ifyg�

i� �

�� ��� �

��� ��� �

��

� �� �� ��

�� f��g�

��

Page 95: Elementary Linear Algebra

�� CHAPTER �� COMPLEX NUMBERS

Complex numbers of the form ifyg where y is a nonzero real number arecalled imaginary numbers�

If two complex numbers are equal we can equate their real and imaginaryparts�

fx�g� ify�g � fx�g� ify�g � x� � x� and y� � y��

if x�� x�� y�� y� are real numbers� Noting that f�g � if�g � f�g gives theuseful special case is

fxg� ifyg � f�g � x � � and y � ��

if x and y are real numbers�The sum and product of two real complex numbers are also real complex

numbers�fxg� fyg � fx� yg� fxgfyg � fxyg�

Also as real complex numbers are scalar matrices their arithmetic is verysimple� They form a �eld under the operations of matrix addition andmultiplication� The additive identity is f�g the additive inverse of fxg isf�xg the multiplicative identity is f�g and the multiplicative inverse of fxgis fx��g� Consequently

fxg � fyg � fxg� ��fyg � fxg� f�yg � fx� yg�fxgfyg � fxgfyg�� � fxgfy��g � fxy��g �

�x

y

��

It is customary to blur the distinction between the real complex numberfxg and the real number x and write fxg as x� Thus we write the complexnumber fxg� ifyg simply as x� iy�

More generally the sum of two complex numbers is a complex number�

�x� � iy� � �x� � iy� � �x� � x� � i�y� � y� � ����

and �using the fact that scalar matrices commute with all matrices undermatrix multiplication and f��gA � �A if A is a matrix the product oftwo complex numbers is a complex number�

�x� � iy� �x� � iy� � x��x� � iy� � �iy� �x� � iy�

� x�x� � x��iy� � �iy� x� � �iy� �iy�

� x�x� � ix�y� � iy�x� � i�y�y�

� �x�x� � f��gy�y� � i�x�y� � y�x�

� �x�x� � y�y� � i�x�y� � y�x� � ����

Page 96: Elementary Linear Algebra

���� CALCULATING WITH COMPLEX NUMBERS ��

The set C of complex numbers forms a �eld under the operations ofmatrix addition and multiplication� The additive identity is � the additiveinverse of x � iy is the complex number ��x � i��y the multiplicativeidentity is � and the multiplicative inverse of the nonzero complex numberx� iy is the complex number u � iv where

u �x

x� � y�and v �

�yx� � y�

�If x� iy �� � then x �� � or y �� � so x� � y� �� ��

From equations ��� and ��� we observe that addition and multiplicationof complex numbers is performed just as for real numbers replacing i� by�� whenever it occurs�

A useful identity satis�ed by complex numbers is

r� � s� � �r � is �r� is �

This leads to a method of expressing the ratio of two complex numbers inthe form x� iy where x and y are real complex numbers�

x� � iy�x� � iy�

��x� � iy� �x� � iy�

�x� � iy� �x� � iy�

��x�x� � y�y� � i��x�y� � y�x�

x�� � y���

The process is known as rationalization of the denominator�

��� Calculating with complex numbers

We can now do all the standard linear algebra calculations over the �eld ofcomplex numbers �nd the reduced rowechelon form of an matrix whose el�ements are complex numbers solve systems of linear equations �nd inversesand calculate determinants�

For example���� � � i �� i

� �� �i

���� � �� � i ��� �i � ���� i

� ��� �i � i��� �i � �� � �i

� �� � ��i �� ��

Page 97: Elementary Linear Algebra

�� CHAPTER �� COMPLEX NUMBERS

Then by Cramer�s rule the linear system

�� � i z � ��� i w � � � �i

�z � ��� �i w � �� �i

has the unique solution

z �

���� � � �i �� i�� �i �� �i

������ � ��i

��� � �i ��� �i � ��� �i ��� i

�� � ��i

����� �i � ��i ��� �i � f����� i � �i��� i g

�� � ��i

���� �i� ��i� ��i� � f�� �i� ��i� �i�g

�� � ��i

���� ��i

�� � ��i

���� � ��i ���� ��i

��� � � ���

����� ���i

��� � � ���

����

���� ���

���i

and similarly w ��������

����

���i�

An important property enjoyed by complex numbers is that every com�plex number has a square root�

THEOREM �����If w is a nonzero complex number then the equation z� � w has preciselytwo solutions z � C �

Proof� Case �� Suppose b � �� Then if a � � z �pa is a solution while

if a � � ip�a is a solution�

Case �� Suppose b �� �� Let z � x� iy� w � a� ib� x� y� a� b � R� Thenthe equation z� � w becomes

�x� iy � � x� � y� � �xyi � a� ib�

Page 98: Elementary Linear Algebra

���� CALCULATING WITH COMPLEX NUMBERS ��

so equating real and imaginary parts gives

x� � y� � a and �xy � b�

Hence x �� � and y � b���x � Consequently

x� ��

b

�x

�� � a�

so �x� � �ax� � b� � � and ��x� � � �a�x� � b� � �� Hence

x� ��a� p��a� � ��b�

��

a�pa� � b�

��

However x� � � so we must take the � sign as a �pa� � b� � �� Hence

x� �a�

pa� � b�

�� x � �

sa�

pa� � b�

��

Then y is determined by y � b���x �

EXAMPLE ����� Solve the equation z� � � � i�

Solution� Put z � x� iy� Then the equation becomes

�x� iy � � x� � y� � �xyi � �� i�

so equating real and imaginary parts gives

x� � y� � � and �xy � ��

Hence x �� � and y � b���x � Consequently

x� ��

�x

�� � ��

so �x� � �x� � � � �� Hence

x� ��� p�� � ��

��

��p��

Hence

x� �� �

p�

�and x � �

s� �

p�

��

Page 99: Elementary Linear Algebra

�� CHAPTER �� COMPLEX NUMBERS

Then

y ��

�x� � �p

�p� �

p��

Hence the solutions are

z � ���s

� �p�

��

ip�p� �

p�

A �

EXAMPLE ����� Solve the equation z� � �p� � i z � � � ��

Solution� Because every complex number has a square root the familiarformula

z ��b�

pb� � �ac

�a

for the solution of the general quadratic equation az� � bz � c � � can beused where now a��� � � b� c � C � Hence

z ���p� � i �

q�p� � i � � �

���p� � i �

q�� � �

p�i� � � �

���p� � i �

p�� � �

p�i

��

Now we have to solve w� � �� � �p�i� Put w � x � iy� Then w� �

x� � y� � �xyi � �� � �p�i and equating real and imaginary parts gives

x� � y� � �� and �xy � �p�� Hence y �

p��x and so x� � ��x� � ��� So

x� � �x� � � � � and �x� � � �x� � � � �� Hence x� � � � � and x � ���Then y � �p�� Hence �� �p�i � � �� � �

p�i and the formula for z now

becomes

z ��p�� i� �� �

p�i

��� p� � �� �

p� i

�or

�� �p�� �� �p� i

��

EXAMPLE ����� Find the cube roots of ��

Page 100: Elementary Linear Algebra

���� GEOMETRIC REPRESENTATION OF C ��

Solution� We have to solve the equation z� � � or z� � � � �� Nowz� � � � �z � � �z� � z � � � So z� � � � �� z � � � � or z� � z � � � ��But

z� � z � � � �� z ���� p�� � �

�����p�i

��

So there are � cube roots of � namely i and ����p�i ���We state the next theorem without proof� It states that every non

constant polynomial with complex number coe�cients has a root in the�eld of complex numbers�

THEOREM ����� �Gauss� If f�z � anzn � an��z

n�� � � � �� a�z � a�where an �� � and n � � then f�z � � for some z � C �It follows that in view of the factor theorem which states that if a � F isa root of a polynomial f�z with coe�cients from a �eld F then z � a is afactor of f�z that is f�z � �z � a g�z where the coe�cients of g�z alsobelong to F � By repeated application of this result we can factorize anypolynomial with complex coe�cients into a product of linear factors withcomplex coe�cients�

f�z � an�z � z� �z � z� � � ��z � zn �

There are available a number of computational algorithms for �nding goodapproximations to the roots of a polynomial with complex coe�cients�

��� Geometric representation of C

Complex numbers can be represented as points in the plane using the cor�respondence x � iy � �x� y � The representation is known as the Argand

diagram or complex plane� The real complex numbers lie on the xaxiswhich is then called the real axis while the imaginary numbers lie on theyaxis which is known as the imaginary axis� The complex numbers withpositive imaginary part lie in the upper half plane while those with negativeimaginary part lie in the lower half plane�

Because of the equation

�x� � iy� � �x� � iy� � �x� � x� � i�y� � y� �

complex numbers add vectorially using the parallellogram law� Similarlythe complex number z� � z� can be represented by the vector from �x�� y� to �x�� y� where z� � x� � iy� and z� � x� � iy�� �See Figure ����

Page 101: Elementary Linear Algebra

�� CHAPTER �� COMPLEX NUMBERS

��

�z� � z�

z� � z�

z�

z�

�������������

����

�����

���������

����R

����R

����

����

��������

Figure ���� Complex addition and subraction�

The geometrical representation of complex numbers can be very usefulwhen complex number methods are used to investigate properties of trianglesand circles� It is very important in the branch of calculus known as ComplexFunction theory where geometric methods play an important role�

We mention that the line through two distinct points P� � �x�� y� andP� � �x�� y� has the form z � �� � t z� � tz�� t � R where z � x � iy isany point on the line and zi � xi� iyi� i � �� �� For the line has parametricequations

x � ��� t x� � tx�� y � ��� t y� � ty�

and these can be combined into a single equation z � ��� t z� � tz��Circles have various equation representations in terms of complex num�

bers as will be seen later�

��� Complex conjugate

DEFINITION ����� �Complex conjugate� If z � x � iy the complex

conjugate of z is the complex number de�ned by z � x� iy� Geometricallythe complex conjugate of z is obtained by re�ecting z in the real axis �seeFigure ��� �

The following properties of the complex conjugate are easy to verify�

Page 102: Elementary Linear Algebra

���� COMPLEX CONJUGATE ��

��

xy

z

z

�����

ZZZZ

Figure ���� The complex conjugate of z� z�

�� z� � z� � z� � z��

�� �z � � z�

�� z� � z� � z� � z��

�� z�z� � z� z��

�� ���z � ��z�

�� �z��z� � z��z��

�� z is real if and only if z � z�

�� With the standard convention that the real and imaginary parts aredenoted by Re z and Im z we have

Re z �z � z

�� Im z �

z � z

�i�

�� If z � x � iy then zz � x� � y��

THEOREM ����� If f�z is a polynomial with real coe�cients then itsnonreal roots occur in complexconjugate pairs i�e� if f�z � � thenf�z � ��

Proof� Suppose f�z � anzn � an��z

n�� � � � � � a�z � a� � � wherean� � � � � a� are real� Then

� � � � f�z � anzn � an��zn�� � � � �� a�z � a�

� an zn � an�� zn�� � � � �� a� z � a�

� anzn � an��z

n�� � � � �� a�z � a�

� f�z �

Page 103: Elementary Linear Algebra

�� CHAPTER �� COMPLEX NUMBERS

EXAMPLE ����� Discuss the position of the roots of the equation

z� � ��in the complex plane�

Solution� The equation z� � �� has real coe�cients and so its roots comein complex conjugate pairs� Also if z is a root so is �z� Also there areclearly no real roots and no imaginary roots� So there must be one root win the �rst quadrant with all remaining roots being given by w� �w and�w� In fact as we shall soon see the roots lie evenly spaced on the unitcircle�

The following theorem is useful in deciding if a polynomial f�z has amultiple root a� that is if �z� a m divides f�z for some m � �� �The proofis left as an exercise�

THEOREM ����� If f�z � �z � a mg�z where m � � and g�z is apolynomial then f ��a � � and the polynomial and its derivative have acommon root�

From theorem ����� we obtain a result which is very useful in the explicitintegration of rational functions �i�e� ratios of polynomials with real coe��cients�

THEOREM ����� If f�z is a nonconstant polynomial with real coe��cients then f�z can be factorized as a product of real linear factors andreal quadratic factors�

Proof� In general f�z will have r real roots z�� � � � � zr and �s nonrealroots zr��� zr��� � � � � zr�s� zr�s occurring in complexconjugate pairs bytheorem ������ Then if an is the coe�cient of highest degree in f�z wehave the factorization

f�z � an�z � z� � � ��z � zr ���z � zr�� �z � zr�� � � ��z � zr�s �z � zr�s �

We then use the following identity for j � r � �� � � � � r � s which in turnshows that paired terms give rise to real quadratic factors�

�z � zj �z � zj � z� � �zj � zj z � zjzj

� z� � �Re zj � �x�j � y�j �

where zj � xj � iyj �

A wellknown example of such a factorization is the following�

Page 104: Elementary Linear Algebra

���� MODULUS OF A COMPLEX NUMBER ��

��

jzjx

y

z

�����

Figure ���� The modulus of z� jzj�

EXAMPLE ����� Find a factorization of z��� into real linear and quadraticfactors�

Solution� Clearly there are no real roots� Also we have the preliminaryfactorization z� � � � �z� � i �z� � i � Now the roots of z� � i are easilyveri�ed to be ��� � i �

p� so the roots of z� � i must be ��� � i �

p��

In other words the roots are w � �� � i �p� and w� �w� �w� Grouping

conjugatecomplex terms gives the factorization

z� � � � �z � w �z � w �z � w �z � w

� �z� � �zRew � ww �z� � �zRew � ww

� �z� �p�z � � �z� �

p�z � � �

��� Modulus of a complex number

DEFINITION ����� �Modulus� If z � x � iy the modulus of z is thenonnegative real number jzj de�ned by jzj �

px� � y�� Geometrically the

modulus of z is the distance from z to � �see Figure ��� �More generally jz��z�j is the distance between z� and z� in the complex

plane� For

jz� � z�j � j�x� � iy� � �x� � iy� j � j�x� � x� � i�y� � y� j�

p�x� � x� � � �y� � y� ��

The following properties of the modulus are easy to verify using the identityjzj� � zz�

�i jz�z�j � jz�jjz�j�

Page 105: Elementary Linear Algebra

��� CHAPTER �� COMPLEX NUMBERS

�ii jz��j � jzj���

�iii

����z�z����� � jz�j

jz�j �

For example to prove �i �

jz�z�j� � �z�z� z�z� � �z�z� z� z�

� �z�z� �z�z� � jz�j�jz�j� � �jz�jjz�j ��

Hence jz�z�j � jz�jjz�j�

EXAMPLE ����� Find jzj when z ��� � i �

�� � �i ��� �i �

Solution�

jzj �j� � ij�

j� � �ijj�� �ij

��p�� � �� �p

�� � ��p�� � ��� �

��p

��p��

THEOREM ����� �Ratio formulae� If z lies on the line through z� andz��

z � ��� t z� � tz�� t � R�we have the useful ratio formulae�

�i

����z � z�z � z�

���� �

���� t

�� t

���� if z �� z�

�ii

���� z � z�z� � z�

���� � jtj�

Circle equations� The equation jz � z�j � r where z� � C and r �

� represents the circle centre z� and radius r� For example the equationjz � �� � �i j � � represents the circle �x� � � � �y � � � � ��

Another useful circle equation is the circle of Apollonius �����z � a

z � b

���� � ��

Page 106: Elementary Linear Algebra

���� MODULUS OF A COMPLEX NUMBER ���

��

x

y

Figure ���� Apollonius circles� jz��ijjz��ij �

�� �

�� �

�� �

�� �

�� �

�� �

�� �

�� �

where a and b are distinct complex numbers and � is a positive real number� �� �� �If � � � the above equation represents the perpendicular bisectorof the segment joining a and b�

An algebraic proof that the above equation represents a circle runs asfollows� We use the following identities�

�i jz � aj� � jzj� � �Re �za � jaj��ii Re �z� � z� � Re z� � Re z��iii Re �tz � tRe z if t � R�

We have����z � a

z � b

���� � � jz � aj� � ��jz � bj�

jzj� � �Re fzag� jaj� � ���jzj� � �Re fzbg� jbj� ��� �� jzj� � �Re fz�a� ��b g � ��jbj�� jaj�

jzj� � �Re

�z

�a� ��b

�� ��

���

��jbj� � jaj��� ��

jzj� � �Re

�z

�a� ��b

�� ��

���

����a� ��b

�� ��

�����

���jbj� � jaj�

�� ���

����a� ��b

�� ��

�����

Page 107: Elementary Linear Algebra

��� CHAPTER �� COMPLEX NUMBERS

Now it is easily veri�ed that

ja� ��bj� � ��� �� ���jbj�� jaj� � ��ja� bj��So we obtain ����z � a

z � b

���� � � ����z �

�a� ��b

�� ��

������

���ja� bj�j�� ��j�

����z �

�a� ��b

�� ��

����� � �ja� bjj�� ��j �

The last equation represents a circle centre z� radius r where

z� �a� ��b

�� ��and r �

�ja� bjj�� ��j �

There are two special points on the circle of Apollonius the points z� andz� de�ned by

z� � a

z� � b� � and

z� � a

z� � b� ���

or

z� �a� �b

�� �and z� �

a� �b

� � �� ����

It is easy to verify that z� and z� are distinct points on the line through a

and b and that z� �z��z�

� � Hence the circle of Apollonius is the circle basedon the segment z�� z� as diameter�

EXAMPLE ����� Find the centre and radius of the circle

jz � �� ij � �jz � �� �ij�Solution� Method �� Proceed algebraically and simplify the equation

jx� iy � �� ij � �jx� iy � �� �ijor

jx� � � i�y � � j � �jx� � � i�y � � j�Squaring both sides gives

�x� � � � �y � � � � ���x� � � � �y � � � �

which reduces to the circle equation

x� � y� � ��

�x� ��

�y � �� � ��

Page 108: Elementary Linear Algebra

���� ARGUMENT OF A COMPLEX NUMBER ���

Completing the square gives

�x� ��

� � � �y � �

� � �

���

��

��

��

� �� ���

��

so the centre is ���� �� and the radius is

q�� �

Method �� Calculate the diametrical points z� and z� de�ned above byequations ����

z� � �� i � ��z� � �� �i

z� � �� i � ���z� � �� �i �

We �nd z� � � � �i and z� � ��� � �i ��� Hence the centre z� is given by

z� �z� � z�

��

��

��

�i

and the radius r is given by

r � jz� � z�j ��������

��

�i

�� �� � �i

���� �������

�� �

�i

���� �p��

��

��� Argument of a complex number

Let z � x � iy be a nonzero complex number r � jzj �px� � y�� Then

we have x � r cos �� y � r sin � where � is the angle made by z with thepositive xaxis� So � is unique up to addition of a multiple of �� radians�

DEFINITION ����� �Argument� Any number � satisfying the abovepair of equations is called an argument of z and is denoted by argz� Theparticular argument of z lying in the range�� � � � is called the principalargument of z and is denoted by Arg z �see Figure ��� �

We have z � r cos � � ir sin � � r�cos � � i sin � and this representationof z is called the polar representation or modulus�argument form of z�

EXAMPLE ����� Arg� � � Arg ��� � � Arg i � �� Arg ��i � ��

� �

We note that y�x � tan � if x �� � so � is determined by this equation upto a multiple of �� In fact

Arg z � tan�� y

x� k��

Page 109: Elementary Linear Algebra

��� CHAPTER �� COMPLEX NUMBERS

��

�����

x

yr

z

Figure ���� The argument of z� arg z � ��

where k � � if x � �� k � � if x � �� y � �� k � �� if x � �� y � ��

To determine Arg z graphically it is simplest to draw the triangle formedby the points �� x� z on the complex plane mark in the positive acute angle� between the rays �� x and �� z and determine Arg z geometrically usingthe fact that � � tan���jyj�jxj as in the following examples�

EXAMPLE ����� Determine the principal argument of z for the followigcomplex numbers�

z � � � �i� �� � �i� ��� �i� �� �i�

Solution� Referring to Figure ��� we see that Arg z has the values

�� � � �� �� � �� ���

where � � tan�� �� �

An important property of the argument of a complex number states thatthe sum of the arguments of two nonzero complex numbers is an argumentof their product�

THEOREM ����� If �� and �� are arguments of z� and z� then �� � ��is an argument of z�z��

Proof� Let z� and z� have polar representations z� � r��cos �� � i sin �� and z� � r��cos �� � i sin �� � Then

z�z� � r��cos �� � i sin �� r��cos �� � i sin ��

� r�r��cos �� cos �� � sin �� sin �� � i�cos �� sin �� � sin �� cos ��

� r�r��cos ��� � �� � i sin ��� � �� �

Page 110: Elementary Linear Algebra

���� ARGUMENT OF A COMPLEX NUMBER ���

x

y

� � �i

������

x

y

�� � �i

�ZZ

ZZ�

x

y

��� �i

���

���

x

y

�� �i

�ZZZZ

Figure ���� Argument examples�

which is the polar representation of z�z� as r�r� � jz�jjz�j � jz�z�j� Hence�� � �� is an argument of z�z��

An easy induction gives the following generalization to a product of ncomplex numbers�

COROLLARY ����� If ��� � � � � �n are arguments for z�� � � � � zn respectivelythen �� � � � �� �n is an argument for z� � � �zn�Taking �� � � � �� �n � � in the previous corollary gives

COROLLARY ����� If � is an argument of z then n� is an argument forzn�

THEOREM ����� If � is an argument of the nonzero complex numberz then �� is an argument of z���

Proof� Let � be an argument of z� Then z � r�cos ��i sin � where r � jzj�Hence

z�� � r���cos � � i sin � ��

� r���cos � � i sin �

� r���cos��� � i sin��� �

Page 111: Elementary Linear Algebra

��� CHAPTER �� COMPLEX NUMBERS

Now r�� � jzj�� � jz��j so �� is an argument of z���

COROLLARY ����� If �� and �� are arguments of z� and z� then �����is an argument of z��z��

In terms of principal arguments we have the following equations�

�i Arg �z�z� � Arg z��Arg z� � �k���ii Arg �z�� � �Arg z � �k���iii Arg �z��z� � Arg z��Arg z� � �k���iv Arg �z� � � �zn � Arg z� � � � ��Arg zn � �k���v Arg �zn � nArg z � �k��

where k�� k�� k�� k�� k� are integers�

In numerical examples we can write �i for example as

Arg �z�z� � Arg z� � Arg z��

EXAMPLE ����� Find the modulus and principal argument of

z �

p� � i

� � i

��

and hence express z in modulusargument form�

Solution� jzj � jp� � ij�j� � ij� �

��

�p� �

� �����

Arg z � ��Arg

p� � i

� � i

� ���Arg�p� � i �Arg �� � i

� ������ �

�������

Hence Arg z �����

��

�� �k� where k is an integer� We see that k � � and

hence Arg z � ��� � Consequently z � ����

�cos �

�� � i sin ���

��

DEFINITION ����� If � is a real number then we de�ne ei� by

ei� � cos � � i sin ��

More generally if z � x� iy then we de�ne ez by

ez � exeiy �

Page 112: Elementary Linear Algebra

���� DE MOIVRE�S THEOREM ���

For example

ei�

� � i� ei� � ��� e� i�

� � �i�The following properties of the complex exponential function are left as

exercises�

THEOREM ����� �i ez�ez� � ez��z� �ii ez� � � �ezn � ez������zn �iii ez �� �

�iv �ez �� � e�z �v ez��ez� � ez��z� �vi ez � ez �

THEOREM ����� The equation

ez � �

has the complete solution z � �k�i� k �Z�Proof� First we observe that

e�k�i � cos ��k� � i sin ��k� � ��

Conversely suppose ez � �� z � x� iy� Then ex�cos y � i sin y � �� Henceex cos y � � and ex sin y � �� Hence sin y � � and so y � n�� n �Z� Thenex cos �n� � � so ex��� n � � from which follows ��� n � � as ex � ��Hence n � �k� k �Zand ex � �� Hence x � � and z � �k�i�

��� De Moivre�s theorem

The next theorem has many uses and is a special case of theorem ������ii �Alternatively it can be proved directly by induction on n�

THEOREM ���� �De Moivre� If n is a positive integer then

�cos � � i sin � n � cos n� � i sin n��

As a �rst application we consider the equation zn � ��

THEOREM ���� The equation zn � � has n distinct solutions namely

the complex numbers k � e�k�in � k � �� �� � � � � n � �� These lie equally

spaced on the unit circle jzj � � and are obtained by starting at � movinground the circle anticlockwise incrementing the argument in steps of ��

n ��See Figure ���

We notice that the roots are the powers of the special root � e��in �

Page 113: Elementary Linear Algebra

��� CHAPTER �� COMPLEX NUMBERS

���

n��

���������

����

�����

HHHHHHHHj

���n

���n

���n

jzj � �

Figure ���� The nth roots of unity�

Proof� With k de�ned as above

nk ��e�k�in

n� e

�k�in

n � ��

by De Moivre�s theorem� However jkj � � and arg k � �k�n so the com�

plex numbers k� k � �� �� � � � � n � � lie equally spaced on the unit circle�Consequently these numbers must be precisely all the roots of zn � �� Forthe polynomial zn � � being of degree n over a �eld can have at most ndistinct roots in that �eld�

