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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
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
� 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
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
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� � � � � � ��
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�
�
� 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 ���� ��� ���� �� ��� �� ��� ��
���� 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 �
� 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 � � ��
���� 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�
� 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
����
� � � � � �� � � � � �� � � � � �� � � � � �
����
���� 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�
� 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�
���� 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� ��
�� 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�
���� 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 � ��
�� 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� � ��
���� 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�
�� 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
��
���� 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�
�� 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�
���� 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
�� � � �
� � ��� � �
�� �
�� 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 � ���
���� 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 � ��
�� 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�
���� 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
� � � �
�� �
��
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 ��
�
�� 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�
���� 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�
�� 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
����� �
���� 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 ��
�� 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
���� 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�
� 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
�
���� 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�
� 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�
���� 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 �
�� � � ��� � ��� �
�� �
� 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��
��
���� 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���
� 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�
���� 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�
� 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��
���� 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
�
�� 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 � ��
���� 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������ �
�� � � �
� � ��� � �
�� �
�� 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������
���� 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�
�� 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�
���� 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
�� 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 � ��
���� 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 �
�����
�
�� 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�
���� 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��
�� 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�
���� 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�� �
����
� � � �� � � �� � �
��
� � � �
������
�� 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� � ���
���� 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 � ������
�� 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�
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�
��
�� 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 � ��
���� 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 �
�� 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� � ��
���� 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�
�� 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 �
���� 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�
�� 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�
���� 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�
�� 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 �
���� 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�
�� 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�
���� 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 � �
�� 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 �
����
� � �� � � � �� � � � �� � � � � �
�� �
���� 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�
�� 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�
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 ���
�
�� 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
��
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�
� 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��
��
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�
������ � ��
�� 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��
��
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 �����
�� 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 �
���� � �� �
����� �
���� �� �
����� �
���� �� �
����� �� � �� �
� �� �� ��
��
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��
������
�� 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���������
� � �� � � �
���������
��������
� � �� �� �� � �� ��� � �
��������
�
� ��
��������
� � �� � �� ��� � �
��������
� ��
��������
� � �� � �� ��� � �
��������
� �
��������
� � �� � �� � �� ��
��������
� �
��������
� � �� � �� � � ��
��������� ���
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 �� ��
�� 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
�� � �
� ��� � �
�
��
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 � �
� 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��
� �
���� 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�� � ��
��
�� � � �
�� � �� ��
���
�� 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��
���� 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�
�� 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����
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�
��
�� 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� � ����
���� 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 �� ��
�� 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�
���� 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�
��
�� 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 ��
���� 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 ����
�� 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�
���� 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 �
�� 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�
���� 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�
��� 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
���� � ��
���� 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
�� ��
�����
�
��� 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 � �� � ��
���� 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��
��� 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 ��� � �� �
���� 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��� �
��� 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 �
���� 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 �
��� 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 �
���� 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 ��� � ��
��� 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�
���� 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�
��� 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��
���� 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�
��� 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�
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 ����
��
��� 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
���� 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�
��� 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
��
�
��
���� 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� � � �����
�� 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
��
���� 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
��� ��
�� �
�
��� 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� ����
���� 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
����� �
��� 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���
���� 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��
��� 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
���� 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�
��� 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 ��� �� ���
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�
��
��� 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���
���� 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 ��
��� 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�
���� 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
���
��� 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 � ��
���� 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�
�
�� 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���
� ��
���� 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 � � � ��
��� 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 � ��
���� 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
��
��� 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
�
���� 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�
��� 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�
���� 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�
��� 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
��
���� 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 � � � �� ������
�� 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 � � � ��
���� 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 � � � ��
��� CHAPTER �� IDENTIFYING SECOND DEGREE EQUATIONS
�ii� x� � y� � xy � x� �y � � � ��
�iii� x� � �xy � �y� � x� �y � � � ��
�Answers� �i� �x � y � � � � and x � y � � � �� �ii� �x � y � � � ���iii� x� �y � � � � and x� �y � � � ���
������� �
���������������
������
��� ����������
�� ���� ����� � ���� � ��� ������ ��� ��� � ��������� ���� ���� �� ��� �� � ���� ������� � ���� � ���� ��� ���������� ������� �� � � �������� ��������� ��� ������� ���� �� �� ���� ����� ������ �������� � ������� �� ������
������ � ���� �� �� ����� � �� ����� ��� �� �������
���� � ���� �� � ���� ��� �� ��� �� � ���� ��� �� �� ���� �� ��� ����
���� ��
��� � �� � ! ��� � �� � ! ��� � �� ��
����� ��� � ������� � ��� ���� �� ��������� ��� � ����
��� �� �� � � ���� ��� �� ��� � � ���� ��� �� � ���
����
�� �� � ��
�� � ���� � ��
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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 ���� ����
���
��
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�
���
���� 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�
���
�� � N� Magnenat�Thalman and D� Thalmann� Stateoftheartin Com�
puter Animation� ����� Springer�Verlag Tokyo�
���� W�K� Nicholson� Elementary Linear Algebra� ����� PWS�Kent� Boston�
���
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� ���
��
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� ��
��
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� ���
��
SOLUTIONS TO PROBLEMS
ELEMENTARY
LINEAR ALGEBRA
K� R� MATTHEWS
DEPARTMENT OF MATHEMATICS
UNIVERSITY OF QUEENSLAND
First Printing� ����
CONTENTS
PROBLEMS ��� �������������������������������������������� �
PROBLEMS ��� �������������������������������������������� ��
PROBLEMS ��� �������������������������������������������� ��
PROBLEMS ��� �������������������������������������������� ��
PROBLEMS ��� �������������������������������������������� ��
PROBLEMS ��� �������������������������������������������� ��
PROBLEMS ��� �������������������������������������������� �
PROBLEMS ��� �������������������������������������������� ��
PROBLEMS ��� �������������������������������������������� �
i
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The augmented matrix has been converted to reduced row�echelon formand we read o� the unique solution x ��� y ��
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From the last matrix we see that the original system is inconsistent�
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�����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�
�
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�
�
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
������
�
�
������� � � � � ��� � � � � �����
��� � � ����
� � � � � �� 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�
�
��� � � �� �� �� � �� � 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�
������ � � � �� � � � �� � � � �� � � � �
����� �
�
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�
��� � � � �� � � �� � � �
���
�
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�
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� �
��� �
�
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 ��
��
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�
��
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�
��
�� 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��
��
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��
�
�
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�
�
��
��� 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��
��
� 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�
�
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�
��
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�
��
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��
��
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
��� �
��
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���
��
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�� �
��� � � ���
� � ���� �� ��
��� �
��
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�
�����
� � � �� � � ��� � � �� � � �
���������
� �� � �� � �� �� � � �� � � �
�����
��
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�
��
���
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 �
� �
��
��
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
��
��
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
��
��
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 �
�� � �� � �
� ��� ����
��� �
�����
��
��
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�
��
��� 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
��
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�
�
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� �
��� �����
���
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��
�
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��
�
�� 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��
�
�� 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� � ��
�
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
�
��� �� �� 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�
�
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��
��
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 � ��
��
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�
��
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 � ��
�
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���
��
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�
�
��������
����������
����������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
�������
�
�������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 ���������
� � � �� � � �� � �� � � �� ��
����������
���������
� � � � � � ��� ��� ��� ���� �� �� ���
����������
�
�������� � ��
��� ��� ����� �� ���
������� � ���
�������� � ��� � �
�� �� ���
�������� ���
�������� �� � � �
�� �� ���
������� � ������
����� �� �� ���
����� � ����
�� 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 �
�������
�
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�
��� � � �� �� � �� �� � �
���
�
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�
��
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
��
�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��
��
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 � ��
��
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 ����
�
�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
�����
��
� 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��
�
��� �������� 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��
��
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
��
�
��
� � �� � 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������
��
�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��
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 ���
�� ��
�
�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
�
��
�
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��
�
�� � � � 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
�
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
��� � �� �
�
��� 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�
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�
�
�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�
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��
��
�� 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 �
��� �� �
��
��
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���
��
��
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�
��
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
��
��
�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
��
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�
��
�� �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 � ��
��
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 �
�� �� �
��
��
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� ��
��
��
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
��
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
���
��
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
� ���� ������� ��� ��� �� � ��
� ���� ������� ���� ���
��
� ���� ��������� ����� ��
� ��� ������ ���
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 �
��� � � �
� � �� � �
�� �
��
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���
�
� �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�
�
��
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�
�
�
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
��
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 � � � �
��
�� � �� 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�� � � ��
��
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� � �� � � � ��
��
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 � ���
��
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 �
��� �
���
����
�
��� �
��� �
��� ��������
��� �
��
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 ����
��
�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 � �������� ������ ������
��
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 �� � �� �� � ��
�
�
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 � ���
�
��� 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�
��
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
��
�
� � 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� �� ��
��
�������
�
�
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�
��
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�
���
��� 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
���
�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 � ��
���
�����
����������
����
����
���
��
��
����������
CCCCCCCC
CCCCCCCC
C
D
B
A
E
���