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Complete Solutions 7(a) 1

Complete Solutions to Exercise 7(a)MATLAB: To find the eigenvalues of a matrix A the command is eig(A). To find eigenvaluesand eigenvectors enter the command [V, d]=eig(A) this shows the eigenvectors down thecolumns of V and the eigenvalues along the diagonal entries of d.

1. We substitute the given matrix, A, into det A I 0 :(a) We have

7 3det 7 4 0

0 4

7 4 0

Thus the eigenvalues are 4, 7 .

Let u xy

be the eigenvector for 7 . We have

7 7 3 0

0 4 7 0

0 3 0

0 11 0

x

y

x

y

Multiplying out the matrices gives3 0

11 0

y

y

Thus y=0 and x is any real number apart from zero. A particular value of x can be 1. So a

particular eigenvector for 7 is u 10

.

Similarly let v be an eigenvector for 4:

7 4 3 0

0 4 4 0

11 3 0

0 0 0

x

y

x

y

Multiplying out the first row yields

11 3 0

3

11

x y

x y

If 1y then x 3 11 , thus3 11

1

v or using smallest positive integers gives3

11

.

(b)

2 2

5 2det 5 1 8

4 1

5 1 8 5 4 8 4 5 8

Putting this quadratic to zero and solving


2 2

1 2

4 5 8 4 3 0

3 1 0 gives 1, 3

Letx

y

u be the eigenvector for 1:

5 1 2 0

4 1 1 0

4 2 0

4 2 0

x

y

x

y

Multiplying out the matrix4 2 0

4 2 0

x y

x y

Solving these gives x 1, y 2 . Thus1

2

u is an eigenvector for 1 .

Let v xy

be the eigenvector for 3 :

5 3 2 0

4 1 3 0

2 2 0

4 4 0

x

y

x

y

Multiplying gives2 2 0

4 4 0

x y

x y

Solving these gives x y 1 . An eigenvector corresponding to 3 is1

1

.

(c) Substituting the given matrix into det A I yields

by (6.1)

2 2 2

1 4det 1 1 2 4

2 1

1 1 8

1 8 1 8 9

Solving the equation 2 9 0 gives2 9

9 3, 3

Letx

y

u be the eigenvector for 3 .


1 3 4 0

2 1 3 0

2 4 0

2 4 0

x

y

x

y

Solving gives x 2, y 1. An eigenvector is2

1

corresponding to 3 .

Letx

y

v be an eigenvector for 3 :

1 3 4 0

2 1 3 0

4 4 0

2 2 0

x

y

x

y

Hence x y 1 . The eigenvector1

1

corresponds to 3 .

2. Let

x

y

z

w be the eigenvector for the eigenvalue 5 of the matrix in EXAMPLE 5.

Substituting this, 5 , into A I w 0 gives

1 5 0 4 0

0 4 5 0 0

3 5 3 5 0

6 0 4 0

0 9 0 0

3 5 2 0

x

y

z

x

y

z

Multiplying the matrices6 4 0 (†)

9 0 (††)3 5 2 0 (†††)

x z

y

x y z

From (††) we have y 0 . Substituting this into †††

3 2 0

3 2

x z

x z

2

3x z

Let z a where a 0 , thus the general eigenvector is

2 3

0

1

a

.

Let v be the eigenvector for 4 :


1 4 0 4 0

0 4 4 0 0

3 5 3 4 0

3 0 4 0

0 0 0 0

3 5 7 0

x

y

z

x

y

z

The general eigenvector is20

9

15

s

v where s is not zero

3. The eigenvalues are determined by

det 0 A I

Substituting the given3 1

1 3

A and subtracting along the leading diagonal gives

2

2

3 1det det

1 3

3 3 1

9 6 1

6 8

A I

Factorizing and equating the last line to zero:

2

1 2

6 8 2 4 0

2, 4

Next we find the eigenvectors belonging to 1 2 . Using A I u O with 1 2 ,

x

y

u and0

0

O gives

3 2 1 Taking Away 2 along the

21 3 2 Leading Diagonal

1 1 0

1 1 0

x

y

x

y

A I u

Expanding out yields 0x y x y . Let y s where 0s then x s and

1

1

ss

s

u . Thus the eigenvector for 1 2 is1

1s

u .