The more general equation zn � a where a � C a �� � can be reducedto the previous case�

Let � be argument of z so that a � jajei�� Then if w � jaj��ne i�n wehave

wn ��jaj��ne i�n

n� �jaj��n n

�ei�

n

n� jajei� � a�

So w is a particular solution� Substituting for a in the original equationwe get zn � wn or �z�w n � �� Hence the complete solution is z�w �

Page 114: Elementary Linear Algebra

���� DE MOIVRE�S THEOREM ���

z�

z�

zn��

PPPPPPPPPq

���������

����

������

���n

jzj � �jaj ��n

Figure ���� The roots of zn � a�

e�k�in � k � �� �� � � � � n� � or

zk � jaj��ne i�n e �k�in � jaj��ne i����k��

n � ����

k � �� �� � � � � n� �� So the roots are equally spaced on the circle

jzj � jaj��n

and are generated from the special solution having argument equal to �arga �nby incrementing the argument in steps of ���n� �See Figure ����

EXAMPLE ���� Factorize the polynomial z� � � as a product of reallinear and quadratic factors�

Solution� The roots are �� e��i� � e

���i� � e

��i� � e

���i� using the fact that non

real roots come in conjugatecomplex pairs� Hence

z� � � � �z � � �z � e��i� �z � e

���i� �z � e

��i� �z � e

���i� �

Now

�z � e��i� �z � e

���i� � z� � z�e

��i� � e

���i� � �

� z� � �z cos ��� � ��

Page 115: Elementary Linear Algebra

��� CHAPTER �� COMPLEX NUMBERS

Similarly

�z � e��i� �z � e

���i� � z� � �z cos ��

� � ��

This gives the desired factorization�

EXAMPLE ���� Solve z� � i�

Solution� jij � � and Arg i � �� � �� So by equation ��� the solutions are

zk � jij���e i����k��� � k � �� �� ��

First k � � gives

z� � ei�

� � cos�

�� i sin

��

p�

��

i

��

Next k � � gives

z� � e��i� � cos

��

�� i sin

��

���p��

�i

��

Finally k � � gives

z� � e�i� � cos

��

�� i sin

��

�� �i�

We �nish this chapter with two more examples of De Moivre�s theorem�

EXAMPLE ���� If

C � � � cos � � � � �� cos �n� � ��

S � sin � � � � �� sin �n� � ��

prove that

C �sin n�

sin ��

cos �n����� and S �

sin n��

sin ��

sin �n����� �

if � �� �k�� k �Z�

Page 116: Elementary Linear Algebra

���� PROBLEMS ���

Solution�

C � iS � � � �cos � � i sin � � � � �� �cos �n� � � � i sin �n� � �

� � � ei� � � � �� ei�n����

� � � z � � � �� zn��� where z � ei�

��� zn

�� z� if z �� �� i�e� � �� �k��

��� ein�

�� ei��

ein�

� �e�in�

� � ein�

ei�

� �e�i�

� � ei�

� ei�n��� ��sin n�

sin ��

� �cos �n� � �� � i sin �n� � �� sin n�

sin ��

The result follows by equating real and imaginary parts�

EXAMPLE ���� Express cos n� and sin n� in terms of cos � and sin �using the equation cos n� � sin n� � �cos � � i sin � n�

Solution� The binomial theorem gives

�cos � � i sin � n � cosn � ��n�

�cosn�� ��i sin � �

�n�

�cosn�� ��i sin � � � � � �

� �i sin � n�

Equating real and imaginary parts gives

cos n� � cosn � � �n�� cosn�� � sin� � � � � �sin n� �

�n�

�cosn�� � sin � � �n�� cosn�� � sin� � � � � � �

�� PROBLEMS

�� Express the following complex numbers in the form x� iy� x� y real�

�i ��� � i ���� �i � �ii � � �i

�� �i� �iii

�� � �i �

�� i�

�Answers� �i ��� � ��i� �ii ���� �

���i� �iii �

� �i� ��

�� Solve the following equations�

Page 117: Elementary Linear Algebra

��� CHAPTER �� COMPLEX NUMBERS

�i iz � ��� ��i z � �z � �i�

�ii �� � i z � ��� i w � ��i�� � �i z � �� � i w � � � �i�

�Answers��i z � � ��� � i

�� � �ii z � �� � �i� w � ��� � �i

� ��

�� Express � � �� � i � �� � i � � � � �� �� � i �� in the form x� iy� x� yreal� �Answer� �� � ��� i��

�� Solve the equations� �i z� � ��� �i� �ii z� � �� � i z � �� �i � ��

�Answers� �i z � ���� �i � �ii z � �� i� � � �i��

�� Find the modulus and principal argument of each of the followingcomplex numbers�

�i � � i� �ii ��� � i

� � �iii �� � �i� �iv ����� � i

p� �

�Answers� �i p��� tan�� �

� � �ii p��� � �� � tan�� �

� � �iii p�� � �

tan�� ���

�� Express the following complex numbers in modulus�argument form�

�i z � �� � i �� � ip� �p�� i �

�ii z ��� � i ���� i

p� �

�p� � i �

�Answers�

�i z � �p��cos ��

�� � i sin ���� � �ii z � ����cos ���

�� � i sin ����� ��

�� �i If z � ��cos �� �i sin �

� and w � ��cos � �i sin �

�nd the polarform of

�a zw� �b zw � �c

wz � �d

z�

w� �

�ii Express the following complex numbers in the form x� iy�

�a �� � i ��� �b ���ip�

��

�Answers� �i � �a ��cos ���� � i sin ��

�� � �b ���cos

��� � i sin �

�� �

�c ���cos � �

�� � i sin � ��� � �d ��

� �cos����� � i sin ���

�� �

�ii � �a ���� �b �i��

Page 118: Elementary Linear Algebra

���� PROBLEMS ���

�� Solve the equations�

�i z� � � � ip�� �ii z� � i� �iii z� � ��i� �iv z� � �� �i�

�Answers� �i z � � �p��i�p�

� �ii ik�cos �� � i sin �

� � k � �� �� �� �� �iii

z � �i� �p�� i�p�� i� �iv z � ik�

� �cos �

� � i sin �� � k � �� �� �� ���

�� Find the reduced rowechelon form of the complex matrix�� � � i �� � �i �

� � i �� � i �� � �i �� � i � � i

�� �

�Answer�

�� � i �

� � �� � �

����

��� �i Prove that the line equation lx�my � n is equivalent to

pz � pz � �n�

where p � l � im�

�ii Use �ii to deduce that re�ection in the straight line

pz � pz � n

is described by the equation

pw � pz � n�

�Hint� The complex number l� im is perpendicular to the givenline��

�iii Prove that the line jz�aj � jz�bj may be written as pz�pz � nwhere p � b� a and n � jbj� � jaj�� Deduce that if z lies on the

Apollonius circle jz�ajjz�bj � � then w the re�ection of z in the line

jz � aj � jz � bj lies on the Apollonius circle jz�ajjz�bj �

�� �

��� Let a and b be distinct complex numbers and � � � � ��

�i Prove that each of the following sets in the complex plane rep�resents a circular arc and sketch the circular arcs on the samediagram�

Page 119: Elementary Linear Algebra

��� CHAPTER �� COMPLEX NUMBERS

Argz � a

z � b� �� ��� � � �� �� ��

Also show that Argz � a

z � b� � represents the line segment joining

a and b while Argz � a

z � b� � represents the remaining portion of

the line through a and b�

�ii Use �i to prove that four distinct points z�� z�� z�� z� are con�cyclic or collinear if and only if the cross�ratio

z� � z�z� � z�

�z� � z�z� � z�

is real�

�iii Use �ii to derive Ptolemy�sTheorem� Four distinct pointsA� B� C� Dare concyclic or collinear if and only if one of the following holds�

AB �CD �BC �AD � AC �BDBD �AC �AD �BC � AB �CDBD �AC � AB �CD � AD �BC�

Page 120: Elementary Linear Algebra

Chapter �

EIGENVALUES AND

EIGENVECTORS

��� Motivation

We motivate the chapter on eigenvalues by discussing the equation

ax� � �hxy � by� � c�

where not all of a� h� b are zero� The expression ax� � �hxy � by� is calleda quadratic form in x and y and we have the identity

ax� � �hxy � by� ��x y

� � a h

h b

��x

y

�� X tAX�

where X �

�xy

�and A �

�a hh b

�� A is called the matrix of the quadratic

form�

We now rotate the x� y axes anticlockwise through � radians to newx�� y� axes� The equations describing the rotation of axes are derived asfollows�

Let P have coordinates �x� y� relative to the x� y axes and coordinates�x�� y�� relative to the x�� y� axes� Then referring to Figure ����

��

Page 121: Elementary Linear Algebra

��� CHAPTER �� EIGENVALUES AND EIGENVECTORS

��

���������

��

��

��

��I

���������

��������R

��������������

x

y

x�y�

P

Q

R

O

Figure ���� Rotating the axes�

x � OQ � OP cos �� � ��

� OP �cos � cos�� sin � sin ��

� �OP cos�� cos � � �OP sin�� sin �

� OR cos � � PR sin �

� x� cos � � y� sin ��

Similarly y � x� sin � � y� cos ��We can combine these transformation equations into the single matrix

equation� �x

y

��

�cos � � sin �sin � cos �

��x�y�

��

or X � PY where X �

�x

y

�� Y �

�x�y�

�and P �

�cos � � sin �sin � cos �

��

We note that the columns of P give the directions of the positive x� and y�axes� Also P is an orthogonal matrix � we have PP t � I� and so P�� � P t�The matrix P has the special property that detP � ��

A matrix of the type P �

�cos � � sin �sin � cos �

�is called a rotation matrix�

We shall show soon that any �� � real orthogonal matrix with determinant

Page 122: Elementary Linear Algebra

���� MOTIVATION ���

equal to � is a rotation matrix�We can also solve for the new coordinates in terms of the old ones��

x�y�

�� Y � P tX �

�cos � sin �

� sin � cos �

� �x

y

��

so x� � x cos � � y sin � and y� � �x sin � � y cos �� Then

X tAX � �PY �tA�PY � � Y t�P tAP �Y�

Now suppose as we later show that it is possible to choose an angle � sothat P tAP is a diagonal matrix say diag���� ���� Then

X tAX ��x� y�

� � �� ��

� �x�y�

�� ��x

�� � ��y

�� �����

and relative to the new axes the equation ax� � �hxy � by� � c becomes��x

��� ��y

��� c which is quite easy to sketch� This curve is symmetrical

about the x� and y� axes with P� and P� the respective columns of P giving the directions of the axes of symmetry�

Also it can be veri�ed that P� and P� satisfy the equations

AP� � ��P� and AP� � ��P��

These equations force a restriction on �� and ��� For if P� �

�u�v�

� the

�rst equation becomes�a h

h b

� �u�v�

�� ��

�u�v�

�or

�a� �� h

h b� ��

��u�v�

��

��

Hence we are dealing with a homogeneous system of two linear equations intwo unknowns having a non�trivial solution �u�� v��� Hence���� a� �� h

h b� ��

���� � �

Similarly �� satis�es the same equation� In expanded form �� and ��satisfy

�� � �a� b��� ab� h� � �

This equation has real roots

� �a� b�

p�a� b�� � ��ab� h��

��

a� b�p�a� b�� � �h�

������

�The roots are distinct if a �� b or h �� � The case a � b and h � needsno investigation as it gives an equation of a circle��

The equation ����a�b���ab�h� � is called the eigenvalue equation

of the matrix A�

Page 123: Elementary Linear Algebra

��� CHAPTER �� EIGENVALUES AND EIGENVECTORS

��� De�nitions and examples

DEFINITION ����� �Eigenvalue� eigenvector�

Let A be a complex square matrix� Then if � is a complex number andX a non�zero complex column vector satisfying AX � �X we call X aneigenvector of A while � is called an eigenvalue of A� We also say that Xis an eigenvector corresponding to the eigenvalue ��

So in the above example P� and P� are eigenvectors corresponding to ��and �� respectively� We shall give an algorithm which starts from the

eigenvalues of A �

�a hh b

�and constructs a rotation matrix P such that

P tAP is diagonal�As noted above if � is an eigenvalue of an n � n matrix A with

corresponding eigenvector X then �A � �In�X � with X �� sodet �A� �In� � and there are at most n distinct eigenvalues of A�

Conversely if det �A� �In� � then �A� �In�X � has a non�trivialsolutionX and so � is an eigenvalue ofA withX a corresponding eigenvector�

DEFINITION ����� �Characteristic equation� polynomial�The equation det �A � �In� � is called the characteristic equation of Awhile the polynomial det �A��In� is called the characteristic polynomial ofA� The characteristic polynomial of A is often denoted by chA����

Hence the eigenvalues of A are the roots of the characteristic polynomialof A�

For a �� � matrix A �

�a bc d

� it is easily veri�ed that the character�

istic polynomial is ��� �traceA���detA where traceA � a�d is the sumof the diagonal elements of A�

EXAMPLE ����� Find the eigenvalues ofA �

�� �� �

�and �nd all eigen�

vectors�

Solution� The characteristic equation of A is �� � ��� � � or

��� ����� �� � �

Hence � � � or �� The eigenvector equation �A� �In�X � reduces to

��� � �� �� �

� �xy

��

��

Page 124: Elementary Linear Algebra

���� DEFINITIONS AND EXAMPLES ���

or

��� ��x� y �

x� ��� ��y � �

Taking � � � gives

x� y �

x� y � �

which has solution x � �y� y arbitrary� Consequently the eigenvectors

corresponding to � � � are the vectors

��yy

� with y �� �

Taking � � � gives

�x� y �

x� y � �

which has solution x � y� y arbitrary� Consequently the eigenvectors corre�

sponding to � � � are the vectors

�y

y

� with y �� �

Our next result has wide applicability�

THEOREM ����� Let A be a �� � matrix having distinct eigenvalues ��and �� and corresponding eigenvectors X� and X�� Let P be the matrixwhose columns are X� and X� respectively� Then P is non�singular and

P��AP �

��� ��

��

Proof� Suppose AX� � ��X� and AX� � ��X�� We show that the systemof homogeneous equations

xX� � yX� �

has only the trivial solution� Then by theorem ���� the matrix P ��X�jX�� is non�singular� So assume

xX� � yX� � � �����

Then A�xX� � yX�� � A � so x�AX�� � y�AX�� � � Hence

x��X� � y��X� � � �����

Page 125: Elementary Linear Algebra

�� CHAPTER �� EIGENVALUES AND EIGENVECTORS

Multiplying equation ��� by �� and subtracting from equation ��� gives

��� � ���yX� � �

Hence y � as ������� �� and X� �� � Then from equation ��� xX� � and hence x � �

Then the equations AX� � ��X� and AX� � ��X� give

AP � A�X�jX�� � �AX�jAX�� � ���X�j��X��

� �X�jX��

��� ��

�� P

��� ��

��

so

P��AP �

��� ��

��

EXAMPLE ����� Let A �

�� �� �

�be the matrix of example ������ Then

X� �

����

�and X� �

���

�are eigenvectors corresponding to eigenvalues

� and � respectively� Hence if P �

��� �� �

� we have

P��AP �

�� �

��

There are two immediate applications of theorem ������ The �rst is to thecalculation of An� If P��AP �diag ���� ��� then A � Pdiag ���� ���P

��

and

An �

�P

��� ��

�P��

�n

� P

��� ��

�nP�� � P

��n�

�n�

�P���

The second application is to solving a system of linear di�erential equations

dx

dt� ax� by

dy

dt� cx� dy�

where A �

�a bc d

�is a matrix of real or complex numbers and x and y

are functions of t� The system can be written in matrix form as �X � AX where

X �

�xy

�and �X �

��x�y

��

�dxdtdydt

��

Page 126: Elementary Linear Algebra

���� DEFINITIONS AND EXAMPLES ���

We make the substitution X � PY where Y �

�x�y�

�� Then x� and y�

are also functions of t and

�X � P �Y � AX � A�PY �� so �Y � �P��AP �Y �

��� ��

�Y�

Hence �x� � ��x� and �y� � ��y��These di�erential equations are well�known to have the solutions x� �

x�� �e��t and x� � x�� �e

��t where x�� � is the value of x� when t � �

�If dxdt

� kx where k is a constant then

d

dt

�e�ktx

�� �ke�ktx� e�kt

dx

dt� �ke�ktx� e�ktkx � �

Hence e�ktx is constant so e�ktx � e�k�x� � � x� �� Hence x � x� �ekt��

However

�x�� �y�� �

�� P��

�x� �y� �

� so this determines x�� � and y�� � in

terms of x� � and y� �� Hence ultimately x and y are determined as explicitfunctions of t using the equation X � PY �

EXAMPLE ����� Let A �

�� ��� �

�� Use the eigenvalue method to

derive an explicit formula for An and also solve the system of di�erentialequations

dx

dt� �x� �y

dy

dt� �x� y�

given x � � and y � �� when t � �

Solution� The characteristic polynomial ofA is ������� which has distinct

roots �� � �� and �� � ��� We �nd corresponding eigenvectorsX� �

���

and X� �

���

�� Hence if P �

�� �� �

� we have P��AP � diag ���� ����

Hence

An �Pdiag ���� ���P��

n� Pdiag �����n� ����n�P��

�� �� �

� �����n ����n

��� ��

�� �

Page 127: Elementary Linear Algebra

��� CHAPTER �� EIGENVALUES AND EIGENVECTORS

� ����n�� �� �

��� �n

� �� ��

�� �

� ����n�� �� �n

� �� �n

� �� ��

�� �

� ����n��� �� �n �� � �� �n

�� �� �n �� � �� �n

��

To solve the di�erential equation system make the substitution X �PY � Then x � x� � �y�� y � x� � �y�� The system then becomes

�x� � �x�

�y� � ��y��

so x� � x�� �e�t� y� � y�� �e

��t� Now

�x�� �y�� �

�� P��

�x� �y� �

��

�� ��

�� �

�����

��

����

��

so x� � ���e�t and y� � �e��t� Hence x � ���e�t � ���e��t� � ���e�t ���e��t� y � ���e�t � ���e��t� � ���e�t � ��e��t�

For a more complicated example we solve a system of inhomogeneous

recurrence relations�

EXAMPLE ����� Solve the system of recurrence relations

xn�� � �xn � yn � �

yn�� � �xn � �yn � ��

given that x� � and y� � ���

Solution� The system can be written in matrix form as

Xn�� � AXn �B�

where

A �

�� ��

�� �

�and B �

����

��

It is then an easy induction to prove that

Xn � AnX� � �An�� � � � ��A� I��B� ����

Page 128: Elementary Linear Algebra

���� DEFINITIONS AND EXAMPLES ���

Also it is easy to verify by the eigenvalue method that

An ��

�� � �n �� �n

�� �n � � �n

��

�U �

�n

�V�

where U �

�� �� �

�and V �

�� ��

�� �

�� Hence

An�� � � � �� A� I� �n

�U �

��n�� � � � �� � � ��

�V

�n

�U �

��n�� � ��

�V�

Then equation �� gives

Xn �

��

�U �

�n

�V

��

��

��

�n

�U �

��n�� � ��

�V

�����

��

which simpli�es to �xnyn

��

���n� �� �n�����n� � �n���

��

Hence xn � ��n� � � �n��� and yn � ��n� � �n����

REMARK ����� If �A � I���� existed �that is if det �A � I�� �� or

equivalently if � is not an eigenvalue of A� then we could have used theformula

An�� � � � �� A� I� � �An � I���A� I����� �����

However the eigenvalues ofA are � and � in the above problem so formula ���cannot be used there�

Our discussion of eigenvalues and eigenvectors has been limited to � � �matrices� The discussion is a more complicated for matrices of size greaterthan two and is best left to a second course in linear algebra� Neverthelessthe following result is a useful generalization of theorem ������ The readeris referred to ��� page � � for a proof�

THEOREM ����� Let A be an n � n matrix having distinct eigenvalues��� � � � � �n and corresponding eigenvectors X�� � � � � Xn� Let P be the matrixwhose columns are respectively X�� � � � � Xn� Then P is non�singular and

P��AP �

���� �� � � � �� � � � ���

������

��� � � � �n

����� �

Page 129: Elementary Linear Algebra

��� CHAPTER �� EIGENVALUES AND EIGENVECTORS

Another useful result which covers the case where there are multiple eigen�values is the following �The reader is referred to ��� pages ������ for aproof��

THEOREM ����� Suppose the characteristic polynomial of A has the fac�torization

det ��In � A� � ��� c��n� � � � ��� ct�

nt �

where c�� � � � � ct are the distinct eigenvalues of A� Suppose that for i ��� � � � � t we have nullity �ciIn�A� � ni� For each i choose a basisXi�� � � � � Xini

for the eigenspace N�ciIn �A�� Then the matrix

P � �X��j � � � jX�n� j � � � jXt�j � � � jXtnt�

is non�singular and P��AP is the following diagonal matrix

P��AP �

���� c�In� � � � c�In� � � � ���

������

��� � � � ctInt

����� �

�The notation means that on the diagonal there are n� elements c� followedby n� elements c�� � � nt elements ct��

��� PROBLEMS

�� Let A �

�� ���

�� Find a non�singular matrix P such that P��AP �

diag ��� �� and hence prove that

An ��n � �

�A�

�� �n

�I��

�� If A �

� �� �� �� ��

� prove that An tends to a limiting matrix

���� ������ ���

as n���

Page 130: Elementary Linear Algebra

���� PROBLEMS ��

�� Solve the system of di�erential equations

dx

dt� �x� �y

dy

dt� x� �y�

given x � �� and y � �� when t � �

�Answer� x � �et � �e��t� y � �et � �e��t��

�� Solve the system of recurrence relations

xn�� � �xn � yn

yn�� � �xn � �yn�

given that x� � � and y� � ��

�Answer� xn � �n����� �n�� yn � �n���� � �n���

� Let A �

�a bc d

�be a real or complex matrix with distinct eigenvalues

��� �� and corresponding eigenvectors X�� X�� Also let P � �X�jX���

�a� Prove that the system of recurrence relations

xn�� � axn � byn

yn�� � cxn � dyn

has the solution �xnyn

�� ��n�X� � ��n�X��

where � and � are determined by the equation���

�� P��

�x�y�

��

�b� Prove that the system of di�erential equations

dx

dt� ax� by

dy

dt� cx� dy

has the solution �xy

�� �e��tX� � �e��tX��

Page 131: Elementary Linear Algebra

��� CHAPTER �� EIGENVALUES AND EIGENVECTORS

where � and � are determined by the equation

���

�� P��

�x� �y� �

��

�� Let A �

�a bc d

�be a real matrix with non�real eigenvalues � � a�ib

and � � a � ib with corresponding eigenvectors X � U � iV andX � U � iV where U and V are real vectors� Also let P be the realmatrix de�ned by P � �U jV �� Finally let a � ib � rei� where r � and � is real�

�a� Prove that

AU � aU � bV

AV � bU � aV�

�b� Deduce that

P��AP �

�a �bb a

��

�c� Prove that the system of recurrence relations

xn�� � axn � byn

yn�� � cxn � dyn

has the solution�xnyn

�� rnf��U � �V � cosn� � ��U � �V � sinn�g�

where � and � are determined by the equation

���

�� P��

�x�y�

��

�d� Prove that the system of di�erential equations

dx

dt� ax� by

dy

dt� cx� dy

Page 132: Elementary Linear Algebra

���� PROBLEMS ���

has the solution�xy

�� eatf��U � �V � cos bt� ��U � �V � sin btg�

where � and � are determined by the equation��

�� P��

�x� �y� �

��

�Hint� Let

�x

y

�� P

�x�y�

�� Also let z � x� � iy�� Prove that

�z � �a� ib�z

and deduce that

x� � iy� � eat��� i���cos bt� i sin bt��

Then equate real and imaginary parts to solve for x�� y� andhence x� y��

�� �The case of repeated eigenvalues�� Let A �

�a b

c d

�and suppose

that the characteristic polynomial of A ��� �a� d��� �ad� bc� hasa repeated root �� Also assume that A �� �I�� Let B � A� �I��

�i� Prove that �a� d�� � �bc � �

�ii� Prove that B� � �

�iii� Prove that BX� �� for some vector X�� indeed show that X�

can be taken to be

��

�or

� �

��

�iv� Let X� � BX�� Prove that P � �X�jX�� is non�singular

AX� � �X� and AX� � �X� �X�

and deduce that

P��AP �

�� � �

��

�� Use the previous result to solve system of the di�erential equations

dx

dt� �x� y

dy

dt� �x� �y�

Page 133: Elementary Linear Algebra

��� CHAPTER �� EIGENVALUES AND EIGENVECTORS

given that x � � � y when t � �

�To solve the di�erential equation

dx

dt� kx � f�t�� k a constant�

multiply throughout by e�kt thereby converting the left�hand side todxdt�e�ktx���

�Answer� x � ��� �t�e�t� y � �� � �t�e�t��

�� Let

A �

� ��� ���

��� ��� ������ ��� ���

�� �

�a� Verify that det ��I� �A� the characteristic polynomial of A isgiven by

��� �������

���

�b� Find a non�singular matrix P such that P��AP � diag ��� � �

���

�c� Prove that

An ��

� � � �

� � �� � �

�� �

� � �n

� � � ���� �� ��� �� �

��

if n � ��

� � Let

A �

� � ��

� ���� ��

�� �

�a� Verify that det ��I� �A� the characteristic polynomial of A isgiven by

��� ������ ���

�b� Find a non�singular matrix P such that P��AP � diag ��� �� ���

Page 134: Elementary Linear Algebra

Chapter �

Identifying second degree

equations

��� The eigenvalue method

In this section we apply eigenvalue methods to determine the geometricalnature of the second degree equation

ax� � �hxy � by� � �gx� �fy � c � �� �����

where not all of a� h� b are zero�

Let A �

�a hh b

�be the matrix of the quadratic form ax���hxy�by��

We saw in section �� equation �� that A has real eigenvalues �� and ��given by

�� �a� b�p�a� b�� � �h�

�� �� �

a� b�p�a� b�� � �h�

��

We show that it is always possible to rotate the x� y axes to x�� x� axes whosepositive directions are determined by eigenvectors X� and X� correspondingto ��and �� in such a way that relative to the x�� y� axes equation ��� takesthe form

a�x� � b�y� � �g�x� �f �y � c � �� �����

Then by completing the square and suitably translating the x�� y� axesto new x�� y� axes equation ��� can be reduced to one of several standardforms each of which is easy to sketch� We need some preliminary de�nitions�

��

Page 135: Elementary Linear Algebra

��� CHAPTER �� IDENTIFYING SECOND DEGREE EQUATIONS

DEFINITION ����� �Orthogonal matrix� An n � n real matrix P iscalled orthogonal if

P tP � In�

It follows that if P is orthogonal then detP � ��� Fordet �P tP � � detP t detP � �detP ���

so �detP �� �det In � �� Hence detP � ���If P is an orthogonal matrix with detP � � then P is called a proper

orthogonal matrix�

THEOREM ����� If P is a �� � orthogonal matrix with detP � � then

P �

�cos � � sin �sin � cos �

for some ��

REMARK ����� Hence by the discusssion at the beginning of Chapter if P is a proper orthogonal matrix the coordinate transformation�

xy

�� P

�x�y�

represents a rotation of the axes with new x� and y� axes given by therepective columns of P �

Proof� Suppose that P tP � I� where � �detP � �� Let

P �

�a bc d

��

Then the equation

P t � P�� ��

�adjP

gives �a cb d

��

�d �b

�c a

�Hence a � d� b � �c and so

P �

�a �cc a

��

where a� � c� � �� But then the point �a� c� lies on the unit circle soa � cos � and c � sin � where � is uniquely determined up to multiples of���

Page 136: Elementary Linear Algebra

���� THE EIGENVALUE METHOD ���

DEFINITION ����� �Dot product�� If X �

�a

b

�and Y �

�c

d

� then

X � Y the dot product of X and Y is de�ned by

X � Y � ac� bd�

The dot product has the following properties�

�i� X � �Y � Z� � X � Y �X � Z��ii� X � Y � Y �X �

�iii� �tX� � Y � t�X � Y ��

�iv� X �X � a� � b� if X �

�ab

��

�v� X � Y � X tY �

The length of X is de�ned by

jjX jj �pa� � b� � �X �X�����

We see that jjX jj is the distance between the origin O � ��� �� and the point�a� b��

THEOREM ����� �Geometrical interpretation of the dot product�Let A � �x�� y�� and B � �x�� y�� be points each distinct from the origin

O � ��� ��� Then if X �

�x�y�

�and Y �

�x�y�

� we have

X � Y � OA �OB cos ��

where � is the angle between the rays OA and OB�

Proof� By the cosine law applied to triangle OAB we have

AB� � OA� � OB� � �OA �OB cos �� �����

Now AB� � �x� � x��� � �y� � y���� OA� � x�� � y�� � OB� � x�� � y�� �

Substituting in equation ��� then gives

�x� � x��� � �y� � y��

� � �x�� � y��� � �x�� � y���� �OA �OB cos ��

Page 137: Elementary Linear Algebra

��� CHAPTER �� IDENTIFYING SECOND DEGREE EQUATIONS

which simpli�es to give

OA �OB cos � � x�x� � y�y� � X � Y�

It follows from theorem ����� that if A � �x�� y�� and B � �x�� y�� are

points distinct from O � ��� �� and X �

�x�y�

�and Y �

�x�y�

� then

X � Y � � means that the rays OA and OB are perpendicular� This is thereason for the following de�nition�

DEFINITION ����� �Orthogonal vectors� VectorsX and Y are calledorthogonal if

X � Y � ��

There is also a connection with orthogonal matrices�

THEOREM ����� Let P be a �� � real matrix� Then P is an orthogonalmatrix if and only if the columns of P are orthogonal and have unit length�

Proof� P is orthogonal if and only if P tP � I�� Now if P � �X�jX�� thematrix P tP is an important matrix called the Gram matrix of the columnvectors X� and X�� It is easy to prove that

P tP � �Xi �Xj � �

�X� �X� X� �X�

X� �X� X� �X�

��

Hence the equation P tP � I� is equivalent to�X� �X� X� �X�

X� �X� X� �X�

��

�� �� �

��

or equating corresponding elements of both sides�

X� �X� � �� X� �X� � �� X� �X� � ��

which says that the columns of P are orthogonal and of unit length�

The next theorem describes a fundamental property of real symmetricmatrices and the proof generalizes to symmetric matrices of any size�

THEOREM ����� If X� andX� are eigenvectors corresponding to distincteigenvalues �� and �� of a real symmetric matrix A then X� and X� areorthogonal vectors�

Page 138: Elementary Linear Algebra

���� THE EIGENVALUE METHOD ���

Proof� Suppose

AX� � ��X�� AX� � ��X�� �����

where X� and X� are non�zero column vectors At � A and �� �� ���

We have to prove that X t�X� � �� From equation ���

X t�AX� � ��X

t�X� �����

and

X t�AX� � ��X

t�X�� ����

From equation ��� taking transposes

�X t�AX��

t � ���Xt�X��

t

so

X t�A

tX� � ��Xt�X��

Hence

X t�AX� � ��X

t�X�� �����

Finally subtracting equation �� from equation ��� we have

��� � ���Xt�X� � �

and hence since �� �� ��

X t�X� � ��

THEOREM ����� Let A be a real � � � symmetric matrix with distincteigenvalues �� and ��� Then a proper orthogonal �� � matrix P exists suchthat

P tAP � diag ���� ����

Also the rotation of axes �xy

�� P

�x�y�

�diagonalizes� the quadratic form corresponding to A�

X tAX � ��x�� � ��y

���

Page 139: Elementary Linear Algebra

��� CHAPTER �� IDENTIFYING SECOND DEGREE EQUATIONS

Proof� Let X� and X� be eigenvectors corresponding to �� and ��� Thenby theorem ����� X� and X� are orthogonal� By dividing X� and X� bytheir lengths �i�e� normalizing X� and X�� if necessary we can assume thatX� and X� have unit length� Then by theorem ����� P � �X�jX�� is anorthogonal matrix� By replacing X� by �X� if necessary we can assumethat detP � �� Then by theorem ���� we have

P tAP � P��AP �

��� �� ��

��

Also under the rotation X � PY

X tAX � �PY �tA�PY � � Y t�P tAP �Y � Y t diag ���� ���Y

� ��x�� � ��y

���

EXAMPLE ����� Let A be the symmetric matrix

A �

��� �� �

��

Find a proper orthogonal matrix P such that P tAP is diagonal�

Solution� The characteristic equation of A is �� � � �� �� � � or

��� ����� �� � ��

Hence A has distinct eigenvalues �� � � and �� � �� We �nd correspondingeigenvectors

X� �

� ���

�and X� �

���

��

Now jjX�jj � jjX�jj �p��� So we take

X� ��p��

� ���

�and X� �

�p��

���

��

Then if P � �X�jX�� the proof of theorem ����� shows that

P tAP �

�� �� �

��

However detP � �� so replacing X� by �X� will give detP � ��

Page 140: Elementary Linear Algebra

���� THE EIGENVALUE METHOD ���

y2

x2

2 4-2-4

2

4

-2

-4

x

y

Figure ���� ��x� � ��xy � �y� � �x� ��y � �� � ��

REMARK ����� �A shortcut� Once we have determined one eigenvec�

tor X� �

�ab

� the other can be taken to be

� �ba

� as these these vectors

are always orthogonal� Also P � �X�jX�� will have detP � a� � b� � ��

We now apply the above ideas to determine the geometric nature ofsecond degree equations in x and y�

EXAMPLE ����� Sketch the curve determined by the equation

��x� � ��xy � �y� � �x� ��y � �� � ��

Solution� With P taken to be the proper orthogonal matrix de�ned in theprevious example by

P �

���p�� ��

p��

���p�� ��p��

��

then as theorem ����� predicts P is a rotation matrix and the transformation

X �

�xy

�� PY � P

�x�y�

Page 141: Elementary Linear Algebra

�� CHAPTER �� IDENTIFYING SECOND DEGREE EQUATIONS

or more explicitly

x ��x� � �y�p

��� y �

��x� � �y�p��

� �����

will rotate the x� y axes to positions given by the respective columns of P ��More generally we can always arrange for the x� axis to point either intothe �rst or fourth quadrant��

Now A �

��� �� �

�is the matrix of the quadratic form

��x� � ��xy � �y��

so we have by Theorem �����

��x� � ��xy � �y� � �x�� � �y���

Then under the rotation X � PY our original quadratic equation becomes

�x�� � �y�� ��p��

��x� � �y��� ��p��

���x� � �y�� � �� � ��

or

�x�� � �y�� ���p��x� �

p��y� � �� � ��

Now complete the square in x� and y��

�x�� �

�p��x�

�� �

�y�� �

�p��y�

�� �� � ��

�x� �

�p��

��� �

�y� �

�p��

��

� �

��p��

��� �

��p��

��� ��

� ��� ��� �

Then if we perform a translation of axes to the new origin �x�� y�� ��� �p

��� � �p

����

x� � x� ��p��� y� � y� �

�p���

equation �� reduces to�x�� � �y�� � ���

orx����

y���

� ��

Page 142: Elementary Linear Algebra

���� THE EIGENVALUE METHOD ���

x

y

Figure ����x�

a��y�

b�� �� � � b � a� an ellipse�

This equation is now in one of the standard forms listed below as Figure ���and is that of a whose centre is at �x�� y�� � ��� �� and whose axes ofsymmetry lie along the x�� y� axes� In terms of the original x� y coordinateswe �nd that the centre is �x� y� � ���� ��� Also Y � P tX so equations ���can be solved to give

x� ��x� � �y�p

��� y� �

�x� � �y�p��

Hence the y��axis is given by

� � x� � x� ��p��

��x� �yp

���

�p���

or �x� �y � � � �� Similarly the x� axis is given by �x� �y � � � ��This ellipse is sketched in Figure ����

Figures ��� ��� ��� and ��� are a collection of standard second degreeequations� Figure ��� is an ellipse� Figures ��� are hyperbolas �in both these

examples the asymptotes are the lines y � � b

ax�� Figures ��� and ���

represent parabolas�

EXAMPLE ����� Sketch y� � �x� ��y � � � ��

Page 143: Elementary Linear Algebra

��� CHAPTER �� IDENTIFYING SECOND DEGREE EQUATIONS

x

y

x

y

Figure ���� �i�x�

a�� y�

b�� �� �ii�

x�

a�� y�

b�� ��� � � b� � � a�

x

y

x

y

Figure ���� �i� y� � �ax� a � �� �ii� y� � �ax� a � ��

Page 144: Elementary Linear Algebra

���� THE EIGENVALUE METHOD ��

x

y

x

y

Figure ���� �iii� x� � �ay� a � �� �iv� x� � �ay� a � ��

Solution� Complete the square�

y� � ��y � ��� �x� �� � �

�y � ��� � �x� �� � ��x� ���

or y�� � �x� under the translation of axes x� � x� �� y� � y� �� Hence weget a parabola with vertex at the new origin �x�� y�� � ��� �� i�e� �x� y� ����� ���

The parabola is sketched in Figure ���

EXAMPLE ����� Sketch the curve x� � �xy � �y� � �y � � ��

Solution� We have x� � �xy � �y� � X tAX where

A �

�� ��

�� �

��

The characteristic equation of A is ����� � � so A has distinct eigenvalues�� � � and �� � �� We �nd corresponding unit length eigenvectors

X� ��p�

��

���� X� �

�p�

���

��

Then P � �X�jX�� is a proper orthogonal matrix and under the rotation ofaxes X � PY or

x �x� � �y�p

y ���x� � y�p

��

Page 145: Elementary Linear Algebra

��� CHAPTER �� IDENTIFYING SECOND DEGREE EQUATIONS

x1

y1

4 8 12-4-8

4

8

12

-4

-8

x

y

Figure ��� y� � �x� ��y � � � ��

we havex� � �xy � �y� � ��x

�� � ��y

�� � �x���

The original quadratic equation becomes

�x�� �

p�p����x� � y��� � �

��x�� ��p�x�� �

p�y� � � �

��x� � �p��� � ���

p�y� �

p��y� � �

p���

or �x�� � � �p�y� where the x�� y� axes have been translated to x�� y� axes

using the transformation

x� � x� � �p�� y� � y� � �

p��

Hence the vertex of the parabola is at �x�� y�� � ��� �� i�e� �x�� y�� �� �p

�� �p�� or �x� y� � ���

�� ���� The axis of symmetry of the parabola is the

line x� � � i�e� x� � ��p�� Using the rotation equations in the form

x� �x � �yp

Page 146: Elementary Linear Algebra

���� A CLASSIFICATION ALGORITHM ���

x2

y2

2 4-2-4

2

4

-2

-4

x

y

Figure ���� x� � �xy � �y� � �y � � ��

y� ��x� yp

��

we havex� �yp

��

�p�� or x� �y � ��

The parabola is sketched in Figure ����

��� A classi�cation algorithm

There are several possible degenerate cases that can arise from the generalsecond degree equation� For example x��y� � � represents the point ��� ���x� � y� � �� de�nes the empty set as does x� � �� or y� � ��� x� � �de�nes the line x � �� �x� y�� � � de�nes the line x� y � �� x� � y� � �de�nes the lines x � y � �� x � y � �� x� � � de�nes the parallel linesx � ��� �x� y�� � � likewise de�nes two parallel lines x� y � ���

We state without proof a complete classi�cation � of the various cases

�This classi�cation forms the basis of a computer program which was used to produce

the diagrams in this chapter� I am grateful to Peter Adams for his programming assistance�

Page 147: Elementary Linear Algebra

��� CHAPTER �� IDENTIFYING SECOND DEGREE EQUATIONS

that can possibly arise for the general second degree equation

ax� � �hxy � by� � �gx� �fy � c � �� ������

It turns out to be more convenient to �rst perform a suitable translation ofaxes before rotating the axes� Let

� �

������a h gh b f

g f c

������ � C � ab� h�� A � bc� f�� B � ca� g��

If C �� � let

� �

����� g hf b

����C

� �

����� a gh f

����C

� ������

CASE �� � � ��

����� C �� �� Translate axes to the new origin ��� � where � and aregiven by equations �����

x � x� � �� y � y� � �

Then equation ���� reduces to

ax�� � �hx�y� � by�� � ��

�a� C � �� Single point �x� y� � ��� ��

�b� C � �� Two nonparallel lines intersecting in �x� y� � ��� ��

The lines are

y �

x� ��

�h �p�Cb

if b �� ��

x � � andy �

x� �� � a

�h� if b � ��

����� C � ��

�a� h � ��

�i� a � g � ��

�A� A � �� Empty set�

�B� A � �� Single line y � �f�b�

Page 148: Elementary Linear Algebra

���� A CLASSIFICATION ALGORITHM ���

�C� A � �� Two parallel lines

y ��f �p�A

b

�ii� b � f � ��

�A� B � �� Empty set�

�B� B � �� Single line x � �g�a��C� B � �� Two parallel lines

x ��g �p�B

a

�b� h �� ��

�i� B � �� Empty set�

�ii� B � �� Single line ax� hy � �g��iii� B � �� Two parallel lines

ax� hy � �g �p�B�

CASE �� � �� ��

����� C �� �� Translate axes to the new origin ��� � where � and aregiven by equations �����

x � x� � �� y � y� � �

Equation ���� becomes

ax�� � �hx�y� � by�� � ��

C� ������

CASE ����i� h � �� Equation ���� becomes ax�� � by�� ���C �

�a� C � �� Hyperbola�

�b� C � � and a� � �� Empty set�

�c� C � � and a� � ��

�i� a � b� Circle centre ��� � radiusq

g��f��aca �

�ii� a �� b� Ellipse�

Page 149: Elementary Linear Algebra

��� CHAPTER �� IDENTIFYING SECOND DEGREE EQUATIONS

CASE ����ii� h �� ��

Rotate the �x�� y�� axes with the new positive x��axis in the directionof

��b� a� R���� �h��where R �

p�a� b�� � �h��

Then equation ���� becomes

��x�� � ��y

�� � ��

C� ������

where�� � �a� b�R���� �� � �a� b�R����

Here ���� � C�

�a� C � �� Hyperbola�

Here �� � � � �� and equation ���� becomes

x��u�

� y��v�

���j�j �

where

u �

sj�jC��

� v �

sj�j�C�� �

�b� C � � and a� � �� Empty set�

�c� C � � and a� � �� Ellipse�

Here ��� ��� a� b have the same sign and �� �� �� and equa�tion ���� becomes

x��u�

�y��v�

� ��

where

u �

r�

�C�� � v �r

�C�� �

����� C � ��

�a� h � ��

�i� a � �� Then b �� � and g �� �� Parabola with vertex��A�gb

� �f

b

��

Page 150: Elementary Linear Algebra

���� A CLASSIFICATION ALGORITHM ���

Translate axes to �x�� y�� axes�

y�� � ��g

bx��

�ii� b � �� Then a �� � and f �� �� Parabola with vertex��g

a��B�fa

��

Translate axes to �x�� y�� axes�

x�� � ��f

ay��

�b� h �� �� Parabola� Let

k �ga� bf

a� b�

The vertex of the parabola is���akf � hk� � hac�

d�a�k� � ac� �kg�

d

��

Now translate to the vertex as the new origin then rotate to�x�� y�� axes with the positive x��axis along �sa� �sh� wheres � sign �a��

�The positive x��axis points into the �rst or fourth quadrant��Then the parabola has equation

x�� ���stpa� � h�

y��

where t � �af � gh���a� b��

REMARK ����� If � � � it is not necessary to rotate the axes� Insteadit is always possible to translate the axes suitably so that the coe�cients ofthe terms of the �rst degree vanish�

EXAMPLE ����� Identify the curve

�x� � xy � y� � y � � � �� ������

Page 151: Elementary Linear Algebra

�� CHAPTER �� IDENTIFYING SECOND DEGREE EQUATIONS

Solution� Here

� �

������� �

��

��

�� �� � ��

������ � ��

Let x � x� � �� y � y� � and substitute in equation ���� to get

��x� � ��� � �x� � ���y� � �� �y� � �� � ��y� � �� � � �� ������

Then equating the coe�cients of x� and y� to � gives

��� � �

�� � � � � ��

which has the unique solution � � ��

�� � �

�� Then equation ���� simpli�es

to�x�� � x�y� � y�� � � � ��x� � y���x� � y���

so relative to the x�� y� coordinates equation ���� describes two lines� �x��y� � � or x� � y� � �� In terms of the original x� y coordinates these linesbecome ��x� �

��� �y� �

�� � � and �x� �

�� � �y� �

�� � � i�e� �x� y�� � �

and x� y � � � � which intersect in the point

�x� y� � ��� � � ���

���

���

EXAMPLE ����� Identify the curve

x� � �xy � y� � ��x� �y � � � �� �����

Solution� Here

� �

������� � �� � �� � �

������ � ��

Let x � x� � �� y � y� � and substitute in equation ��� to get

�x��������x�����y�����y���

����x�������y����� � �� ������

Then equating the coe�cients of x� and y� to � gives the same equation

��� � � � � ��

Take � � �� � ��� Then equation ���� simpli�es to

x�� � �x�y� � y�� � � � �x� � y����

and in terms of x� y coordinates equation ��� becomes

�x� y � ��� � �� or x� y � � � ��

Page 152: Elementary Linear Algebra

���� PROBLEMS ���

��� PROBLEMS

�� Sketch the curves

�i� x� � �x� �y � � � ��

�ii� y� � ��x� �y � �� � ��

�� Sketch the hyperbola�xy � �y� � �

and �nd the equations of the asymptotes�

�Answer� y � � and y � �

�x��

�� Sketch the ellipse�x� � �xy � �y� � �

and �nd the equations of the axes of symmetry�

�Answer� y � �x and x � ��y���� Sketch the conics de�ned by the following equations� Find the centre

when the conic is an ellipse or hyperbola asymptotes if an hyperbolathe vertex and axis of symmetry if a parabola�

�i� �x� � y� � ��x� �y � � � ��

�ii� �x� � �xy � �y� � �p�x� �

p�y � � � ��

�iii� �x� � y� � �xy � ��y � � � ��

�iv� ��x� � ��xy � ��y� � ��x� ��y � � � ��

�Answers� �i� hyperbola centre ��� ��� asymptotes �x � �y � �� ��� �x� �y � ��

�ii� ellipse centre ���p���

�iii� parabola vertex ���

�� �

�� axis of symmetry �x� y � � � ��

�iv� hyperbola centre �� �

�� �

�� asymptotes �x � y � � � � and

��x� �y � � � ���

�� Identify the lines determined by the equations�

�i� �x� � y� � �xy � �x� �y � � � ��

Page 153: Elementary Linear Algebra

��� CHAPTER �� IDENTIFYING SECOND DEGREE EQUATIONS

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Page 193: Elementary Linear Algebra

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Page 194: Elementary Linear Algebra

Chapter �

FURTHER READING

Matrix theory has many applications to science� mathematics� economicsand engineering� Some of these applications can be found in the books

��� �� �� �� ��� ��� �� �� �� ����

For the numerical side of matrix theory� �� is recommended� Its bibliography

is also useful as a source of further references�

For applications to�

�� Graph theory� see � � ����

�� Coding theory� see ��� ����

�� Game theory� see �����

�� Statistics� see ����

�� Economics� see ����

� Biological systems� see �����

� Markov non�negative matrices� see ���� ��� ��� � ��

�� The general equation of the second degree in three variables� see �����

�� A�ne and projective geometry� see ���� ��� ����

�� Computer graphics� see ���� ����

���

Page 195: Elementary Linear Algebra

��

Page 196: Elementary Linear Algebra

Bibliography

��� B� Noble� Applied Linear Algebra� ����� Prentice Hall� NJ�

��� B� Noble and J�W� Daniel� Applied Linear Algebra� third edition� �����

Prentice Hall� NJ�

��� R�P� Yantis and R�J� Painter� Elementary Matrix Algebra with Appli�

cation� second edition� ����� Prindle� Weber and Schmidt� Inc� Boston�Massachusetts�

��� T�J� Fletcher� Linear Algebra through its Applications� ����� Van Nos�

trand Reinhold Company� New York�

��� A�R� Magid� Applied Matrix Models� ���� John Wiley and Sons� NewYork�

�� D�R� Hill and C�B� Moler� Experiments in Computational Matrix Alge�

bra� ����� Random House� New York�

� � N� Deo� Graph Theory with Applications to Engineering and Computer

Science� ���� Prentice�Hall� N� J�

��� V� Pless� Introduction to the Theory of ErrorCorrecting Codes� �����

John Wiley and Sons� New York�

��� F�A� Graybill� Matrices with Applications in Statistics� �����

Wadsworth� Belmont Ca�

��� A�C� Chiang� Fundamental Methods of Mathematical Economics� sec�

ond edition� ���� McGraw�Hill Book Company� New York�

���� N�J� Pullman�Matrix Theory and its Applications� �����Marcel DekkerInc� New York�

���

Page 197: Elementary Linear Algebra

���� J�M� Geramita and N�J� Pullman� An Introduction to the Application

of Nonnegative Matrices to Biological Systems� ���� Queen�s Papers

in Pure and Applied Mathematics �� Queen�s University� Kingston�Canada�

���� M� Pearl� Matrix Theory and Finite Mathematics� ����� McGraw�HillBook Company� New York�

���� J�G� Kemeny and J�L� Snell� Finite Markov Chains� ����� Van NostrandReinhold� N�J�

���� E�R� Berlekamp� Algebraic Coding Theory� ����� McGraw�Hill Book

Company� New York�

��� G� Strang� Linear Algebra and its Applications� ����� Harcourt Brace

Jovanovich� San Diego�

�� � H� Minc� Nonnegative Matrices� ����� John Wiley and Sons� New York�

���� G�C� Preston and A�R� Lovaglia� Modern Analytic Geometry� �����

Harper and Row� New York�

���� J�A� Murtha and E�R� Willard� Linear Algebra and Geometry� �����

Holt� Rinehart and Winston� Inc� New York�

��� L�A� Pipes� Matrix Methods for Engineering� ����� Prentice�Hall� Inc�N� J�

���� D� Gans� Transformations and Geometries� ����� Appleton�Century�Crofts� New York�

���� J�N� Kapur� Transformation Geometry� ����� A�liated East�West

Press� New Delhi�

���� G�C� Reid� Postscript Language Tutorial and Cookbook� ����� Addison�

Wesley Publishing Company� New York�

���� D� Hearn and M�P� Baker� Computer Graphics� ����� Prentice�Hall�

Inc� N� J�

���� C�G� Cullen� Linear Algebra with Applications� ����� Scott� Foresmanand Company� Glenview� Illinois�

��� R�E� Larson and B�H� Edwards� Elementary Linear Algebra� ����� D�C�Heath and Company� Lexington� Massachusetts Toronto�

���

Page 198: Elementary Linear Algebra

�� � N� Magnenat�Thalman and D� Thalmann� Stateoftheartin Com�

puter Animation� ����� Springer�Verlag Tokyo�

���� W�K� Nicholson� Elementary Linear Algebra� ����� PWS�Kent� Boston�

���

Page 199: Elementary Linear Algebra

Index

�� � determinant� ��

algorithm� Gauss�Jordan� �

angle between vectors� ���

asymptotes� ���

basis� left�to�right algorithm� ��

Cauchy�Schwarz inequality� ��

centroid� ���

column space� ��

complex number� �

complex number� imaginary num�

ber�

complex number� imaginary part�

complex number� rationalization�

complex number� real� �

complex number� real part� �

complex numbers� Apollonius� cir�

cle� �

complex numbers� Argand diagram�

complex numbers� argument� ��

complex numbers� complex conju�

gate� �

complex numbers� complex expo�

nential� ��

complex numbers� complex plane�

complex numbers� cross�ratio� ���

complex numbers� De Moivre� ��

complex numbers� lower half plane�

complex numbers� modulus�

complex numbers� modulus�argument

form� ��

complex numbers� polar represen�

tation� ��

complex numbers� ratio formulae�

complex numbers� square root� �

complex numbers� upper half plane�

coordinate axes� ���

coordinate planes� ���

cosine rule� ���

determinant� ��

determinant� cofactor� ��

determinant� diagonal matrix� ��

determinant� Laplace expansion� ��

determinant� lower triangular� ��

determinant� minor� ��

determinant� recursive de nition�

��

determinant� scalar matrix� ��

determinant� Surveyor�s formula�

��

determinant� upper triangular� ��

di�erential equations� ��

direction of a vector� ���

distance� ���

distance to a plane� ���

��

Page 200: Elementary Linear Algebra

dot product� ���� ���

eigenvalue� ���

eigenvalues� characteristic equation�

���

eigenvector� ���

ellipse� ���

equation� linear� �

equations� consistent system of� ��

��

equations� Cramer�s rule� �

equations� dependent unknowns� ��

equations� homogeneous system of�

��

equations� homogeneous� non�trivial

solution� ��

equations� homogeneous� trivial so�

lution� ��

equations� inconsistent system of�

equations� independent unknowns�

��

equations� system of linear� �

factor theorem� �

eld� �

eld� additive inverse� �

eld� multiplicative inverse� �

Gauss� theorem� �

hyperbola� ���

imaginary axis� �

independence� left�to�right test� �

inversion� ��

Joachimsthal� ���

least squares� ��

least squares� normal equations� ��

least squares� residuals� ��

length of a vector� ���� ���

linear combination� ��

linear dependence� ��

linear equations� Cramer�s rule� ��

linear transformation� ��

linearly independent� ��

mathematical induction� ��

matrices� row�equivalence of� �

matrix� ��

matrix� addition� ��

matrix� additive inverse� ��

matrix� adjoint� ��

matrix� augmented� �

matrix� coe�cient� ��

matrix� coe�cient � �

matrix� diagonal� �

matrix� elementary row� ��

matrix� elementary row operations�

matrix� equality� ��

matrix� Gram� ���

matrix� identity� ��

matrix� inverse� ��

matrix� invertible� ��

matrix� Markov� ��

matrix� non�singular� ��

matrix� non�singular diagonal� �

matrix� orthogonal � ��

matrix� power� ��

matrix� product� ��

matrix� proper orthogonal� ��

matrix� reduced row�echelon form�

matrix� row�echelon form� �

matrix� scalar multiple� ��

matrix� singular� ��

matrix� skew�symmetric� ��

matrix� subtraction� ��

matrix� symmetric� ��

��

Page 201: Elementary Linear Algebra

matrix� transpose� ��

matrix� unit vectors� ��

matrix� zero� ��

modular addition� �

modular multiplication� �

normal form� ��

orthogonal matrix� ���

orthogonal vectors� ���

parabola� ���

parallel lines� ���

parallelogram law� ��

perpendicular vectors� ���

plane� ���

plane through � points� ���� ���

position vector� ���

positive octant� ���

projection on a line� ���

rank� ��

real axis� �

recurrence relations� ��

re�ection equations� �

rotation equations� ��

row space� ��

scalar multiplication of vectors� ��

scalar triple product� ���

skew lines� ���

subspace� ��

subspace� basis� ��

subspace� dimension� ��

subspace� generated� ��

subspace� null space� ��

Three�dimensional space� ���

triangle inequality� ��

unit vectors� ���

vector cross�product� ���

vector equality� ��� ���

vector� column� ��

vector� of constants� ��

vector� of unknowns� ��

vectors� parallel vectors� ���

��

Page 202: Elementary Linear Algebra

SOLUTIONS TO PROBLEMS

ELEMENTARY

LINEAR ALGEBRA

K� R� MATTHEWS

DEPARTMENT OF MATHEMATICS

UNIVERSITY OF QUEENSLAND

First Printing� ����

Page 203: Elementary Linear Algebra

CONTENTS

PROBLEMS ��� �������������������������������������������� �

PROBLEMS ��� �������������������������������������������� ��

PROBLEMS ��� �������������������������������������������� ��

PROBLEMS ��� �������������������������������������������� ��

PROBLEMS ��� �������������������������������������������� ��

PROBLEMS ��� �������������������������������������������� ��

PROBLEMS ��� �������������������������������������������� �

PROBLEMS ��� �������������������������������������������� ��

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��� R�� R� �R�

R� ���R�

��� � � �� � �� � �

����

�� �a�

��� � � � �� � �� � �� �� �

��� R�� R� � �R�

R� � R� � R�

��� � � � �� � �� �� �� �� ���

���

R� � R� �R�

R� � R� �R�

��� � � � ��

� � �� �� � � ��

���R� �

��� R�

��� � � � �� � �� �� � � �

���

R� � R� � �R�

R� � R� �R�

��� � � � ��

� � � ���

� � � ��

����

The augmented matrix has been converted to reduced row�echelon formand we read o� the unique solution x ��� y ��

� � z �� �

�b�

��� � � �� � ��

� �� � � ��� � ��� �� �

��� R� � R� � �R�

R� � R� �R�

��� � � �� � ��

� �� �� �� ���� ��� � ��

���

R� � R� �R�

��� � � �� � ��� �� �� �� ���� � � � �

����

From the last matrix we see that the original system is inconsistent�

Page 205: Elementary Linear Algebra

�c�

�����

� �� � �� �� � �

�� �� � �� �� �� �

�����R� � R�

������ �� � �� �� � �

�� �� � �� �� �� �

�����

R� � R� � �R�

R� � R� � �R�

R� � R� � �R�

�����

� �� � �� � � ��

�� � � ��� � � ��

�����

R� � R� R�

R� � R� � R�

R� � R� � �R�

������ � � ��

�� � � ��

�� � � �� � � �

������

The augmented matrix has been converted to reduced row�echelon formand we read o� the complete solution x ��

� � �z� y ��� � �z� with z

arbitrary�

��

��� � �� � a

� � �� b

�� �� �� c

���R� � R� �R�

��� � �� � a

� � � b� a

�� �� �� c

���

R� � R�

��� � � � b� a

� �� � a

�� �� �� c

��� R� � R� � �R�

R� � R� �R�

��� � � � b� a

� �� �� ��b �a� � ��� �b� �a c

���

R� � R� R�

R� ���� R�

��� � � � b� a

� � ����

�b��a�

� � � �b� �a c

���

R� � R� � �R�

��� � � ��

��b�a�

� � ����

�b��a�

� � � �b� �a c

����

From the last matrix we see that the original system is inconsistent if�b� �a c � �� If �b� �a c �� the system is consistent and the solutionis

x �b a�

�z� y

��b� �a�

��

�z�

where z is arbitrary�

��

��� � � �

t � t

� t � �

��� R�� R� � tR�

R� � R� � �� t�R�

��� � � �

� �� t �� �� t �� t

���

R� � R� �R�

��� � � �� �� t �� � �� t

��� B�

Page 206: Elementary Linear Algebra

Case �� t � �� No solution�

Case �� t �� B

��� � � �

� �� �� � �

����

��� � � �

� � �� � �

��� �

We read o� the unique solution x �� y ��

�� Method ��������� � � �� �� � �� � �� �� � � ��

�����

R� � R� �R�

R� � R� �R�

R� � R� �R�

������� � � �� �� � �� � �� �� � � ��

�����

������ � � ��� � � ��� � � ��� � � ��

�����R� � R� � R� �R� �R�

������ � � ��� � � ��� � � ��� � � �

����� �

Hence the given homogeneous system has complete solution

x� x�� x� x�� x� x��

with x� arbitrary�

Method �� Write the system as

x� x� x� x� �x�

x� x� x� x� �x�

x� x� x� x� �x�

x� x� x� x� �x��

Then it is immediate that any solution must satisfy x� x� x� x��Conversely� if x�� x�� x�� x� satisfy x� x� x� x�� we get a solution�

�� ��� � �� �� �

�R� � R�

�� �� �

�� � �

R� � R� � ��� ��R�

�� �� �� ��� ���

� B�

Page 207: Elementary Linear Algebra

Case �� ��� ��� � �� That is ���� ����� �� � � or � � �� �� Here B is

row equivalent to

�� �� �

��

R� ��

�������R�

�� �� �� �

�R�� R� � ��� ��R�

�� �� �

��

Hence we get the trivial solution x �� y ��

Case �� � �� Then B

�� ��� �

�and the solution is x y� with y

arbitrary�

Case �� � �� Then B

�� �� �

�and the solution is x �y� with y

arbitrary�

� �� � � �� �� � ��

�R� �

�R�

�� �

���

��

� �� � ��

R� � R� � �R�

�� �

���

��

� ��� ��

� ���

R� ���

R�

�� �

���

��

� � �� �

R� � R� ��

�R�

�� � �

� �� � �

� �

��

Hence the solution of the associated homogeneous system is

x� ��

�x�� x� �

�x� � x��

with x� and x� arbitrary�

��

A

������

�� n � � � � �� �� n � � � ����

��� � � ����

� � � � � �� n

������

R� � R� �Rn

R� � R� �Rn

���Rn�� � Rn�� �Rn

�������n � � � � n

� �n � � � n���

��� � � ����

� � � � � �� n

������

Page 208: Elementary Linear Algebra

������� � � � � ��� � � � � �����

��� � � ����

� � � � � �� n

������Rn � Rn � Rn�� � � � �R�

������� � � � � ��� � � � � �����

��� � � ����

� � � � � �

������ �

The last matrix is in reduced row�echelon form�Consequently the homogeneous system with coe�cient matrix A has the

solutionx� xn� x� xn� � � � � xn�� xn�

with xn arbitrary�Alternatively� writing the system in the form

x� � � � xn nx�

x� � � � xn nx����

x� � � � xn nxn

shows that any solution must satisfy nx� nx� � � � nxn� so x� x� � � � xn� Conversely if x� xn� � � � � xn�� xn� we see that x�� � � � � xn is asolution�

��� Let A

�a b

c d

�and assume that ad� bc � ��

Case �� a � ���a b

c d

�R� �

�aR�

�� b

a

c d

�R� � R� � cR�

�� b

a

� ad�bca

R� �a

ad�bcR�

�� b

a

� �

�R� � R� �

baR�

�� �� �

��

Case �� a �� Then bc � � and hence c � ��

A

�� b

c d

�R� � R�

�c d

� b

��

�� d

c

� �

��

�� �� �

��

So in both cases� A has reduced row�echelon form equal to

�� �� �

��

��� We simplify the augmented matrix of the system using row operations�

Page 209: Elementary Linear Algebra

��� � � �� �� �� � �� � a� � �� a �

��� R� � R� � �R�

R� � R� � �R�

��� � � �� �

� �� �� ���� �� a� � � a� ��

���

R�� R� �R�

R� ���� R�

R� � R� � �R�

��� � � �� �

� � �� ���

� � a� � �� a � �

��� R�� R� � �R�

��� � � � �

�� � �� ��

�� � a� � �� a� �

��� �

Denote the last matrix by B�

Case �� a� � �� � �� i�e� a � ��� Then

R� ��

a���R�

R� � R� �R�

R� � R� �R�

���� � � �a���

��a��

� � � ��a�����a��

� � � �a��

���

and we get the unique solution

x a ��

��a ��� y

��a ��

��a ��� z

a ��

Case �� a ��� Then B

��� � � � �

�� � �� ��

�� � � �

���� so our system is inconsistent�

Case �� a �� Then B

��� � � � �

�� � �� ��

�� � � �

���� We read o� that the system is

consistent� with complete solution x �� � z� y ��

� �z� where z isarbitrary�

��� We reduce the augmented array of the system to reduced row�echelonform� �

����� � � � �� � � � �� � � � �� � � � �

����� R� � R� R�

������ � � � �� � � � �� � � � �� � � � �

�����

R� � R� R�

�����

� � � � �� � � � �� � � � �� � � � �

����� R� � R� R�

R� � R�

������ � � � �� � � � �� � � � �� � � � �

����� �

Page 210: Elementary Linear Algebra

The last matrix is in reduced row�echelon form and we read o� the solutionof the corresponding homogeneous system�

x� �x� � x� x� x�

x� �x� � x� x� x�

x� �x� x��

where x� and x� are arbitrary elements of Z�� Hence there are four solutions�

x� x� x� x� x�� � � � �� � � � �� � � � �� � � � �

��� �a� We reduce the augmented matrix to reduced row�echelon form�

��� � � � �

� � � �� � � �

��� R� � �R�

��� � � � �

� � � �� � � �

���

R� � R� R�

R�� R� �R�

��� � � � �� � � �� � � �

��� R� � �R�

��� � � � �� � � �� � � �

���

R�� R� �R�

R�� R� �R�

��� � � � �� � � �� � � �

��� R� � R� �R�

R� � R� �R�

��� � � � �

� � � �� � � �

��� �

Consequently the system has the unique solution x �� y �� z ��

�b� Again we reduce the augmented matrix to reduced row�echelon form�

��� � � � �� � � �� � � �

��� R� � R�

��� � � � �� � � �� � � �

���

R� � R� R�

R�� R� �R�

��� � � � �� � � �� � � �

��� R� � �R�

��� � � � �� � � �� � � �

���

Page 211: Elementary Linear Algebra

R� � R� �R�

R� � R� R�

��� � � � �� � � �� � � �

��� �

We read o� the complete solution

x �� �z � �z

y �� �z � �z�

where z is an arbitrary element of Z��

��� Suppose that ���� � � � � �n� and ���� � � � � �n� are solutions of the systemof linear equations

nXj �

aijxj bi� � � i � m�

ThennX

j �

aij�j bi andnX

j �

aij�j bi

for � � i � m�Let �i ��� t��i t�i for � � i � m� Then ���� � � � � �n� is a solution of

the given system� For

nXj �

aij�j nX

j �

aijf��� t��j t�jg

nX

j �

aij��� t��j nX

j �

aijt�j

��� t�bi tbi

bi�

��� Suppose that ���� � � � � �n� is a solution of the system of linear equations

nXj �

aijxj bi� � � i � m� ���

Then the system can be rewritten as

nXj �

aijxj nX

j �

aij�j � � � i � m�

Page 212: Elementary Linear Algebra

or equivalentlynX

j �

aij�xj � �j� �� � � i � m�

So we havenX

j �

aijyj �� � � i � m�

where xj � �j yj � Hence xj �j yj � � � j � n� where �y�� � � � � yn� isa solution of the associated homogeneous system� Conversely if �y�� � � � � yn�is a solution of the associated homogeneous system and xj �j yj � � �j � n� then reversing the argument shows that �x�� � � � � xn� is a solution ofthe system � �

��� We simplify the augmented matrix using row operations� working to�wards row�echelon form���� � � �� � �a � � � b

� � � a � a

��� R� � R� � aR�

R� � R� � �R�

��� � � �� � �

� �� a � a �� a b� a

� �� � a� � a � �

���

R� � R�

R� � �R�

��� � � �� � �

� � �� �� a �� a

� �� a � a �� a b� a

���

R� � R� �a� ��R�

��� � � �� � �

� � �� �� a �� a

� � �� �a ��� a��a� �� �a� �a b� �

��� B�

Case �� a � �� Then �� �a � � and

B �

���

� � �� � �� � �� �� a �� a

� � � a���

�a���a�b�����a

��� �

Hence we can solve for x� y and z in terms of the arbitrary variable w�

Case �� a �� Then

B

��� � � �� � �� � �� � �� � � � b� �

��� �

Page 213: Elementary Linear Algebra

Hence there is no solution if b � �� However if b �� then

B

��� � � �� � �

� � �� � �� � � � �

����

��� � � � � �� � �� � �� � � � �

���

and we get the solution x �� �z� y �z � w� where w is arbitrary�

��� �a� We �rst prove that � � � � �� Observe that the elements

� �� � �� � a� � b

are distinct elements of F by virtue of the cancellation law for addition� Forthis law states that �x �y � x y and hence x � y � �x � �y�

Hence the above four elements are just the elements �� �� a� b in someorder� Consequently

�� �� �� �� �� a� �� b� � � a b

�� � � �� �� � a b� � �� � a b��

so � � � � � after cancellation�Now � � � � �� ���� ��� so we have x� �� where x � ��

Hence x �� Then a a a�� �� a � � ��Next a b �� For a b must be one of �� �� a� b� Clearly we can�t

have a b a or b� also if a b �� then a b a a and hence b a�hence a b �� Then

a � a �a b� �a a� b � b b�

Similarly b � a� Consequently the addition table for F is

� � a b� � � a b� � � b aa a b � �b b a � �

We now �nd the multiplication table� First� ab must be one of �� a� b�however we can�t have ab a or b� so this leaves ab ��

Next a� b� For a� must be one of �� a� b� however a� a� a � ora �� also

a� �� a� � � �� �a� ���a �� �� �a� ��� �� a ��

��

Page 214: Elementary Linear Algebra

hence a� b� Similarly b� a� Consequently the multiplication table for Fis

� � � a b� � � � �� � � a ba � a b �b � b � a

�b� We use the addition and multiplication tables for F �

A

��� � a b a

a b b �� � � a

��� R� � R� aR�

R� � R� R�

��� � a b a

� � a a

� b a �

���

R� � R�

��� � a b a

� b a �� � a a

��� R� � aR�

R� � bR�

��� � a b a

� � b �� � � �

���

R� � R� aR�

��� � � a a

� � b �� � � �

��� R�� R� aR�

R� � R� bR�

��� � � � �� � � b

� � � �

��� �

The last matrix is in reduced row�echelon form�

��

Page 215: Elementary Linear Algebra

Section ���

�� Suppose B �

��� a b

c d

e f

��� and that AB � I�� Then

��� � �� � �

� ��� a b

c d

e f

��� �

�� �� �

��

��a� e �b� f

c� e d� f

��

Hence�a� e � �c� e � �

��b� f � �d� f � �

e � a� �c � �e � ��a� ��

�f � b

d � �� f � �� b�

B �

��� a b

�a� � �� b

a� � b

��� �

Next

�BA��B � �BA��BA�B � B�AB��AB� � BI�I� � BI� � B

� Let pn denote the statement

An � ��n���� A� ����n�

� I��

Then p� asserts that A � ������ A� �����

� I�� which is true� So let n � � andassume pn� Then from ���

An�� � A �An � An��n���

� A� ����n�� I�

o� ��n���

� A� � ����n�� A

� ��n���� �A� �I�� �

����n�� A � ��n���������n�

� A� ��n�������� I�

� ����n��n���� A � ����n���

� I�

� ��n������ A� ����n���

� I��

Hence pn�� is true and the induction proceeds�

��

Page 216: Elementary Linear Algebra

�� The equation xn�� � axn � bxn�� is seen to be equivalent to�xn��xn

��

�a b

� �

� �xnxn��

orXn � AXn���

where Xn �

�xn��xn

�and A �

�a b

� �

�� Then

Xn � AnX�

if n � �� Hence by Question ��xn��xn

��

���n � ��

�A�

��� �n�

�I�

�x�x�

��n � ��

� ��� �

��

����n� �� ���n

�� �x�x�

��� ��n � ��� � ���n

� ��n � �������n���

���n�

����x�x�

Hence equating the ��� �� elements gives

xn ���n � ��

�x� �

��� �n�

�x� if n � �

� Note� �� � �� � a� d and ���� � ad� bc�Then

��� � ���kn � ����kn�� � ��� � �����n��� � �n��� �� � � � �� ���

n��� � �n��� �

�������n��� � �n��� �� � � � �� ���n��� � �n��� �

� ��n� � �n��� �� � � � �� ���n��� �

���n��� �� � � � �� ���n��� � �n��

���n��� �� � � � �� ���n��� �

� �n� � �n��� �� � � � �� ���n��� � �n� � kn��

��

Page 217: Elementary Linear Algebra

If �� � �� we see

kn � �n��� � �n��� �� � � � �� ���n��� � �n���

� �n��� � �n��� �� � � � �� ���n��� � �n���

� n�n���

If �� �� �� we see that

��� � ���kn � ��� � �����n��� � �n��� �� � � � �� ���

n��� � �n��� �

� �n� � �n��� �� � � � �� ���n���

���n��� �� � � � �� ���n��� � �n��

� �n� � �n� �

Hence kn ��n

���n

����� �

We have to proveAn � knA� ����kn��I�� �

n���

A� � A� also k�A� ����k�I� � k�A� �����I�

� A�

Let n � � and assume equation � holds� Then

An�� � An �A � �knA � ����kn��I��A

� knA� � ����kn��A�

Now A� � �a� d�A� �ad� bc�I� � ��� � ���A� ����I�� Hence

An�� � kn��� � ���A� ����I� � ����kn��A

� fkn��� � ���� ����kn��gA� ����knI�

� kn��A� ����knI��

and the induction goes through�

�� Here ��� �� are the roots of the polynomial x� � �x� � � �x� ���x� ���So we can take �� � �� �� � ��� Then

kn ��n � ����n�� ���� �

�n � ����n��

Page 218: Elementary Linear Algebra

Hence

An �

�n � ����n��

�A � ����

�n�� � ����n

�I�

��n � ����n��

�� �� �

�� �

�n�� � ����n

��� �� �

��

which is equivalent to the stated result�

�� In terms of matrices we have�Fn��Fn

��

�� �� �

� �FnFn��

�for n � ��

�Fn��Fn

��

�� �� �

�n�F�

F�

��

�� �� �

�n���

��

Now ��� �� are the roots of the polynomial x� � x� � here�

Hence �� ���

p�

� and �� ���

p�

� and

kn �

���

p�

n�� � ���

p�

n����

p�

� ����

p�

���

p�

n�� � ���

p�

n��p�

Hence

An � knA � ����kn��I�

� knA � kn��I�

So �Fn��Fn

�� �knA� kn��I��

���

� kn

���

�� kn��

���

��

�kn � kn��

kn

��

Hence

Fn � kn �

���

p�

n�� �

���

p�

n��

p�

��

Page 219: Elementary Linear Algebra

��� From Question � we know that�xnyn

��

�� r

� �

�n �a

b

��

Now by Question with A �

�� r

� �

An � knA� ����kn��I�

� knA� ��� r�kn��I��

where �� � � �pr and �� � � � p

r are the roots of the polynomialx� � �x� ��� r� and

kn ��n� � �n��pr

Hence�xnyn

�� �knA� ��� r�kn��I��

�a

b

��kn knr

kn kn

������ r�kn�� �

� ��� r�kn��

�� �a

b

�kn � ��� r�kn�� knr

kn kn � ��� r�kn��

� �a

b

�a�kn � ��� r�kn��� � bknr

akn � b�kn � ��� r�kn���

��

Hence in view of the fact that

kn

kn���

�n� � �n�

�n��� � �n���

��n���� f����gn�

�n��� ��� f����gn��� � ��� as n���

we have �xnyn

��

a�kn � ��� r�kn��� � bknr

akn � b�kn � ��� r�kn���

�a� kn

kn��� ��� r�� � b kn

kn��r

a kn

kn��� b� kn

kn��� ��� r��

��

Page 220: Elementary Linear Algebra

� a��� � ��� r�� � b��r

a�� � b��� � ��� r��

�a�pr � r� � b�� �

pr�r

a�� �pr� � b�

pr � r�

prfa�� �p

r� � b�� �pr�prg

a�� �pr� � b�

pr � r�

�pr�

Page 221: Elementary Linear Algebra

Section ���

�� �AjI�� �

�� �

�� �

����� � �� �

�R� � R� � �R�

�� �� ��

����� � �� �

R� ����R�

�� �� �

����� � ����� ����

�R� � R� � �R�

�� �� �

����� ���� ��������� ����

��

Hence A is nonsingular and A�� �

����� ��������� ����

��

MoreoverE�����E������E����A � I��

soA�� � E�����E������E�����

Hence

A � fE����g��fE������g

��fE�����g�� � E�����E����E�����

�� Let D � �dij� be an m�m diagonal matrix and let A � �ajk� be an m�nmatrix� Then

DA�ik �nX

j��

dijajk � diiaik �

as dij � � if i �� j� It follows that the ith row of DA is obtained bymultiplying the ith row of A by dii�

Similarly postmultiplication of a matrix by a diagonal matrix D resultsin a matrix whose columns are those of A multiplied by the respectivediagonal elements of D�

In particular

diag a�� � � � � an�diag b�� � � � � bn� � diag a�b�� � � � � anbn��

as the lefthand side can be regarded as premultiplication of the matrixdiag b�� � � � � bn� by the diagonal matrix diag a�� � � � � an��

Finally suppose that each of a�� � � � � an is nonzero� Then a��� � � � � � a��n

all exist and we have

diag a�� � � � � an�diag a��� � � � � � a��n � � diag a�a

��� � � � � � ana

��n �

� diag �� � � � � �� � In�

��

Page 222: Elementary Linear Algebra

Hence diag a�� � � � � an� is nonsingular and its inverse is diag a��� � � � � � a��n ��Next suppose that ai � �� Then diag a�� � � � � an� is rowequivalent to a

matix containing a zero row and is hence singular�

�� �AjI�� �

��� � � �

� � �� � �

�������� � �� � �� � �

��� R� � R�

��� � � � � � �

� � � � � �� � � � � �

���

R� � R� � �R�

��� � � � � � �

� � � � � �� � �� � �� �

��� R� � R�

��� � � � � � �� � �� � �� �� � � � � �

���

R� ���R�

��� � � � � � �� � �� � �� �� � � ��� � �

��� R� � R� � �R�

��� � � �� � � ��� � �� � �� �� � � ��� � �

���

R� � R� � ��R�

R� � R� � �R�

��� � � � ��� � ��

� � � ��� �� �� � � ��� � �

����

Hence A is nonsingular and A�� �

��� ��� � ��

��� �� ���� � �

����

Also

E����E������E�����E�����E��E�����E��A � I��

Hence

A�� � E����E������E�����E�����E��E�����E���

soA � E��E����E��E���E����E�����E������

��

A �

��� � � k

� �� �� � ��

����

��� � � k� �� �� �k� �� ��� �k

����

��� � � k� �� �� �k� � ��� �k

��� � B�

Hence if ��� �k �� � i�e� if k �� �� we see that B can be reduced to I�and hence A is nonsingular�

��

Page 223: Elementary Linear Algebra

If k � �� then B �

��� � � ��� �� ��� � �

��� � B and consequently A is singu�

lar as it is rowequivalent to a matrix containing a zero row�

�� E����

�� �

�� ��

��

�� �� �

�� Hence as in the previous question �

� ��� ��

�is singular�

�� Starting from the equation A� � �A� ��I� � � we deduce

AA� �I�� � ���I� � A� �I��A�

Hence AB � BA � I� where B � ���� A � �I��� Consequently A is non

singular and A�� � B�

�� We assume the equation A� � �A� � �A� I��

ii� A� � A�A � �A� � �A� I��A � �A� � �A� � A

� ��A� � �A� I��� �A� � A � �A� � �A� �I��

iii� A� � �A� � �A � I�� Hence

AA� � �A� �I�� � I� � A� � �A� �I��A�

Hence A is nonsingular and

A�� � A� � �A� �I�

��� �� �� �

� � ��� � �

��� �

�� i� If B� � � then

In �B�In �B � B�� � InIn �B � B���BIn � B �B��

� In � B �B��� B � B� � B��

� In �B� � In � � � In�

Similarly In � B �B��In � B� � In�Hence A � In �B is nonsingular and A�� � In �B �B��

��

Page 224: Elementary Linear Algebra

It follows that the system AX � b has the unique solution

X � A��b � In � B �B��b � b�Bb �B�b�

ii� Let B �

��� � r s

� � t� � �

���� Then B� �

��� � � rt

� � �� � �

��� and B� � �� Hence

from the preceding question

I� �B��� � I� �B � B�

��� � � �

� � �� � �

����

��� � r s� � t� � �

����

��� � � rt

� � �� � �

���

��� � r s� rt

� � t� � �

��� �

�� i� Suppose that A� � �� Then if A�� exists we deduce that A��AA� �A��� which gives A � � and this is a contradiction as the zero matrix issingular� We conclude that A does not have an inverse�

ii�� Suppose that A� � A and that A�� exists� Then

A��AA� � A��A�

which gives A � In� Equivalently if A� � A and A �� In then A does nothave an inverse�

��� The system of linear equations

x� y � z � a

z � b

�x� y � �z � c

is equivalent to the matrix equation AX � B where

A �

��� � � ��� � �� � �

��� � X �

��� xyz

��� � B �

��� abc

��� �

��

Page 225: Elementary Linear Algebra

By Question � A�� exists and hence the system has the unique solution

X �

��� �� �� �

� � ��� � �

������ ab

c

��� �

��� �a � �b� c

�a� �b� c

b

��� �

Hence x � �a � �b� c� y � �a� �b� c� z � b�

���

A � E���E��E���� � E���E��

������ � � �� � � �� � � �� � � �

�����

� E���

�����

� � � �� � � �� � � �� � � �

����� �

�����

� � � �� � � �� � � �� � � �

����� �

Also

A�� � E���E��E�������

� E�������E���� E����

��

� E�����E��E�����

� E�����E��

�����

� � � �� � � �� � ��� �� � � �

�����

� E�����

�����

� � � �� � � �� � ��� �� � � �

�����

������ � � �� � � �� � ��� �� �� � �

����� �

��� All matrices in this question are over Z���

��

Page 226: Elementary Linear Algebra

a�

�����

� � � �� � � �� � � �� � � �

���������

� � � �� � � �� � � �� � � �

������

������ � � �� � � �� � � �� � � �

���������

� � � �� � � �� � � �� � � �

�����

�����

� � � �� � � �� � � �� � � �

���������

� � � �� � � �� � � �� � � �

������

������ � � �� � � �� � � �� � � �

���������

� � � �� � � �� � � �� � � �

�����

�����

� � � �� � � �� � � �� � � �

���������

� � � �� � � �� � � �� � � �

����� �

Hence A is nonsingular and

A�� �

������ � � �� � � �� � � �� � � �

����� �

b� A �

�����

� � � �� � � �� � � �� � � �

����� R� � R� �R�

������ � � �� � � �� � � �� � � �

����� so A is singular�

���

a�

��� � � ��� � �� � �

�������� � �� � �� � �

���

R� ���R�

R�� R� �R�

R�� R� �R�

R�� R�

��� � � �� � �� � �

�������� � ���� � ���� � ����

���

R� � R� �R�

��� � � �� � �� � �

�������� � ���� � ���� �� ��

��� �

Hence A�� exists and

A�� �

��� � � ���

� � ���� �� ��

��� �

��

Page 227: Elementary Linear Algebra

b�

��� � � �

� � �� � �

�������� � �� � �� � �

��� R� � R� � �R�

R� � R�

R� � R�

��� � � �� � �� � �

�������� � �� � �� �� �

���

R� � R� � �R�

��� � � �

� � �� � �

�������� � �� � �� �� ��

���

R� ���R�

��� � � �� � �� � �

�������� � �� � �

��� �� ��

���

R� � R� �R�

��� � � �� � �� � �

����������� � �

� � ���� �� ��

��� �

Hence A�� exists and

A�� �

��� ���� � �

� � ���� �� ��

��� �

c�

��� � � ��

� � �� � �

��� R� �

��R�

R� ���R�

��� � � ��

� � �� � �

��� R� � R� � R�

��� � � ��

� � �� � �

��� �

Hence A is singular by virtue of the zero row�

d�

��� � � �

� �� �� � �

�������� � �� � �� � �

��� R� �

��R�

R� ���� R�

R� ���R�

��� � � �� � �� � �

���������� � �� ���� �� � ���

��� �

Hence A�� exists and A�� � diag ���� ����� �����

Of course this was also immediate from Question ���

e�

������ � � �� � � �� � � �� � � �

���������

� � � �� � � �� � � �� � � �

����� R� � R� � �R�

������ � � �� � � �� � � �� � � �

���������

� �� � �� � � �� � � �� � � �

�����

R� � R� � �R�

�����

� � � �� � � ��� � � �� � � �

���������

� �� � �� � �� �� � � �� � � �

�����

��

Page 228: Elementary Linear Algebra

R� � R� � �R�

R� � R� � �R�

R�� R� �R�

R� ���R�

�����

� � � �� � � �� � � �� � � �

���������

� �� � ��� � �� �� � � ��� � � ���

����� �

Hence A�� exists and

A�� �

�����

� �� � ��� � �� �� � � ��� � � ���

����� �

f���� � � �� � �� � �

��� R� � R� � �R�

R� � R� � �R�

��� � � �� �� ��� �� ��

��� R� � R� � R�

��� � � �

� �� ��� � �

��� �

Hence A is singular by virtue of the zero row�

��� Suppose that A is nonsingular� Then

AA�� � In � A��A�

Taking transposes throughout gives

AA���t � I tn � A��A�t

A���tAt � In � AtA���t�

so At is nonsingular and At��� � A���t�

��� Let A �

�a bc d

� where ad� bc � �� Then the equation

A� � a� d�A� ad� bc�I� � �

reduces to A� � a � d�A � � and hence A� � a � d�A� From the lastequation if A�� exists we deduce that A � a� d�I� or�

a bc d

��

�a� d �� a� d

��

Hence a � a � d� b � �� c � �� d � a � d and a � b � c � d � � whichcontradicts the assumption that A is nonsingular�

��

Page 229: Elementary Linear Algebra

���

A �

��� � a b

�a � c�b �c �

��� R� � R� � aR�

R� � R� � bR�

��� � a b

� � � a� c� ab� ab� c � � b�

���

R� ��

��a�R�

��� � a b

� � c�ab��a�

� ab� c � � b�

���

R� � R� � ab� c�R�

���� a b

� � c�ab��a�

� � � � b� � �c�ab�c�ab��a�

��� � B�

Now

� � b� �c� ab�c� ab�

� � a�� � � b� �

c� � ab��

� � a�

�� � a� � b� � c�

� � a��� ��

Hence B can be reduced to I� using four more row operations and conse�quently A is nonsingular�

��� The proposition is clearly true when n � �� So let n � � and assumeP��AP �n � P��AnP � Then

P��AP �n�� � P��AP �nP��AP �

� P��AnP �P��AP �

� P��AnPP���AP

� P��AnIAP

� P��AnA�P

� P��An��P

and the induction goes through�

��� Let A �

���� ������ ���

�and P �

�� �

�� �

�� Then P�� � �

�� ��� �

��

We then verify that P��AP �

����� �� �

�� Then from the previous ques�

tion

P��AnP � P��AP �n �

����� �� �

�n�

������n �

� �n

��

������n �

� �

��

��

Page 230: Elementary Linear Algebra

Hence

An � P

������n �

� �

�P�� �

�� �

�� �

� ������n �

� �

��

�� ��� �

��

������n �

������n �

��� ��� �

��

�������n � � ��������n � ��������n � � ������n � �

��

�� �� �

��

������n

�� ��

�� �

��

Notice that An � ��

�� �� �

�as n � �� This problem is a special case of

a more general result about Markov matrices�

��� Let A �

�a b

c d

�be a matrix whose elements are nonnegative real

numbers satisfying

a � �� b � �� c � �� d � �� a � c � � � b� d�

Also let P �

�b �c ��

�and suppose that A �� I��

i� detP � �b � c � �b� c�� Now b � c � �� Also if b � c � � then wewould have b � c � � and hence d � a � � resulting in A � I�� HencedetP � � and P is nonsingular�

Next

P��AP ���

b� c

��� ���c b

� �a b

c d

� �b �c ��

���

b� c

��a � c �b� d

�ac� bc �cb� bd

� �b �c ��

���

b� c

��� ��

�ac� bc �cb� bd

� �b �c ��

���

b� c

��b� c �

�ac� bc�b� �cb� bd�c �ac� bc� cb� bd

��

��

Page 231: Elementary Linear Algebra

Now

�acb� b�c� c�b� bdc � �cba� c� � bcb� d�

� �cb� bc � ��

Also

�a� d� ��b� c� � �ab� ac� db� dc� b� c

� �ac� b�� a� � c�� d�� bd

� �ac� bc� cb� bd�

Hence

P��AP ���

b� c

��b� c� �

� �a � d� ��b� c�

��

�� �� a� d� �

��

ii� We next prove that if we impose the extra restriction that A ��

�� �� �

then ja� d� �j � �� This will then have the following consequence�

A � P

�� �� a� d� �

�P��

An � P

�� �� a� d� �

�nP��

� P

�� �� a� d� ��n

�P��

� P

�� �� �

�P��

�b �c ��

��� �� �

���

b� c

��� ���c b

���

b� c

�b �c �

� ��� ���c b

���

b� c

��b �b�c �c

��

b� c

�b bc c

��

��

Page 232: Elementary Linear Algebra

where we have used the fact that a� d� ��n � � as n���

We �rst prove the inequality ja� d� �j � ��

a� d� � � � � d� � � d � �

a� d� � � � � �� � � ���

Next if a� d� � � � we have a� d � �� so a � � � d and hence c � � � b contradicting our assumption that A �� I�� Also if a � d � � � �� then

a� d � �� so a � � � d and hence c � � � b and hence A �

�� �� �

��

��� The system is inconsistent� We work towards reducing the augmentedmatrix� �

�� � �� �� �

�����������

��� R� � R� � R�

R� � R� � �R�

��� � �

� ��� ��

����������

���

R� � R� �R�

��� � �� ��� �

���������

��

��� �

The last row reveals inconsistency�The system in matrix form is AX � B where

A �

��� � �

� �� �

��� � X �

�xy

�� B �

��� �

���

��� �

The normal equations are given by the matrix equation

AtAX � AtB�

Now

AtA �

�� � �� � �

� ��� � �� �� �

��� �

��� ���� ��

AtB �

�� � �� � �

� ��� ����

��� �

�����

��

��

Page 233: Elementary Linear Algebra

Hence the normal equations are

��x� ��y � ��

��x� ��y � ���

These may be solved for example by Cramer�s rule�

x �

����� �� ���� ��

���������� �� ���� ��

������

��

�� �

y �

����� �� ���� ��

���������� �� ���� ��

��������

��

��� Substituting the coordinates of the �ve points into the parabola equationgives the following equations�

a � �

a� b� c � �

a � �b� �c � ��

a � �b� �c � �

a� �b� ��c � ��

The associated normal equations are given by��� � �� ���� �� ����� ��� ���

������ a

bc

��� �

��� ��

�����

��� �

which have the solution a � ���� b � ��� c � ��

��� Suppose that A is symmetric i�e� At � A and that AB is de�ned� Then

BtAB�t � BtAtBt�t � BtAB�

so BtAB is also symmetric�

��

Page 234: Elementary Linear Algebra

��� Let A be m� n and B be n�m where m � n� Then the homogeneoussystem BX � � has a nontrivial solution X as the number of unknownsis greater than the number of equations� Then

AB�X � ABX� � A� � �

and the m�m matrix AB is therefore singular as X �� ��

��� i� Let B be a singular n� n matrix� Then BX � � for some nonzerocolumn vector X � Then AB�X � ABX� � A� � � and hence AB is alsosingular�

ii� Suppose A is a singular n� n matrix� Then At is also singular andhence by i� so is BtAt � AB�t� Consequently AB is also singular

��

Page 235: Elementary Linear Algebra

Section ���

�� �a� Let S be the set of vectors �x� y� satisfying x � �y� Then S is a vectorsubspace of R�� For

�i� ��� �� � S as x � �y holds with x � � and y � ��

�ii� S is closed under addition� For let �x�� y�� and �x�� y�� belong to S�Then x� � �y� and x� � �y�� Hence

x� x� � �y� �y� � ��y� y��

and hence�x� x�� y� y�� � �x�� y�� �x�� y��

belongs to S�

�iii� S is closed under scalar multiplication� For let �x� y� � S and t � R�Then x � �y and hence tx � ��ty�� Consequently

�tx� ty� � t�x� y� � S�

�b� Let S be the set of vectors �x� y� satisfying x � �y and �x � y� Then S isa subspace of R�� This can be proved in the same way as �a� or alternativelywe see that x � �y and �x � y imply x � �x and hence x � � � y� HenceS � f��� ��g the set consisting of the zero vector� This is always a subspace�

�c� Let S be the set of vectors �x� y� satisfying x � �y �� Then S doesn�tcontain the zero vector and consequently fails to be a vector subspace�

�d� Let S be the set of vectors �x� y� satisfying xy � �� Then S is notclosed under addition of vectors� For example ��� �� � S and ��� �� � S but��� �� ��� �� � ��� �� �� S�

�e� Let S be the set of vectors �x� y� satisfying x � � and y � �� Then S isnot closed under scalar multiplication� For example ��� �� � S and �� � Rbut ������� �� � ���� �� �� S�

�� Let X� Y� Z be vectors in Rn� Then by Lemma ����

hX Y� X Z� Y Zi � hX� Y� Zi�

as each of X Y� X Z� Y Z is a linear combination of X� Y� Z�

Page 236: Elementary Linear Algebra

Also

X ��

��X Y �

��X Z��

��Y Z��

Y ��

��X Y ��

��X Z�

��Y Z��

Z ���

��X Y �

��X Z�

��Y Z��

sohX� Y� Zi � hX Y� X Z� Y Zi�

HencehX� Y� Zi � hX Y� X Z� Y Zi�

� Let X� �

�����

����

����� � X� �

�����

����

����� and X� �

�����

���

������ We have to decide if

X�� X�� X� are linearly independent that is if the equation xX� yX� zX� � � has only the trivial solution� This equation is equivalent to thefolowing homogeneous system

x �y z � �

�x y z � �

x y z � �

�x �y z � ��

We reduce the coe�cient matrix to reduced row�echelon form������

� � �� � �� � �� �

������

�����

� � �� � �� � �� � �

�����

and consequently the system has only the trivial solution x � �� y � �� z ��� Hence the given vectors are linearly independent�

�� The vectors

X� �

��� �

����

��� � X� �

��� ��

��

��� � X� �

��� �����

���

Page 237: Elementary Linear Algebra

are linearly dependent for precisely those values of � for which the equationxX�yX�zX� � � has a non�trivial solution� This equation is equivalentto the system of homogeneous equations

�x� y � z � �

�x �y � z � �

�x� y �z � ��

Now the coe�cient determinant of this system is�������� �� ��

�� � ���� �� �

������� � �� ������ ���

So the values of � which make X�� X�� X� linearly independent are those �satisfying � �� �� and � �� ��

�� Let A be the following matrix of rationals�

A �

�����

� � � � �� � � � � � � � � �� �� � ��

����� �

Then A has reduced row�echelon form

B �

�����

� � � � ��� � � � �� � � � �� � � �

����� �

From B we read o� the following�

�a� The rows of B form a basis for R�A�� �Consequently the rows of Aalso form a basis for R�A���

�b� The �rst four columns of A form a basis for C�A��

�c� To �nd a basis for N�A� we solve AX � � and equivalently BX � ��From B we see that the solution is

x� � x�

x� � �

x� � �x�

x� � � x��

Page 238: Elementary Linear Algebra

with x� arbitrary� Then

X �

�������

x��

�x�� x�

x�

������� � x�

�������

��

��� �

������� �

so ��� �� ��� � � ��t is a basis for N�A��

�� In Section ��� problem �� we found that the matrix

A �

�����

� � � � �� � � � �� � � � �� � � � �

�����

has reduced row�echelon form

B �

�����

� � � � �� � � � �� � � � �� � � � �

����� �

From B we read o� the following�

�a� The three non�zero rows of B form a basis for R�A��

�b� The �rst three columns of A form a basis for C�A��

�c� To �nd a basis for N�A� we solve AX � � and equivalently BX � ��From B we see that the solution is

x� � �x� � x� � x� x�

x� � �x� � x� � x� x�

x� � �x� � x��

with x� and x� arbitrary elements of Z�� Hence

X �

�������

x� x�x� x�x�x�x�

������� � x�

�������

�����

�������

�������

�����

������� �

Hence ��� �� �� �� ��t and ��� �� �� �� ��t form a basis for N�A��

Page 239: Elementary Linear Algebra

�� Let A be the following matrix over Z��

A �

�����

� � � � � � � � � �� � � � � � � � �

����� �

We �nd that A has reduced row�echelon form B�

B �

�����

� � � � � �� � � � � �� � � � � �� � � � �

����� �

From B we read o� the following�

�a� The four rows of B form a basis for R�A�� �Consequently the rows ofA also form a basis for R�A��

�b� The �rst four columns of A form a basis for C�A��

�c� To �nd a basis for N�A� we solve AX � � and equivalently BX � ��From B we see that the solution is

x� � ��x� � �x� � x� x�

x� � ��x� � �x� � x� x�

x� � �

x� � � x� � �x��

where x� and x� are arbitrary elements of Z�� Hence

X � x�

���������

�����

��������� x�

���������

������

����������

so � � �� �� �� �� ��t and ��� �� �� �� �� ��t form a basis for R�A��

Page 240: Elementary Linear Algebra

�� Let F � f�� �� a� bg be a �eld and let A be the following matrix over F �

A �

��� � a b a

a b b �� � � a

��� �

In Section ��� problem �� we found that A had reduced row�echelon form

B �

��� � � � �

� � � b

� � � �

��� �

From B we read o� the following�

�a� The rows of B form a basis for R�A�� �Consequently the rows of Aalso form a basis for R�A��

�b� The �rst three columns of A form a basis for C�A��

�c� To �nd a basis for N�A� we solve AX � � and equivalently BX � ��From B we see that the solution is

x� � �

x� � �bx� � bx�

x� � �x� � x��

where x� is an arbitrary element of F � Hence

X � x�

�����

�b

��

����� �

so ��� b� �� ��t is a basis for N�A��

�� Suppose that X�� � � � � Xm form a basis for a subspace S� We have toprove that

X�� X� X�� � � � � X� � � �Xm

also form a basis for S�First we prove the independence of the family� Suppose

x�X� x��X� X�� � � � xm�X� � � �Xm� � ��

Page 241: Elementary Linear Algebra

Then�x� x� � � � xm�X� � � � xmXm � ��

Then the linear independence of X�� � � � � Xm gives

x� x� � � � xm � �� � � � � xm � ��

form which we deduce that x� � �� � � � � xm � ��Secondly we have to prove that every vector of S is expressible as a linear

combination of X�� X� X�� � � � � X� � � �Xm� Suppose X � S� Then

X � a�X� � � � amXm�

We have to �nd x�� � � � � xm such that

X � x�X� x��X� X�� � � � xm�X� � � �Xm�

� �x� x� � � � xm�X� � � � xmXm�

Then

a�X� � � � amXm � �x� x� � � � xm�X� � � � xmXm�

So if we can solve the system

x� x� � � � xm � a�� � � � � xm � am�

we are �nished� Clearly these equations have the unique solution

x� � a� � a�� � � � � xm�� � am � am��� xm � am�

��� Let A �

�a b c

� � �

�� If �a� b� c� is a multiple of ��� �� �� �that is

a � b � c� then rankA � �� For if

�a� b� c� � t��� �� ���

then

R�A� � h�a� b� c�� ��� �� ��i � ht��� �� ��� ��� �� ��i � h��� �� ��i�

so ��� �� �� is a basis for R�A��However if �a� b� c� is not a multiple of ��� �� �� �that is at least two

of a� b� c are distinct� then the left�to�right test shows that �a� b� c� and

Page 242: Elementary Linear Algebra

��� �� �� are linearly independent and hence form a basis for R�A�� Conse�quently rankA � � in this case�

��� Let S be a subspace of Fn with dimS � m� Also suppose thatX�� � � � � Xm are vectors in S such that S � hX�� � � � � Xmi� We have toprove that X�� � � � � Xm form a basis for S� in other words we must provethat X�� � � � � Xm are linearly independent�

However if X�� � � � � Xm were linearly dependent then one of these vec�tors would be a linear combination of the remaining vectors� ConsequentlyS would be spanned by m � � vectors� But there exist a family of m lin�early independent vectors in S� Then by Theorem � �� we would have thecontradiction m � m� ��

��� Let �x� y� z�t � S� Then x �y z � �� Hence x � ��y � z and

��� x

y

z

��� �

��� ��y � z

y

z

��� � y

��� ��

��

��� z

��� �

��

��� �

Hence ���� �� ��t and �� � �� ��t form a basis for S�Next ���� ����� ��� � � so ���� ��� ��t � S�To �nd a basis for S which includes ���� ��� ��t we note that ���� �� ��t

is not a multiple of ���� ��� ��t� Hence we have found a linearly independentfamily of two vectors in S a subspace of dimension equal to �� Consequentlythese two vectors form a basis for S�

� � Without loss of generality suppose that X� � X�� Then we have thenon�trivial dependency relation�

�X� ����X� �X� � � � �Xm � ��

��� �a� Suppose that Xm�� is a linear combination of X�� � � � � Xm� Then

hX�� � � � � Xm� Xm��i � hX�� � � � � Xmi

and hencedim hX�� � � � � Xm� Xm��i � dim hX�� � � � � Xmi�

�b� Suppose that Xm�� is not a linear combination of X�� � � � � Xm� If notall of X�� � � � � Xm are zero there will be a subfamily Xc� � � � � � Xcr which isa basis for hX�� � � � � Xmi�

Page 243: Elementary Linear Algebra

Then as Xm�� is not a linear combination of Xc� � � � � � Xcr it follows thatXc� � � � � � Xcr � Xm�� are linearly independent� Also

hX�� � � � � Xm� Xm��i � hXc� � � � � � Xcr � Xm��i�

Consequently

dim hX�� � � � � Xm� Xm��i � r � � dim hX�� � � � � Xmi ��

Our result can be rephrased in a form suitable for the second part of theproblem�

dim hX�� � � � � Xm� Xm��i � dim hX�� � � � � Xmi

if and only if Xm�� is a linear combination of X�� � � � � Xm�

If X � �x�� � � � � xn�t then AX � B is equivalent to

B � x�A�� � � � xnA�n�

So AX � B is soluble for X if and only if B is a linear combination of thecolumns of A that is B � C�A�� However by the �rst part of this questionB � C�A� if and only if dimC��AjB�� � dimC�A� that is rank �AjB� �rankA�

��� Let a�� � � � � an be elements of F not all zero� Let S denote the set ofvectors �x�� � � � � xn�t where x�� � � � � xn satisfy

a�x� � � � anxn � ��

Then S � N�A� where A is the row matrix �a�� � � � � an�� Now rankA � �as A �� �� So by the �rank nullity� theorem noting that the number ofcolumns of A equals n we have

dimN�A� � nullity �A� � n� rankA � n� ��

��� �a� �Proof of Lemma ����� Suppose that each of X�� � � � � Xr is a linearcombination of Y�� � � � � Ys� Then

Xi �sX

j��

aijYj � �� � i � r��

��

Page 244: Elementary Linear Algebra

Now let X �Pr

i�� xiXi be a linear combination of X�� � � � � Xr� Then

X � x��a��Y� � � � a�sYs�

� � �

xr�ar�Y� � � � arsYs�

� y�Y� � � � ysYs�

where yj � a�jx�� � �arjxr� Hence X is a linear combination of Y�� � � � � Ys�Another way of stating Lemma ���� is

hX�� � � � � Xri � hY�� � � � � Ysi� ���

if each of X�� � � � � Xr is a linear combination of Y�� � � � � Ys�

�b� �Proof of Theorem ����� Suppose that each of X�� � � � � Xr is a linearcombination of Y�� � � � � Ys and that each of Y�� � � � � Ys is a linear combinationof X�� � � � � Xr� Then by �a� equation ��� above

hX�� � � � � Xri � hY�� � � � � Ysi

andhY�� � � � � Ysi � hX�� � � � � Xri�

HencehX�� � � � � Xri � hY�� � � � � Ysi�

�c� �Proof of Corollary ����� Suppose that each of Z�� � � � � Zt is a linearcombination of X�� � � � � Xr� Then each of X�� � � � � Xr� Z�� � � � � Zt is a linearcombination of X�� � � � � Xr�

Also each ofX�� � � � � Xr is a linear combination ofX�� � � � � Xr� Z�� � � � � Ztso by Theorem ����

hX�� � � � � Xr� Z�� � � � � Zti � hX�� � � � � Xri�

�d� �Proof of Theorem � ��� Let Y�� � � � � Ys be vectors in hX�� � � � � Xriand assume that s � r� We have to prove that Y�� � � � � Ys are linearlydependent� So we consider the equation

x�Y� � � � xsYs � ��

��

Page 245: Elementary Linear Algebra

Now Yi �Pr

j�� aijXj for � � i � s� Hence

x�Y� � � � xsYs � x��a��X� � � � a�rXr�

� � �

xr�as�X� � � � asrXr��

� y�X� � � � yrXr� ���

where yj � a�jx� � � � asjxs� However the homogeneous system

y� � �� � � � � yr � �

has a non�trivial solution x�� � � � � xs as s � r and from ��� this results in anon�trivial solution of the equation

x�Y� � � � xsYs � ��

Hence Y�� � � � � Ys are linearly dependent�

��� Let R and S be subspaces of Fn with R � S� We �rst prove

dimR � dim S�

LetX�� � � � � Xr be a basis forR� Now by Theorem ���� becauseX�� � � � � Xr

form a linearly independent family lying in S this family can be extendedto a basis X�� � � � � Xr� � � � � Xs for S� Then

dimS � s � r � dimR�

Next suppose that dimR � dimS� Let X�� � � � � Xr be a basis for R� Thenbecause X�� � � � � Xr form a linearly independent family in S and S is a sub�space whose dimension is r it follows from Theorem ��� that X�� � � � � Xr

form a basis for S� Then

S � hX�� � � � � Xri � R�

��� Suppose that R and S are subspaces of Fn with the property that R�Sis also a subspace of Fn� We have to prove that R � S or S � R� We argueby contradiction� Suppose that R �� S and S �� R� Then there exist vectorsu and v such that

u � R and v �� S� v � S and v �� R�

��

Page 246: Elementary Linear Algebra

Consider the vector u v� As we are assuming R�S is a subspace R�S isclosed under addition� Hence u v � R � S and so u v � R or u v � S�However if u v � R then v � �u v� � u � R which is a contradiction�similarly if u v � S�

Hence we have derived a contradiction on the asumption that R �� S andS �� R� Consequently at least one of these must be false� In other wordsR � S or S � R�

��� Let X�� � � � � Xr be a basis for S��i� First let

Y� � a��X� � � � a�rXr

��� ���

Yr � ar�X� � � � arrXr�

where A � �aij � is non�singular� Then the above system of equations canbe solved for X�� � � � � Xr in terms of Y�� � � � � Yr� Consequently by Theorem ����

hY�� � � � � Yri � hX�� � � � � Xri � S�

It follows from problem �� that Y�� � � � � Yr is a basis for S��ii� We show that all bases for S are given by equations �� So suppose

that Y�� � � � � Yr forms a basis for S� Then because X�� � � � � Xr form a basisfor S we can express Y�� � � � � Yr in terms of X�� � � � � Xr as in � for somematrix A � �aij �� We show A is non�singular by demonstrating that thelinear independence of Y�� � � � � Yr implies that the rows of A are linearlyindependent�

So assume

x��a��� � � � � a�r� � � � xr�ar�� � � � � arr� � ��� � � � � ���

Then on equating components we have

a��x� � � � ar�xr � ����

a�rx� � � � arrxr � ��

Hence

x�Y� � � � xrYr � x��a��X� � � � a�rXr� � � � xr�ar�X� � � � arrXr�

� �a��x� � � � ar�xr�X� � � � �a�rx� � � � arrxr�Xr

� �X� � � � �Xr � ��

Page 247: Elementary Linear Algebra

Then the linear independence of Y�� � � � � Yr implies x� � �� � � � � xr � ���We mention that the last argument is reversible and provides an alter�

native proof of part �i���

��

Page 248: Elementary Linear Algebra

P�

P�

P�

O������������

����

����

������������

����

����

��������

Section ���

�� We �rst prove that the area of a triangle P�P�P�� where the pointsare in anti�clockwise orientation� is given by the formula

������ x� x�y� y�

����������� x� x�y� y�

����������� x� x�y� y�

�������

Referring to the above diagram� we have

AreaP�P�P� � AreaOP�P� �AreaOP�P� � AreaOP�P�

��

����� x� x�y� y�

������ �

����� x� x�y� y�

������ �

����� x� x�y� y�

����� �which gives the desired formula�

We now turn to the area of a quadrilateral� One possible con�gurationoccurs when the quadrilateral is convex as in �gure �a below� The interiordiagonal breaks the quadrilateral into two triangles P�P�P� and P�P�P��Then

AreaP�P�P�P� � AreaP�P�P� � AreaP�P�P�

Page 249: Elementary Linear Algebra

��������

����������

����������HH

HH

P�

P�

P�

P�

�a

����������

����

��

LLLLLLLLLLL

�����

P�

P�

P�

P��b

��

������ x� x�y� y�

����������� x� x�y� y�

����������� x� x�y� y�

������

��

������ x� x�y� y�

����������� x� x�y� y�

����������� x� x�y� y�

������

��

������ x� x�y� y�

����������� x� x�y� y�

����������� x� x�y� y�

����������� x� x�y� y�

�������

after cancellation�Another possible con�guration for the quadrilateral occurs when it is not

convex� as in �gure �b� The interior diagonal P�P� then gives two trianglesP�P�P� and P�P�P� and we can proceed similarly as before�

��

� �

�������a� x b� y c� zx� u y � v z � w

u� a v � b w � c

������� ��������

a b cx� u y � v z � w

u� a v � b w � c

���������������

x y zx� u y � v z � w

u� a v � b w � c

������� �Now�������

a b cx� u y � v z � w

u� a v � b w � c

������� ��������

a b cx y z

u� a v � b w� c

���������������

a b cu v w

u� a v � b w � c

�������

�������a b cx y z

u v w

���������������a b cx y z

a b c

���������������a b cu v w

u v w

��������������a b cu v w

a b c

�������

Page 250: Elementary Linear Algebra

�������a b cx y z

u v w

������� �

Similarly�������x y z

x � u y � v z � wu � a v � b w � c

������� ��������x y z

u v wa b c

������� � �

�������x y z

a b cu v w

������� ��������a b c

x y zu v w

������� �

Hence � � �

�������a b c

x y zu v w

��������

��

�������n� �n� �� �n� ��

�n� �� �n� �� �n� ��

�n� �� �n� �� �n� �

�������C� � C� � C�

C� � C� � C�

�������n� �n� � �n� �

�n� �� �n� � �n� ��n� �� �n� � �n� �

�������C� � C� � C�

�������n� �n� � �

�n � �� �n� � ��n � �� �n� � �

�������R� � R� � R�

R� � R� � R�

�������n� �n� � �

�n� � � ��n� � � �

������� � ���

� �a �������� �� ������ �� ���� ��� ��

������� ��������

� ��� ������ ��� ���� ��� ��

������� � ���

�������� � ������ � ���� � ��

�������� ���

�������� � ���� � � ��

���� � ��

������� � ������

����� � � �� ���� ��

����� � ���������

�b ���������

� � � �� � � �� � �� � � �� ��

����������

���������

� � � � � � ��� ��� ��� ���� �� �� ���

����������

Page 251: Elementary Linear Algebra

�������� � ��

��� ��� ����� �� ���

������� � ���

�������� � ��� � �

�� �� ���

�������� ���

�������� �� � � �

�� �� ���

������� � ������

����� �� �� ���

����� � ����

�� detA �

�������� � ��� � � � ��

������� ��������� � �� � ��� � �

������� ������ � ��� �

����� � ����

Hence A is non�singular and

A�� ��

���adjA �

���

��� C�� C�� C��

C�� C�� C��

C�� C�� C��

��� �

���

��� ��� � �

�� � ���� �� �

��� �

� �i��������a �b b� c�b �a a� c

a� b a� b b

�������R� � R� � R�

��������a� �b �b� �a b� a

�b �a a� c

a� b a� b b

�������

� �a�b

�������� � ��b �a a� c

a� b a� b b

�������C� � C� � C�

��a�b

�������� � �

��b� a �a a� c

� a� b b

�������� ��a� b�a� b

����� � �a� b b

����� � ���a� b�a� b��

�ii �������b� c b c

c c� a ab a a� b

�������C� � C� � C�

�������c b c

�a c� a ab� a a a � b

�������C� � C� � C�

�������c b ��a c� a �ab� a a �a

������� � �a

�������c b ��a c� a �b� a a �

�������

Page 252: Elementary Linear Algebra

R� � R� � R�

��a

�������c b �

�a c� a �b �c �

������� � ��a

����� c bb �c

����� � �a�c� � b��

�� Suppose that the curve y � ax� � bx � c passes through the points�x�� y�� �x�� y�� �x�� y�� where xi �� xj if i �� j� Then

ax�� � bx� � c � y�

ax�� � bx� � c � y�

ax�� � bx� � c � y��

The coe�cient determinant is essentially a Vandermonde determinant��������x�� x� �x�� x� �x�� x� �

������� ��������x�� x�� x��x� x� x�� � �

������� � �

�������� � �x� x� x�x�� x�� x��

������� � ��x��x��x��x��x��x��

Hence the coe�cient determinant is non�zero and by Cramer�s rule� thereis a unique solution for a� b� c�

�� Let � � detA �

�������� � ��� � k� k �

�������� Then

� �C� � C� � C�

C� � C� � C�

�������� � �� � k � �� k � �

������� ������ � k � �k � �

������ � �k� ��k� � � ��k� � k � � ��k � ��k � ��

Hence detA � � if and only if k � �� or k � ��Consequently if k �� �� and k �� �� then detA �� � and the given system

x� y � z � �

�x� �y � kz � �

x� ky � �z � �

has a unique solution� We consider the cases k � �� and k � � separately�k � �� �

AM �

��� � � �� �� � �� �� �� � �

��� R� � R� � �R�

R� � R� �R�

��� � � �� �� � �� �� � �

���

Page 253: Elementary Linear Algebra

R� � R� � R�

��� � � �� �

� � �� �� � � �

��� �

from which we read o� inconsistency�k � � �

AM �

��� � � �� �� � � �� � � �

��� R� � R� � �R�

R� � R� �R�

��� � � �� �� � �� � �

���

R� � R� �R�

��� � � �� �� � �� � � �

��� �

We read o� the complete solution x � �z� y � �� z� where z is arbitrary�Finally we have to determine the solution for which x�� y�� z� is least�

x� � y� � z� � ��z� � ��� z� � z� � �z� � �z � �

� ��z� �

��z �

� � �

�z �

��

��

��

��

��

� �

�z �

��

��

��

���

��

We see that the least value of x��y��z� is �� ��

���� ��

��and this occurs when

z � ����� with corresponding values x � ����� and y � �� � �

��� ������

�� Let � �

��� � �� ba � �� � �

������� be the coe�cient determinant of the given system�

Then expanding along column � gives

� � �

����� a �� �

������ �

����� � ba �

����� � ���� ���� ab

� �ab� � � ��ab� ���

Hence � � � if and only if ab � ��� Hence if ab �� ��� the given system hasa unique solution�

��

Page 254: Elementary Linear Algebra

If ab � �� we must argue with care�

AM �

��� � �� b �a � � �� � � �

����

��� � �� b �� �a �� ab �� �a� �� ��b ��

���

��� � �� b �

� � ��b��

��

� �a �� ab �� �a

����

��� � �� b �

� � ��b��

��

� � ���ab�

���a�

���

��� � �� b �

� � ��b��

��

� � � ���a�

��� � B�

Hence if � �a �� �� i�e� a �� �� the system has no solution�If a � � �and hence b � � then

B �

��� � �� �� � ��

��

� � � �

����

��� � � ���� ���� � ��

��

� � � �

��� �

Consequently the complete solution of the system is x � �

�� �

�z� y � ��

�� �

�z�

where z is arbitrary� Hence there are in�nitely many solutions�

���

� �

���������

� � � �� � � � � �t� � � � t t

���������

R� � R� � �R�

R� � R� � �R�

R� � R� �R�

���������

� � � �� � � �� � � �t� � � �� t t � �

����������

�������� � �� � �t� � �� t t � �

�������R� � R� � �R�

�������� � �� � �t� �� �� t t� �

��������

����� � �t� ��� t t� �

����� � �t� �

����� � �t� ��� �

����� � �t� ���t� ��

Hence � � � if and only if t � � or t � �

�� Consequently the given matrix

B is non�singular if and only if t �� � and t �� �

��

��� Let A be a �� � matrix with detA �� �� Then

��

Page 255: Elementary Linear Algebra

�i

A adjA � �detAI� ��

�detA det � adjA � det �detA � I� � �detA��

Hence� as detA �� �� dividing out by detA in the last equation gives

det � adjA � �detA��

�ii � Also from equation ���

detAA

adjA � I��

so adjA is non�singular and

� adjA�� ��

detAA�

FinallyA�� adj �A�� � �detA��I�

and multiplying both sides of the last equation by A gives

adj �A�� � A�detA��I� ��

detAA�

��� Let A be a real �� � matrix satisfying AtA � I�� Then

�i At�A� I� � AtA� At � I� � At

� ��At � I� � ��At � I t� � ��A� I�t�

Taking determinants of both sides then gives

detAt det �A� I� � det ���A� I�t

detA det �A� I� � ���� det �A� I�t

� � det �A� I� ���

�ii Also detAAt � det I�� so

detAt detA � � � �detA��

��

Page 256: Elementary Linear Algebra

Hence detA � ����iii Suppose that detA � �� Then equation �� gives

det �A� I� � � det �A� I��

so �� � � det �A� I� � � and hence det �A� I� � ��

��� Suppose that column � is a linear combination of the remaining columns�

A�� � x�A�� � � � �� xnA�n�

Then

detA �

����������

x�a�� � � � �� xna�n a�� � � � a�nx�a�� � � � �� xna�n a�� � � � a�n

������

������

x�an� � � � �� xnann an� � � � ann

�����������

Now detA is unchanged in value if we perform the operation

C� � C� � x�C� � � � � � xnCn �

detA �

����������

� a�� � � � a�n� a�� � � � a�n���

������

���� an� � � � ann

����������� ��

Conversely� suppose that detA � �� Then the homogeneous system AX � �has a non�trivial solution X � �x�� � � � � xn�

t� So

x�A�� � � � �� xnA�n � ��

Suppose for example that x� �� �� Then

A�� �

�x�x�

� � � ��

�xnx�

A�n

and the �rst column of A is a linear combination of the remaining columns�

�� Consider the system

��x� �y � z � �x� �y � z �

��x� y � z � ��

��

Page 257: Elementary Linear Algebra

Let � �

��������� � ��� � ��

�� �� �

������� ��������� � ��� � ��� � ��

������� � �

����� � ��� ��

����� � �� �� ��

Hence the system has a unique solution which can be calculated usingCramer�s rule�

x ���

�� y �

��

�� z �

��

��

where

�� �

�������� � �� � ��

�� �� �

������� � ��

�� �

��������� � ��� ��

�� �� �

������� � � �

�� �

��������� � �� �

�� �� ��

������� � ���

Hence x � ��

��� �� y � ��

��� �� z � ��

��� �

��� In Remark ���� take A � In� Then we deduce

�a detEij � ���

�b detEi�t � t�

�c detEij�t � ��

Now suppose that B is a non�singular n� n matrix� Then we know that Bis a product of elementary row matrices�

B � E� � � �Em�

Consequently we have to prove that

detE� � � �EmA � detE� � � �Em detA�

We prove this by induction on m�First the case m � �� We have to prove detE�A � detE� detA if E� is

an elementary row matrix� This follows form Remark ����

Page 258: Elementary Linear Algebra

�a detEijA � � detA � detEij detA�

�b detEi�tA � t detA � detEi�t detA�

�c detEij�tA � detA � detEij�t detA�

Let m � � and assume the proposition holds for products of m elementaryrow matrices� Then

detE� � � �EmEm��A � det �E� � � �Em�Em��A

� det �E� � � �Em det �Em��A

� det �E� � � �Em detEm�� detA

� det ��E� � � �EmEm�� detA

and the induction goes through�Hence detBA � detB detA if B is non�singular�If B is singular� problem � � Chapter ��� tells us thatBA is also singlular�

However singular matrices have zero determinant� so

detB � � detBA � ��

so the equation detBA � detB detA holds trivially in this case�

� � ���������

a� b� c a � b a aa� b a� b� c a a

a a a� b� c a� ba a a� b a � b� c

���������R� � R� � R�

R� � R� � R�

R� � R� � R�

���������

c �c � �b b� c �b� c �b

� � c �ca a a� b a � b� c

���������C� � C� � C�

���������

c � � �b �b� c �b� c �b� � c �c

a �a a� b a� b� c

���������� c

��������b� c �b� c �b� c �c

�a a� b a� b� c

�������C� � C� � C�

�c

��������b� c �b� c ��b� c� c ��a a� b �a� �b� c

������� � c������ �b� c ��b� c

�a �a� �b� c

�����

��

Page 259: Elementary Linear Algebra

� c���b� c

����� � ���a �a� �b� c

����� � c���b� c�a� �b� c�

��� Let � �

���������

� � u� u� u� u�u� � � u� u� u�u� u� � � u� u�u� u� u� � � u�

���������� Then using the operation

R� � R� �R� �R� � R�

we have

� �

���������

t t t tu� � � u� u� u�u� u� � � u� u�u� u� u� � � u�

����������where t � � � u� � u� � u� � u�

� �� � u� � u� � u� � u�

���������

� � � �u� � � u� u� u�u� u� � � u� u�u� u� u� � � u�

���������The last determinant equals

C� � C� � C�

C� � C� � C�

C� � C� � C�

���������

� � � �u� � � �u� � � �u� � � �

���������� ��

��� Suppose that At � �A� that A �Mn�n�F � where n is odd� Then

detAt � det��A

detA � ���n detA � � detA�

Hence �� � � detA � � and consequently detA � � if � � � �� � in F �

������������

� � � �r � � �r r � �r r r �

����������

C� � C� � C�

C� � C� � C�

C� � C� � C�

���������

� � � �r �� r � �r � �� r �r � � �� r

���������� ��� r��

Page 260: Elementary Linear Algebra

��� �������� a� � bc a�

� b� � ca b�

� c� � ab c�

�������R� � R� � R�

R� � R� � R�

�������� a� � bc a�

� b� � ca� a� � bc b� � a�

� c� � ab� a� � bc c� � a�

�������

����� b� � ca� a� � bc b� � a�

c� � ab� a� � bc c� � a�

������

����� �b� a�b� a � c�b� a �b� a�b� a�b� � a��c� a�c� a � b�c� a �c� a�c� a�c� � a�

������

����� �b� a�b� a� c �b� a�b� a�b�� a��c� a�c� a� b �c� a�c� a�c� � a�

������ �b� a�c� a

����� b� a� c �b� a�b�� a�c� a� b �c� a�c� � a�

������ �b� a�c� a�a� b� c

����� � �b� a�b� � a�� �c� a�c� � a�

����� �Finally����� � �b� a�b� � a�

� �c� a�c� � a�

����� � �c� � ac� � ca� � a�� �b� � ab� � ba� � a�

� �c� � b� � a�c� � b� � a��c� b

� �c� b�c� � cb� b� � a�c� b � a�

� �c� b�c� � cb� b� � ac� ab� a��

��

Page 261: Elementary Linear Algebra

Section ���

��

�i� ��� � i����� �i� � �������� �i� � i���� �i�

� f������� ������i�g� i����� i��i�

� ���� � i� � ���i� �� � �� � �i�

�ii�� � �i

�� �i�

�� � �i��� � �i�

��� �i��� � �i�

���� � �i� � �� � �i���i�

�� � ��

��� � ��i

�������

���

��i�

�iii��� � �i��

�� i�

� � �i� ��i��

�� i

�� � �i� �

�� i��� � �i

�� i

���� � �i���� i�

���� � i

�� ��

��

�i�

�� �i�

iz � ��� �i�z � �z � �i � z�i� �� �i� �� � �i

�� z���� �i� � �i� z ���i� � �i

���i��� �i�

� � ���� � �i

����� i

���

�ii� The coe�cient determinant is����� � � i �� i

� � �i � � i

����� � �� � i��� � i�� ��� i��� � �i� � �� � i �� �

Hence Cramer�s rule applies� there is a unique solution given by

z �

����� ��i �� i

� � �i � � i

������� � i

���� ��i

�� � i� �� � �i

w �

����� � � i ��i� � �i � � �i

������� � i

�� � �i

�� � i�

��� i

��

Page 262: Elementary Linear Algebra

��

� � �� � i� � � � �� �� � i��� ��� � i���� � �

�� � i�� �

��� � i���� � �

i� �i

n�� � i����� �

o�

Now �� � i�� � �i� Hence

�� � i���� � ��i��� � ���i�� � ��������� � �����Hence �i ��� � i����� �

�� �i����� � �� � ���� � ��i�

�� �i� Let z� � � � i and write z�x�iy� where x and y are real� Then

z� � x� � y� � �xyi � � � i�

so x� � y� � � and �xy � �� Hence

y � ���x� x� ����x

��

� � �

so x� � x� � � � � This is a quadratic in x�� Hence x� � � or �� andconsequently x� � �� Hence x � �� y � �� or x � �� and y � �� Hencez � �� �i or z � �� � �i�

�ii� z� � ��� i�z� �� �i � has the solutions z � ��� i� d���� where d isany complex number satisfying

d� � �� � i�� � ��� � �i� � � � i�

Hence by part �i� we can take d � �� �i� Consequently

z �� � i� ��� �i�

�� �� i or � � �i�

�i� The number lies in the �rst quadrant ofthe complex plane�

j� � ij �p�� � �� �

p���

Also Arg �� � i� � �� where tan� � ���and � � � ���� Hence � � tan��������

��

x

y

� � i������

��

Page 263: Elementary Linear Algebra

�ii� The number lies in the third quadrant ofthe complex plane������� � i

���� � j��� ij�

��

q����� � ����� � �

p� � � �

p�

��

Also Arg ����i� � � ��� �� where tan� ����

�� � ��� and � � � ���� Hence � �

tan��������

��

x

y

���i�

� ����

�iii� The number lies in the second quadrant ofthe complex plane�

j � � � �ij �q����� � �� �

p��

Also Arg �����i� � ���� where tan� �� and � � � ���� Hence � � tan����

��

x

y�� � �i

� AAAAAAK

�iv� The number lies in the second quadrant ofthe complex plane�������� � i

p�

����� � j � � � ip�j

��

q����� � �

p��� �

p� � � � ��

Also Arg ���� �p�� i� � � � �� where

tan� �p�� ��� �

p� and � � � ����

Hence � � ����

��

x

y

��� �

p�� i

� JJJJJJJ�

� �i� Let z � �� � i��� �p�i��

p�� i�� Then

jzj � j� � ijj��p�ijj

p�� ij

�p�� � ��

q�� � �

p���

q�p��� � �����

�p�p�p� � �

p��

Page 264: Elementary Linear Algebra

Arg z � Arg �� � i� � Arg �� �p�� � Arg �

p�� i� �mod ���

� �

���

�� �

� �

���

Hence Arg z � ��� and the polar decomposition of z is

z � �p�

�cos

��

��� i sin

��

��

��

�ii� Let z � ���i����ip��

�p��i�

� Then

jzj � j�� � i�j�j��� ip��j�

j�p� � i�j� �

�p���

��

��� ����

Arg z � Arg �� � i�� �Arg ���p�i�� � Arg �

p� � i�� �mod ���

� �Arg �� � i� � �Arg ���p�i�� �Arg�

p� � i�

� ��

�� �

����

�� �

� ����

��� ���

���

Hence Arg z � ����� and the polar decomposition of z is

z � ����cos

���

��� i sin

���

��

��

�� �i� Let z � ��cos �� � i sin �

� � and w � ��cos �� � i sin �

� �� �Both of thesenumbers are already in polar form��

�a� zw � �cos ��� ��� � � i sin ��� �

�� ��

� �cos ���� � i sin ��

�� ��

�b� zw � �

��cos ��� � �

� � � i sin ��� � �� ��

� ���cos

��� � i sin �

����

�c� wz � �

��cos ��� � �

� � � i sin ��� � �� ��

� ���cos �

���� � � i sin ����� ���

�d� z�

w� � ��

�� �cos ���� � ��

� � � i sin ���� � ��� ��

� ��� �cos

����� � i sin ���

�� ��

Page 265: Elementary Linear Algebra

�a� �� � i�� � �i� so

�� � i��� � ��i�� � ��i� � ��i��� � ������ � ���

�b� ���ip��� � �i� so

��� ip

����

���� ip

�����

� ��i��� � ��i�

����i �

i� �i�

� �i� To solve the equation z� � � �p�i� we write � �

p�i in modulus�

argument form�

� �p�i � ��cos

�� i sin

���

Then the solutions are

zk �p�

�cos

� �� � �k�

�� i sin

� �� � �k�

��� k � � ��

Now k � gives the solution

z� �p��cos

� i sin

� �

p�

�p�

��

i

��

p�p��

ip��

Clearly z� � �z���ii� To solve the equation z� � i� we write i in modulus�argument form�

i � cos�

�� i sin

��

Then the solutions are

zk � cos

� �� � �k�

�� i sin

� �� � �k�

�� k � � �� �� ��

Now cos��

���k�

�� cos

��� �

k��

�� so

zk � cos

��

�k�

�� sin

��

�k�

�cos

�� i sin

�k

�cos�

� i sin

� ik�cos�

� i sin

��

Page 266: Elementary Linear Algebra

Geometrically� the solutions lie equi�spaced on the unit circle at arguments

��

��

��

��

��

� � �

��

��

� �

��

���

Also z� � �z� and z� � �z���iii� To solve the equation z� � � i� we rewrite the equation as

�z

��i��

� ��

Then �z

��i�� ��

�� �p�i

�� or

��� p�i�

Hence z � ��i� p� � i or �p� � i�Geometrically� the solutions lie equi�spaced on the circle jzj � �� at

arguments�

��

��

��

��

��

� �

��

��

��

��

�iv� To solve z� � �� �i� we write �� �i in modulus�argument form�

�� �i � �����cos

���

� i sin���

��

Hence the solutions are

zk � ���� cos

���� � �k�

�� i sin

���� � �k�

�� k � � �� �� ��

We see the solutions can also be written as

zk � ����ik�cos

���

� i sin���

� ����ik�cos

�� i sin

��

Geometrically� the solutions lie equi�spaced on the circle jzj � ����� at ar�guments

���

����

��

��

��

�����

� ��

��

���

�����

� ��

��

���

��

Also z� � �z� and z� � �z��

Page 267: Elementary Linear Algebra

�� � � � i �� � �i �

� � i �� � i �� � �i �� � i � � i

� � R� � R� �R�

R� � R� �R�

� � i �� � i �� � i �i �� i

� �

R� � R� � �� � i�R�

R� � R� � iR�

� � i � �i

� � R� � iR�

� � i �

� �

R� � R� �R�

� � i �

� � �

The last matrix is in reduced row�echelon form�

�� �i� Let p � l � im and z � x� iy� Then

pz � pz � �l� im��x� iy� � �l� im��x� iy�

� �lx� liy � imx�my� � �lx� liy � imx�my�

� ��lx�my��

Hence pz � pz � �n� lx�my � n�

�ii� Let w be the complex number which results from re�ecting the com�plex number z in the line lx�my � n� Then because p is perpendicular tothe given line� we have

w � z � tp� t � R� �a�

Also the midpoint w�z� of the segment joining w and z lies on the given line�

so

p

�w� z

�� p

�w � z

�� n�

p

�w� z

�� p

�w � z

�� n� �b�

Taking conjugates of equation �a� gives

w � z � tp� �c�

Then substituting in �b�� using �a� and �c�� gives

p

��w� tp

�� p

��z � tp

�� n

Page 268: Elementary Linear Algebra

and hencepw � pz � n�

�iii� Let p � b� a and n � jbj�� jaj�� Then

jz � aj � jz � bj � jz � aj� � jz � bj�� �z � a��z � a� � �z � b��z � b�

� �z � a��z � a� � �z � b��z � b�

� zz � az � za� aa � zz � bz � zb� bb

� �b� a�z � �b� a�z � jbj�� jaj�� pz � pz � n�

Suppose z lies on the circle���z�az�b

��� and let w be the re�ection of z in the

line pz � pz � n� Then by part �ii�

pw � pz � n�

Taking conjugates gives pw � pz � n and hence

z �n � pw

p�a�

Substituting for z in the circle equation� using �a� gives

� �

������n�pw

p � a

n�pwp � b

������ �����n � pw � pa

n � pw � pb

���� � �b�

However

n � pa � jbj� � jaj� � �b� a�a

� bb� aa� ba� aa

� b�b� a� � bp�

Similarly n� pb � ap� Consequently �b� simpli�es to

� �

����� bp� pw

ap� pw

����� ������ b� w

a� w

����� �����w � b

w � a

���� �which gives

���w�aw�b

��� � �� �

Page 269: Elementary Linear Algebra

��� Let a and b be distinct complex numbers and � � � ���i� When z� lies on the circular arc shown� it subtends a constant angle

�� This angle is given by Arg �z� � a��Arg �z� � b�� However

Arg

�z� � a

z� � b

�� Arg �z� � a��Arg �z� � b� � �k�

� �� �k��

It follows that k � � as � � � � and �� � Arg � � �� Hence

Arg

�z� � a

z� � b

�� ��

Similarly if z� lies on the circular arc shown� then

Arg

�z� � a

z� � b

�� �� � ��� � �� � �� ��

Replacing � by � � �� we deduce that if z� lies on the circular arc shown�then

Arg

�z� � a

z� � b

�� � � ��

while if z� lies on the circular arc shown� then

Arg

�z� � a

z� � b

�� ���

The straight line through a and b has the equation

z � ��� t�a� tb�

Page 270: Elementary Linear Algebra

where t is real� Then � t � � describes the segment ab� Also

z � a

z � b�

t

t� ��

Hence z�az�b is real and negative if z is on the segment a� but is real and

positive if z is on the remaining part of the line� with corresponding values

Arg

�z � a

z � b

�� �� �

respectively�

�ii� Case �a� Suppose z�� z� and z� are not collinear� Then these pointsdetermine a circle� Now z� and z� partition this circle into two arcs� If z�and z� lie on the same arc� then

Arg

�z� � z�z� � z�

�� Arg

�z� � z�z� � z�

��

whereas if z� and z� lie on opposite arcs� then

Arg

�z� � z�z� � z�

�� �

and

Arg

�z� � z�z� � z�

�� �� ��

Hence in both cases

Arg

�z� � z�z� � z�

�z� � z�z� � z�

�� Arg

�z� � z�z� � z�

��Arg

�z� � z�z� � z�

��mod ���

� or ��

In other words� the cross�ratio

z� � z�z� � z�

�z� � z�z� � z�

is real��b� If z�� z� and z� are collinear� then again the cross�ratio is real�

The argument is reversible�

�iii� Assume that A� B� C� D are distinct points such that the cross�ratio

r �z� � z�z� � z�

�z� � z�z� � z�

is real� Now r cannot be or �� Then there are three cases�

Page 271: Elementary Linear Algebra

�i� � r � ��

�ii� r � �

�iii� r ��

Case �i�� Here jrj� j�� rj � �� So����z� � z�z� � z�

� z� � z�z� � z�

�����������

�z� � z�z� � z�

� z� � z�z� � z�

����� � ��

Multiplying both sides by the denominator jz� � z�jjz� � z�j gives aftersimpli�cation

jz� � z�jjz� � z�j� jz� � z�jjz� � z�j � jz� � z�jjz� � z�j�

or�a� AD �BC � AB �CD � BD �AC�

Case �ii�� Here � � jrj � j�� rj� This leads to the equation

�b� BD �AC �AD �BC� � AB � CD�

Case �iii�� Here � � j�� rj � jrj� This leads to the equation

�c� BD �AC �AB � CD � AD �BC�

Conversely if �a�� �b� or �c� hold� then we can reverse the argument to deducethat r is a complex number satisfying one of the equations

jrj� j�� rj � �� � � jrj � j�� rj� � � j�� rj � jrj�

from which we deduce that r is real�

Page 272: Elementary Linear Algebra

Section ���

�� Let A �

�� ��� �

�� Then A has characteristic equation ��� ��� � � �

or ��� ����� �� � �� Hence the eigenvalues of A are �� � � and �� � ���� � �� The corresponding eigenvectors satisfy �A� ��I��X � � or�

� ��� ��

��

���

��

or equivalently x� �y � �� Hence�xy

��

��yy

�� y

���

and we take X� �

���

��

Similarly for �� � � we nd the eigenvector X� �

���

��

Hence if P � �X�jX�� �

�� �� �

� then P is non singular and

P��AP �

�� �� �

��

Hence

A � P

�� �� �

�P��

and consequently

An � P

��n �� �n

�P��

�� �� �

� ��n �� �n

��

�� ��

�� �

��

��n�� ��n �

��� ��

�� �

��

��n�� � � ��n�� � ��n � � ��n � �

��n � �

�A�

�� �n

�I��

��

Page 273: Elementary Linear Algebra

�� Let A �

���� ������ ���

�� Then we nd that the eigenvalues are �� � � and

�� � ���� with corresponding eigenvectors

X� �

���

�and X� �

����

��

Then if P � �X�jX�� P is non singular and

P��AP �

�� �� ����

�and A � P

�� �� ����

�P���

Hence

An � P

�� �� ������n

�P��

� P

�� �� �

�P��

�� ��� �

��� �� �

��

�� �

�� �

��

�� �� �

��� �

�� �

��

�� �� �

��

���� ������ ���

��

�� The given system of di�erential equations is equivalent to �X � AX where

A �

�� ��� ��

�and X �

�x

y

��

The matrix P �

�� �� �

�is a non�singular matrix of eigenvectors corre�

sponding to eigenvalues �� � �� and �� � �� Then

P��AP �

��� �� �

��

��

Page 274: Elementary Linear Algebra

The substitution X � PY where Y � �x�� y��t gives

�Y �

��� �� �

�Y�

or equivalently �x� � ��x� and �y� � y��Hence x� � x����e

��t and y� � y����et� To determine x���� and y����

we note that�x����y����

�� P��

�x���y���

�� �

�� ��

�� �

������

��

���

��

Hence x� � �e��t and y� � �et� Consequently

x � �x� � y� � �e��t � �et and y � �x� � y� � ��e��t � �et�

�� Introducing the vector Xn �

�xnyn

� the system of recurrence relations

xn�� � �xn � yn

yn�� � �xn � �yn�

becomes Xn�� � AXn where A �

�� ��

�� �

�� Hence Xn � AnX� where

X� �

���

��

To nd An we can use the eigenvalue method� We get

An ��

��n � �n �n � �n

�n � �n �n � �n

��

Hence

Xn ��

��n � �n �n � �n

�n � �n �n � �n

� ���

��

��n � �n � ���n � �n��n � �n � ���n � �n�

��

��� �n � �n

�� �n � �n

��

���� �n � �n������ �n � �n���

��

��

Page 275: Elementary Linear Algebra

Hence xn � ����� �n � �n� and yn � �

���� �n � �n��

�� Let A �

�a b

c d

�be a real or complex matrix with distinct eigenvalues

��� �� and corresponding eigenvectors X�� X�� Also let P � �X�jX���

�a� The system of recurrence relations

xn�� � axn � byn

yn�� � cxn � dyn

has the solution�xnyn

�� An

�x�y�

��

�P

��� �� ��

�P��

�n �x�y�

� P

��n� �� �n�

�P��

�x�y�

� �X�jX��

��n� �� �n�

� ���

� �X�jX��

��n���n��

�� �n��X� � �n��X��

where ���

�� P��

�x�y�

��

�b� In matrix form the system is �X � AX whereX �

�xy

�� We substitute

X � PY where Y � �x�� y��t� Then

�X � P �Y � AX � A�PY ��

so

�Y � �P��AP �Y �

��� �� ��

� �x�y�

��

Hence �x� � ��x� and �y� � ��y�� Then

x� � x����e��t and y� � y����e

��t�

��

Page 276: Elementary Linear Algebra

But �x���y���

�� P

�x����y����

��

so �x����y����

�� P��

�x���y���

��

��

��

Consequently x���� � � and y���� � � and

�xy

�� P

�x�y�

�� �X�jX��

��e��t

�e��t

� �e��tX� � �e��tX��

�� Let A �

�a bc d

�be a real matrix with non real eigenvalues � � a� ib

and � � a�ib with corresponding eigenvectorsX � U�iV andX � U�iV where U and V are real vectors� Also let P be the real matrix dened byP � �U jV �� Finally let a � ib � rei� where r � � and � is real�

�a� As X is an eigenvector corresponding to the eigenvalue � we have AX ��X and hence

A�U � iV � � �a� ib��U � iV �

AU � iAV � aU � bV � i�bU � aV ��

Equating real and imaginary parts then gives

AU � aU � bV

AV � bU � aV�

�b�

AP � A�U jV � � �AU jAV � � �aU�bV jbU�aV � � �U jV �

�a b

�b a

�� P

�a b

�b a

��

Hence as P can be shown to be non singular

P��AP �

�a b

�b a

��

��

Page 277: Elementary Linear Algebra

�The fact that P is non singular is easily proved by showing the columns ofP are linearly independent� Assume xU � yV � � where x and y are real�Then we nd

�x� iy��U � iV � � �x� iy��U � iV � � ��

Consequently x�iy � � as U�iV and U�iV are eigenvectors correspondingto distinct eigenvalues a� ib and a� ib and are hence linearly independent�Hence x � � and y � ���

�c� The system of recurrence relations

xn�� � axn � byn

yn�� � cxn � dyn

has solution�xnyn

�� An

�x�y�

� P

�a b

�b a

�n

P���x�y�

� P

�r cos � r sin ��r sin � r cos �

�n ��

� Prn�

cos � sin �� sin � cos �

�n ���

� rn�U jV �

�cosn� sinn�� sin n� cosn�

� ���

� rn�U jV �

�� cosn� � � sin n��� sinn� � � cosn�

� rn f�� cosn� � � sinn��U � ��� sinn� � � cosn��V g

� rn f�cosn����U � �V � � �sinn����U � �V �g �

�d� The system of di�erential equations

dx

dt� ax� by

dy

dt� cx� dy

��

Page 278: Elementary Linear Algebra

is attacked using the substitution X � PY where Y � �x�� y��t� Then

�Y � �P��AP �Y�

so ��x��y�

��

�a b

�b a

� �x�y�

��

Equating components gives

�x� � ax� � by�

�y� � �bx� � ay��

Now let z � x� � iy�� Then

�z � �x� � i �y� � �ax� � by�� � i��bx� � ay��

� �a� ib��x� � iy�� � �a� ib�z�

Hence

z � z���e�a�ib�t

x� � iy� � �x���� � iy�����eat�cos bt� i sin bt��

Equating real and imaginary parts gives

x� � eat fx���� cosbt� y���� sin btg

y� � eat fy���� cosbt� x���� sin btg �

Now if we dene � and � by���

�� P��

�x���y���

��

we see that � � x���� and � � y����� Then�x

y

�� P

�x�y�

� �U jV �

�eat�� cos bt� � sin bt�eat�� cos bt� � sin bt�

� eatf�� cos bt� � sin bt�U � �� cos bt� � sin bt�V g

� eatfcos bt��U � �V � � sin bt��U � �V �g�

��

Page 279: Elementary Linear Algebra

�� �The case of repeated eigenvalues�� Let A �

�a bc d

�and suppose that

the characteristic polynomial of A ��� �a� d��� �ad� bc� has a repeatedroot �� Also assume that A �� �I��

�i�

�� � �a� d��� �ad� bc� � ��� ���

� �� � ���� ���

Hence a � d � �� and ad� bc � �� and

�a� d�� � ��ad� bc��

a� � �ad� d� � �ad� �bc�

a� � �ad� d� � �bc � ��

�a� d�� � �bc � ��

�ii� Let B �A� �I�� Then

B� � �A� �I��� � A� � ��A� ��I�

� A� � �a� d�A� �ad� bc�I��

But by problem � chapter ��� A� � �a � d�A � �ad � bc�I� � � soB� � ��

�iii� Now suppose that B �� �� Then BE� �� � or BE� �� � as BEi is thei th column of B� Hence BX� �� � where X� � E� or X� � E��

�iv� Let X� � BX� and P � �X�jX��� We prove P is non singular bydemonstrating that X� and X� are linearly independent�

Assume xX� � yX� � �� Then

xBX� � yX� � �

B�xBX� � yX�� � B� � �

xB�X� � yBX� � �

x�X� � yBX� � �

yBX� � ��

Hence y � � as BX� �� �� Hence xBX� � � and so x � ��

��

Page 280: Elementary Linear Algebra

Finally BX� � B�BX�� � B�X� � � so �A� �I��X� � � and

AX� � �X�� ���

AlsoX� � BX� � �A� �I��X� � AX� � �X��

HenceAX� � X� � �X�� ���

Then using ��� and ��� we have

AP � A�X�jX�� � �AX�jAX��

� ��X�jX� � �X��

� �X�jX��

�� �� �

��

Hence

AP � P

�� �� �

and hence

P��AP �

�� �� �

��

�� The system of di�erential equations is equivalent to the single matrix

equation �X � AX where A �

�� ��� �

��

The characteristic polynomial of A is �� � ���� �� � �� � ��� so wecan use the previous question with � � �� Let

B � A� �I� �

��� ��� �

��

Then BX� �

����

���

���

� if X� �

���

�� Also let X� � BX�� Then if

P � �X�jX�� we have

P��AP �

�� �� �

��

��

Page 281: Elementary Linear Algebra

Now make the change of variables X � PY where Y �

�x�y�

�� Then

�Y � �P��AP �Y �

�� �� �

�Y�

or equivalently �x� � �x� � y� and �y� � �y��Solving for y� gives y� � y����e�t� Consequently

�x� � �x� � y����e�t�

Multiplying both side of this equation by e��t gives

d

dt�e��tx�� � e��t �x� � �e��tx� � y����

e��tx� � y����t� c�

where c is a constant� Substituting t � � gives c � x����� Hence

e��tx� � y����t� x����

and hencex� � e�t�y����t� x������

However since we are assuming x��� � � � y��� we have�x����y����

�� P��

�x���y���

��

��

�� ��

�� ��

����

��

��

�����

��

�������

��

Hence x� � e�t���t���� and y� �

��e

�t�Finally solving for x and y�

xy

��

��� �� �

��x�y�

��� �� �

���� e�t��� t����

��e

�t

��

��� ����e�t���t �

��� �

��e

�t

�e�t���t ����

��

�e�t��� �t�e�t��t� ��

��

��

Page 282: Elementary Linear Algebra

Hence x � e�t��� �t� and y � e�t��t� ���

�� Let

A �

��� ��� ��� ���� ��� ������ ��� ���

�� �

�a� We rst determine the characteristic polynomial chA����

chA��� � det ��I� �A� �

�� ��� ���� �

��� �� ��� �������� ���� �� ���

���

� �� ��� �������� �� ���

� �

��� �������� �� ���

���

� ���

����

��

��

��

���

��

���

���� �

��

���

� �

���

����

��

� �

��� �

��

��

� ���� ��

���

��

�b� Hence the characteristic polynomial has no repeated roots and we canuse Theorem ����� to nd a non singular matrix P such that

P��AP � diag��� ���

���

We take P � �X�jX�jX�� where X�� X�� X� are eigenvectors correspondingto the respective eigenvalues �� �� �

� �Finding X�� We have to solve �A� I��X � �� we have

A � I� �

��� ���� ��� �

��� ���� ������ ��� ����

���

��� � � ��� � ��� � �

�� �

Hence the eigenspace consists of vectors X � �x� y� z�t satisfying x � z and

��

Page 283: Elementary Linear Algebra

y � z with z arbitrary� Hence

X �

��� zz

z

�� � z

��� ���

��

and we can take X� � ��� �� ��t�Finding X�� We solve AX � �� We have

A �

��� ��� ��� �

��� ��� ������ ��� ���

���

��� � � �� � �� � �

�� �

Hence the eigenspace consists of vectors X � �x� y� z�t satisfying x � �y

and z � � with y arbitrary� Hence

X �

��� �y

y�

�� � y

��� ��

��

��

and we can take X� � ���� �� ��t�Finding X�� We solve �A� �

�I��X � �� We have

A��

�I� �

��� ��� ��� ���� � ������ ��� ���

���

��� � � �� � ��� � �

�� �

Hence the eigenspace consists of vectors X � �x� y� z�t satisfying x � ��zand y � z with z arbitrary� Hence

X �

��� ��z

z�

�� � z

��� ��

��

��

and we can take X� � ���� �� ��t�

Hence we can take P �

��� � �� ��

� � �� � �

���

�c� A � Pdiag��� �� ���P

�� so An � Pdiag��� �� ��n �P

���

��

Page 284: Elementary Linear Algebra

Hence

An �

��� � �� ��� � �� � �

����� � � �

� � �� � �

�n

�� �

��� � � �

� � ���� �� �

��

��

��� � � � �

�n

� � ��n

� � ��n

����� � � �

� � ���� �� �

��

��

��� � � �

�n � � ��n �� �

�n

�� ��n �� �

�n � � ��n

�� ��n �� �

�n � � ��n

��

��

��� � � �� � �� � �

���

� � �n

��� � � ���� �� ��� �� �

�� �

��� Let

A �

��� � � ��

� � ���� �� �

�� �

�a� We rst determine the characteristic polynomial chA����

chA��� �

�� � �� ��� �� � �� � �� �

�� R� � R� � R�

�� � �� ��� �� � �� �� � �� �

� ��� ��

�� � �� ��� �� � �� � �

C� � C� � C� � ��� ��

�� � �� ��� �� � ��� �� � �

� ���� ��

�� � ��� ��� �

� ���� �� f��� ������ �� � �g

� ���� ������� ��� ��� �� � ��

� ���� ������� ���� ���

��

Page 285: Elementary Linear Algebra

� ���� ��������� ����� ��

� ��� ������ ���

We have to nd bases for each of the eigenspaces N�A��I�� and N�A��I���First we solve �A� �I��X � �� We have

A� �I� �

��� � � ��

� � ���� �� �

���

��� � � ��� � �� � �

�� �

Hence the eigenspace consists of vectorsX � �x� y� z�t satisfying x � �y�zwith y and z arbitrary� Hence

X �

��� �y � z

y

z

�� � y

��� ��

��

��� z

��� ���

�� �

so X� � ���� �� ��t and X� � ��� �� ��t form a basis for the eigenspacecorresponding to the eigenvalue ��

Next we solve �A� �I��X � �� We have

A� �I� �

��� �� � ��

� �� ���� �� ��

���

��� � � �

� � �� � �

�� �

Hence the eigenspace consists of vectors X � �x� y� z�t satisfying x � �z

and y � �z with z arbitrary� Hence

X �

��� �z

�zz

�� � z

��� �����

��

and we can take X� � ���� ��� ��t as a basis for the eigenspace correspond�ing to the eigenvalue ��

Then Theorem ����� assures us that P � �X�jX�jX�� is non singular and

P��AP �

��� � � �

� � �� � �

�� �

��

Page 286: Elementary Linear Algebra

x1

y1

4.5 9 13.5-4.5-9

4.5

9

13.5

-4.5

-9

x

y

x1

y1

4 8-4-8

4

8

-4

-8

x

y

Figure �� �a�� x� � �x� �y � � � �� �b�� y� � �x� y � � �

Section ���

�� �i� x���x��y�� � �x�������y���� So the equation x���x��y�� � �becomes

x��� �y� � � ���

if we make a translation of axes x� � � x�� y � � � y��However equation ��� can be written as a standard form

y� � ��

�x���

which represents a parabola with vertex at ��� ��� �See Figure ��a���

�ii� y�� �x�y� � �y� ���� ��x� �� Hence y�� �x�y� � �becomes

y��� �x� � � ��

if we make a translation of axes x� � x�� y � � � y��However equation �� can be written as a standard form

y��� �x��

which represents a parabola with vertex at �� ���� �See Figure ��b���

Page 287: Elementary Linear Algebra

� �xy � y� � X tAX � where A �

�� �

�and X �

�xy

�� The

eigenvalues of A are the roots of �� � � � � � �� namely �� � �� and�� � ��

The eigenvectors corresponding to an eigenvalue � are the non�zero vec�tors �x� y�t satisfying�

�� � � � �

� �x

y

��

���

��

�� � �� gives equations

�x� y � �

x� y � �

which has the solution y � �x� Hence�xy

��

�x

�x

�� x

���

��

A corresponding unit eigenvector is ���p� ��

p�t�

�� � � gives equations

�x� y � �

x� �y � �

which has the solution x � y� Hence�xy

��

�yy

�� y

��

��

A corresponding unit eigenvector is ��p� ��

p�t�

Hence if

P �

��p�

�p�

��p�

�p�

��

then P is an orthogonal matrix� Also as detP � �� P is a proper orthogonalmatrix and the equation �

xy

�� P

�x�y�

��

Page 288: Elementary Linear Algebra

represents a rotation to new x�� y� axes whose positive directions are givenby the respective columns of P � Also

P tAP �

��� �� �

��

Then X tAX � ��x��� y�

�and the original equation �xy� y� � � becomes

��x��� y�

�� �� or the standard form

�x��

�y��

�� ��

which represents an hyperbola�The asymptotes assist in drawing the curve� They are given by the

equations�x�

�y��

�� �� or y� � �x��

Now �x�y�

�� P t

�x

y

��

��p�

��p�

�p�

�p�

� �x

y

��

so

x� �x� yp

� y� �

x� yp

Hence the asymptotes are

x� yp

� ��x� yp

��

which reduces to y � � and y � �x� � �See Figure �a���

� �x� � �xy � y� � X tAX � where A �

�� ��

�and X �

�x

y

�� The

eigenvalues of A are the roots of �� � � � � � � �� namely �� � � and�� � �� Corresponding unit eigenvectors turn out to be ���

p� �

p�t and

���p� ��p�t� Hence if

P �

��p�

��p�

�p�

�p�

��

then P is an orthogonal matrix� Also as detP � �� P is a proper orthogonalmatrix and the equation �

xy

�� P

�x�y�

Page 289: Elementary Linear Algebra

x2

y2

8 16-8-16

8

16

-8

-16

x

y x2

y2

0.95 1.9 2.85-0.95-1.9-2.85

0.95

1.9

2.85

-0.95

-1.9

-2.85

x

y

Figure � �a�� �xy � y� � �� �b�� �x� � �xy � y� � �

represents a rotation to new x�� y� axes whose positive directions are givenby the respective columns of P � Also

P tAP �

�� �� �

��

Then X tAX � �x��� �y�

�and the original equation �x� � �xy � y� � �

becomes �x��� �y�

�� �� or the standard form

x��

��

y��

�� ��

which represents an ellipse as in Figure �b��The axes of symmetry turn out to be y � x and x � �y�

�� We give the sketch only for parts �i�� �iii� and �iv�� We give the workingfor �ii� only� See Figures �a� and ��a� and ��b�� respectively�

�ii� We have to investigate the equation

x� � �xy � �y� � �px� ��

py � � � �� � �

Here x� � �xy � �y� � X tAX � where A �

� �

� �

�and X �

�x

y

��

The eigenvalues of A are the roots of ��� � �� � � �� namely �� � � and

��

Page 290: Elementary Linear Algebra

x1

y1

3 6-3-6

3

6

-3

-6

x

y x2

y2

1.5 3 4.5-1.5-3-4.5

1.5

3

4.5

-1.5

-3

-4.5

x

y

Figure � �a�� �x� � �y� � �x� �y � � � �� �b�� x� � �xy � �y� �px� ��

py � � � �

x2

y2

4.5 9-4.5-9

4.5

9

-4.5

-9

x

y

x2

y2

4.5 9-4.5-9

4.5

9

-4.5

-9

x

y

Figure �� �a�� �x�� y�� �xy� ��y� �� � �� �b�� ��x�� ��xy� �y����x� �y � � � �

��

Page 291: Elementary Linear Algebra

�� � �� Corresponding unit eigenvectors turn out to be ���p� ��p�t and

��p� ��

p�t� Hence if

P �

��p�

�p�

��p�

�p�

��

then P is an orthogonal matrix� Also as detP � �� P is a proper orthogonalmatrix and the equation �

x

y

�� P

�x�y�

represents a rotation to new x�� y� axes whose positive directions are givenby the respective columns of P � Also

P tAP �

�� �� �

��

Moreoverx� � �xy � �y� � �x�

�� �y�

��

To get the coe�cients of x� and y� in the transformed form of equation � ��we have to use the rotation equations

x ��p�x� � y��� y �

�p��x� � y���

Then equation � � transforms to

�x��� �y�

�� �x� � �y� � � � ��

or� on completing the square�

��x� � �� � ��y� � ��� � ��

or in standard formx��

��

y��

�� ��

where x� � x� � and y� � y� � �� Thus we have an ellipse� centre�x�� y�� � ��� ��� or �x�� y�� � ��� ��� or �x� y� � ���

p��

The axes of symmetry are given by x� � � and y� � �� or x� � � �and y� � � � �� or

�p�x� y� � � � and

�p�x� y�� � � ��

��

Page 292: Elementary Linear Algebra

which reduce to x� y � p � � and x� y �p � �� See Figure �b��

� �i� Consider the equation

x� � y� � xy � x� �y � � �� ���

� �

������� � ��

� � ��� �

������� � �

�������� � ��� �� �

������� � �

�������� � �� ��

� �

������� � ��

Let x � x� � �� y � y� � � and substitute in equation ��� to get

�x�������y������ �x�����y������x�������y����� � � ���

Then equating the coe�cients of x� and y� to � gives

��� � � � �

�� � � � � ��

which has the unique solution � � � � � ��� Then equation �� simpli�esto

x��� y�

�� x�y� � � � �x� � y���x� � y���

So relative to the x�� y� coordinates� equation ��� describes two lines� x��y� � � and x�� y� � �� In terms of the original x� y coordinates� these linesbecome �x� � � �y� �� � � and �x� �� �y� �� � �� i�e� x� y � � �and x� y � � � �� which intersect in the point

�x� y� � ��� �� � �� ����

�ii� Consider the equation

�x� � y� � �xy � �x� y � � � �� ���

Here

� �

�������� � � �� �� �

������� � ��

as column � � column �Let x � x� � �� y � y� � � and substitute in equation ��� to get

��x�� ��� � �y� � ��� � ��x� � ���y� � �� � ��x�� ��� �y� � �� � � � ��

��

Page 293: Elementary Linear Algebra

Then equating the coe�cients of x� and y� to � gives

���� �� � � � �

���� � � � ��

or equivalently � �� � � � � �� Take � � � and � � �� Then equation ���simpli�es to

�x��� y�

�� �x�y� � � � � x� � y��

�� ���

In terms of x� y coordinates� equation ��� becomes

� x� �y � ���� � �� or x� y � � � ��

�iii� Consider the equation

x� � �xy � �y� � x� y � � �� ���

Arguing as in the previous examples� we �nd that any translation

x � x� � �� y � y� � �

where � � �� � � � � has the property that the coe�cients of x� and y�will be zero in the transformed version of equation ���� Take � � � and� � ��� Then ��� reduces to

x��� �x�y� � �y�

�� �

�� ��

or �x��y��� � �� Hence x��y� � � �� with corresponding equations

x� y � and x� y � ���

��

Page 294: Elementary Linear Algebra

Section ���

�� The given line has equations

x � � � t���� �� � � � ��t�

y � �� � t�� � �� � �� � t�

z � � t��� � � � � �t�

The line meets the plane y � � in the point �x� �� z�� where � � ��� t� ort � ��� The corresponding values for x and z are and �� respectively�

�� E � �

��B�C�� F � ��� t�A� tE� where

t �AF

AE�

AF

AF � FE�

AF�FE

�AF�FE� � ��

��

Hence

F ��

�A�

��

��B�C�

��

�A�

��B�C�

��

��A�B �C��

�� Let A � ��� �� �� B � ��� ��� ��� C � ��� �� ��� Then we prove�AC�

t�AB for some real t� We have

�AC�

��� ���

��� � �

AB�

��� ������

��� �

Hence�AC� ���� �

AB and consequently C is on the line AB� In fact A isbetween C and B� with AC � AB�

� The points P on the line AB which satisfy AP � �

�PB are given by

P � A� t�AB� where jt���� t�j � ��� Hence t���� t� � ����

The equation t���� t� � �� gives t � �� and hence

P �

��� �

���

����

��� �

��� �

��� ��������

��� �

��

Page 295: Elementary Linear Algebra

Hence P � ����� ���� ����The equation t���� t� � ��� gives t � ���� and hence

P �

��� �

���

���� �

��� �

��� �

��� ��

��������

��� �

Hence P � � ��� ���� �������� An equation for M is P � A� t

�BC� which reduces to

x � � � �t

y � �� �t

z � � � t�

An equation for N is Q � E� s�EF � which reduces to

x � �� �s

y � ��z � �� �s�

To �nd if and whereM and N intersect� we set P � Q and attempt to solvefor s and t� We �nd the unique solution t � �� s � ���� proving that thelines meet in the point

�x� y� z� � �� � �� �� �� � � � � �� ��� ����

�� Let A � ��� � ��� B � ���� � ��� C � ��� �� �� Then

�i�

cos � ABC � ��BA � �BC���BA �BC��

where�BA� ���� ��� ���t and �

BC� � � ��� ���t� Hence

cos � ABC �� � �� � �p

� p�

�� p� p�

��

��

Hence � ABC � ��� radians or ����

��

Page 296: Elementary Linear Algebra

�ii�

cos � BAC � ��AB � �AC���AB �AC��

where�AB� ��� �� ��t and

�AC� �� � � ��t� Hence

cos � BAC �� � � �p� p �

� ��

Hence � ABC � ��� radians or ����

�iii�

cos � ACB � ��CA � �CB���CA �CB��

where�CA� ��� � ���t and �

CB� �� � �� ��t� Hence

cos � ACB ��� � � � �p

�p�

� �p �p�

p �p�

p�

��

Hence � ACB � ��� radians or ����

� By Theorem ����� the closest point P on the line AB to the origin O is

given by P � A� t�AB� where

t �

�AO � �ABAB�

��A� �ABAB�

Now

A� �AB���� ��

��

��� �

��� �

��

��� � ���

Hence t � ���� and

P �

��� ��

��

���� �

��

��� ���

��� �

��� ������

���������

���

and P � �������� ������ ������

��

Page 297: Elementary Linear Algebra

Consequently the shortest distance OP is given bys������

��

���

��

��

��

��

��

p���

���

p�� ��� ��

���

p��p��

Alternatively� we can calculate the distance OP �� where P is an arbitrarypoint on the line AB and then minimize OP ��

P � A� t�AB�

��� ��

��

���� t

��� ���

��� �

��� �� � �t

� � t

� � t

��� �

Hence

OP � � ��� � �t�� � �� � t�� � �� � t��

� ��t� � t� �

� ��

�t� �

��t �

��

� ��

�t � �

��

��

���

���

� ��

�t � �

��

���

���

��

Consequently

OP � � ��� ��

���

for all t� moreover

OP � � ��� ��

���

when t � �����

�� We �rst �nd parametric equations for N by solving the equations

x� y � �z � �

x� �y � z � �

The augmented matrix is �� � �� �� � ��

Page 298: Elementary Linear Algebra

which reduces to �� � ��� ����� � ��� ���

Hence x � ��

�� �

�z� y � �

�� z

�� with z arbitrary� Taking z � � gives a point

A � ���

�� �

�� ��� while z � � gives a point B � ��� �� ���

Hence if C � ��� �� ��� then the closest point on N to C is given by

P � A� t�AB� where t � �

�AC �

�AB��AB��

Now

�AC�

��� ��������

��� and

�AB�

��� �������

��� �

so

t ��

�� �

�� ��

�� ��

�� �� ��

��

�����

��

� ���

��

��

Hence

P �

��� ����

����

����

��

��� �������

��� �

��� ���������

��� �

so P � � ��� ���� ������Also the shortest distance PC is given by

PC �

s���

��

��� �

��

��� ��

��

p���

��

�� The intersection of the planes x � y � �z � and �x� �y � z � � is theline given by the equations

x ��

z� y �

��

z�

where z is arbitrary� Hence the line L has a direction vector ���� �� ��t

or the simpler ��� � �t� Then any plane of the form �x� y � z � d willbe perpendicualr to L� The required plane has to pass through the point��� �� ��� so this determines d�

�� � � � � � � � � d � ���

Page 299: Elementary Linear Algebra

��� The length of the projection of the segment AB onto the line CD isgiven by the formula

j �CD � �AB jCD

Here�CD� ���� � ���t and �

AB� � � � � ��t� so

j�CD � �AB j

CD�

j����� � � �� � � ����� �jp����� � � � �����

�j � �jp

���

��

��

��� A direction vector for L is given by�BC� ��� ��� ��t� Hence the plane

through A perpendicular to L is given by

�x� �y � �z � ���� � � ����� ���� � �� � � ��

The position vectorP of an arbitrary point P on L is given by P � B�t�BC�

or ��� x

yz

��� �

��� �

���� t

��� ����

��� �

or equivalently x � �� t� y � �� �t� z � � �t�To �nd the intersection of line L and the given plane� we substitute the

expressions for x� y� z found in terms of t into the plane equation and solvethe resulting linear equation for t�

���� t�� ���� �t� � �� � �t� � ��

which gives t � ����� Hence P �����

��� ��

��� ���

��

�and

AP �

s��� ���

��

��

���� �

��

��

��� ���

��

��

p����

���

p���� ��

���

p���p��

��� Let P be a point inside the triangle ABC� Then the line through P andparallel to AC will meet the segments AB and BC in D and E� respectively�

��

Page 300: Elementary Linear Algebra

Then

P � ��� r�D� rE� � � r � ��

D � ��� s�B� sA� � � s � ��

E � ��� t�B� tC� � � t � ��

Hence

P � ��� r� f��� s�B� sAg� r f��� t�B� tCg� ��� r�sA� f��� r���� s� � r��� t�gB� rtC

� �A� �B� �C�

where

� � ��� r�s� � � ��� r���� s� � r��� t�� � � rt�

Then � � � � �� � � � � �� � � � � ��� r� � r � �� Also

� � � � � � ��� r�s� ��� r���� s� � r��� t� � rt � ��

��� The line AB is given by P � A� t��� � �t� or

x � � � �t� y � �� � t� z � ��� t�

Then B is found by substituting these expressions in the plane equation

�x� y � z � ���

We �nd t � ���� and consequently

B �

��� �

�� ��� ���

�� ��� ��

��

����

�������

��

��

Then

AB � jj �AB jj � jjt��� �

��� jj

� jtjp�� � � � � �

��p� � �p

��

Page 301: Elementary Linear Algebra

� � Let A � ���� �� ��� B � ��� �� �� C � ��� �� ��� Then the area of

triangle ABC is �

�jj�AB � �

AC jj� Now

�AB � �

AC�

��� ���

����

��� ��

���

��� �

��� �

� ��

��� �

Hence jj �AB ��AC jj � p

����

�� Let A� � ��� �� �� A� � ��� ��� ��� A� � � � ��� ��� Then the pointP � �x� y� z� lies on the plane A�A�A� if and only if

�A�P �� �

A�A� ��

A�A�� � ��

or �������x � � y � � z � �� �� ��� �� ��

������� � �x� y � �z � �� � ��

��� Non�parallel lines L and M in three dimensional space are given byequations

P � A� sX� Q � B� tY�

�i� Suppose�PQ is orthogonal to both X and Y � Now

�PQ� Q� P � �B� tY �� �A� sX� �

�AB �tY � sX�

Hence

��AB �tY � sX� �X � �

��AB �tY � sX� � Y � ��

More explicitly

t�Y �X�� s�X �X� � � �AB �X

t�Y � Y �� s�X � Y � � ��AB �Y�

However the coe�cient determinant of this system of linear equationsin t and s is equal to����� Y �X �X �X

Y � Y �X � Y

����� � ��X � Y �� � �X �X��Y � Y �

� jjX � Y jj� �� ��

��

Page 302: Elementary Linear Algebra

�������

y

z

x

O

���������

��

M

L

����������

����

���

� ����������������

�����

PPPPPPPPPD

C

Q

P

��

��

as X �� �� Y �� � and X and Y are not proportional �L and M arenot parallel��

�ii� P and Q can be viewed as the projections of C andD onto the line PQ�where C andD are arbitrary points on the lines L andM� respectively�Hence by equation ���� � of Theorem ����� we have

PQ � CD�

Finally we derive a useful formula for PQ� Again by Theorem ����

PQ �j �AB � �PQ j

PQ� j �AB ��nj�

where �n � �

PQ

�PQ is a unit vector which is orthogonal to X and Y �

Hence�n � t�X � Y ��

where t � ���jjX � Y jj� Hence

PQ �j �AB ��X � Y �jjjX � Y jj �

�� We use the formula of the previous question�

��

Page 303: Elementary Linear Algebra

Line L has the equation P � A� sX � where

X ��AC�

��� ����

��� �

Line M has the equation Q � B� tY � where

Y ��BD�

��� ���

��� �

Hence X � Y � ���� �� �t and jjX � Y jj � p���

Hence the shortest distance between lines AC and BD is equal to

j �AB ��X � Y �jjjX � Y jj �

���������� ����

��� �

��� ��

����������p

���

�p���

��� Let E be the foot of the perpendicular from A� to the plane A�A�A��Then

volA�A�A�A� ��

�� area�A�A�A�� �A�E�

Now

area�A�A�A� ��

�jj �A�A� �

�A�A� jj�

Also A�E is the length of the projection of A�A� onto the line A�E� See�gure below��

Hence A�E � j �A�A� �X j� where X is a unit direction vector for the line

A�E� We can take

X �

�A�A� �

�A�A�

jj�

A�A� ��

A�A� jj�

Hence

volA�A�A�A� ��

�jj �A�A� �

�A�A� jj j

�A�A� ��

�A�A� �

�A�A��j

jj �A�A� �

�A�A� jj

��

�j �A�A� ��

�A�A� �

�A�A��j

��

�j� �A�A� �

�A�A���

�A�A� j�

���

Page 304: Elementary Linear Algebra

��� We have�CB� ��� � ���t� �

CD� ���� �� ��t� �AD� ��� �� ��t� Hence

�CB � �

CD� �i � �j� �k�

so the vector i � j� k is perpendicular to the plane BCD�Now the plane BCD has equation x� y� z � �� as B � ��� �� �� is on

the plane�Also the line through A normal to plane BCD has equation�

�� xy

z

��� �

��� �

���� t

��� �

��� � �� � t�

��� �

��� �

Hence x � � � t� y � � � t� z � �� � t���We remark that this line meets plane BCD in a point E which is given

by a value of t found by solving

�� � t� � �� � t� � � � t� � ��

So t � ���� and E � ����� ���� �����

The distance from A to plane BCD is

j�� � � �� � � � � �j�� � �� � �

���p�

� �p��

To �nd the distance between lines AD and BC� we �rst note that

���

Page 305: Elementary Linear Algebra

�a� The equation of AD is

P �

��� ��

���� t

��� ���

��� �

��� � � �t

� � �t

��� �

�b� The equation of BC is

Q �

��� �

��

���� s

��� �

��

��� �

��� � � s� � s�� s

��� �

Then�PQ� �� � s� �t� � � s� � � s� �t�t and we �nd s and t by solving

the equations�PQ � �AD� � and

�PQ � �BC� �� or

�� � s � �t�� � �� � s�� � �� � s � �t�� � �

�� � s� �t� � �� � s�� �� � s � �t� � ��

Hence t � ���� � s�Correspondingly� P � ������ �� ��� and Q � ����� �� �����Thus we have found the closest points P and Q on the respective lines

AD and BC� Finally the shortest distance between the lines is

PQ � jj �PQ jj � ��

���

Page 306: Elementary Linear Algebra

�����

����������

����

����

���

��

��

����������

CCCCCCCC

CCCCCCCC

C

D

B

A

E

���