Next we find the eigenvectors belonging to 2 4 . Using A I v O with 2 4 ,

x

y

v and0

0

O gives


3 4 1 Taking Away 4 along

41 3 4 the Leading Diagonal

1 1 0

1 1 0

x

y

x

y

A I v

Expanding out yields0

0

x y y x

x y x y

We have y x . Let y s where 0s then x s and1

1

ss

s

v . Thus the

eigenvector for 2 4 is1

1s

v .

The 2 eigenspaces are given by 2

1

1E s

and 4

1

1E s

. Plotting these on 2

gives

x-3 -2 -1 1 2 3

y

-3

-2

-1

1

2

3

Eigenspace for eigenvalue = 4Eigenspace for eigenvalue = 2

v = 1

1u =

-1

1

4. The eigenvalues are determined by

det 0 A I


7 4


2

2

5 2det det

7 4

5 4 14

5 4 14 Taking Out Minus Signs

4 20 5 14 Expanding

6 Simplifying

A I


4E

2E


2

1 2

6 2 3 0

2, 3

Next we find the eigenvectors belonging to 1 2 . Using A I u O with

1 2 ,x

y

u and0

0

O gives

5 2 2

27 4 2

7 2 0

7 2 0

x

y

x

y

A I u

Expanding out yields7 2 0 7 2x y x y

Let 7y s where 0s then 2x s and we have

2 2

7 7

ss

s

u

Thus the eigenvector for 1 2 is2

7s

u where the simplest eigenvector is2

7

. Next

we find the eigenvectors belonging to 2 3 . Using A I v O with 2 3 ,

x

y

v and0

0

O gives

5 3 2

37 4 3

2 2 0

7 7 0

x

y

x

y

A I v

Expanding out yields2 2 0

7 7 0

x y y x

x y x y

We have y x . Let y s where 0s then x s and1

1

ss

s

v . Thus the

eigenvector for 2 3 is1

1s

v where the simplest eigenvector is1

1

.

The effect of multiplying by the matrix A is

2 Au u where2

7s

u and 3Av v where1

1s

v

The 2 eigenspaces are given by 2

2

7E s

and 3

1

1E s

. Plotting these on 2

gives


x-10 -5 5 10

y

-10

-5

5

10

Eigenspace for eigenvalue = 3

Eigenspace for eigenvalue = -2

v = 1

1

u = 2

7

A basis vector for 2E is2

7

and a basis vector for 3E is1

1

.

5. We are given2 8 4 16

and5 1 10 2

A B .

(i) The eigenvalues of matrix A are determined by

det 0 A I


5 1


2

2

2 8det det

5 1

2 1 40


2 2 40 Expanding

42 Simplifying

A I


2

1 2

42 7 6 0

7, 6

(ii) We have

det 0 B I


10 2

B and subtracting along the leading diagonal gives

3E

2E


2

2

4 16det det

10 2

4 2 160


2 8 160 Expanding

2 168 Simplifying

B I


2

3 4

2 168 14 12 0

14, 12

(iii) Note that matrix B is twice matrix A, that is 2B A which transforms over to theeigenvalues because

3 1 4 214 2 7 2 and 12 2 6 2

3 1 4 22 and 2 Prediction is given by question 4.6. We need to prove if the matrix rB A where r be a real number and is the eigenvalueof matrix A then the eigenvalue of B is r .Proof.Let be an eigenvalue of A with eigenvector u. By definition of eigenvalues (7.1) wehave Au u . We are given that rB A so evaluating Bu gives

r r r r Bu A u Au u u

Since rBu u we conclude that r is an eigenvalue of B with eigenvector u. This

completes our proof.■

7. We need to prove that the n n matrix,

0 0

0 0

O

, only has the zero eigenvalues.

Proof. Consider the zero matrix,

0 0

0 0

O A

, as the matrix A in the characteristic

equation det 0 A I :

By Proposition (6.19)

0 0 0

0 0 0 Taking Away alongdet det

0 0 the Leading Diagonal

0 0 0

0 0

0 0det 0

0 0

0 0

n

A I

The Proposition (6.19) is from chapter 6 and claims:


(6.19). The determinant of a triangular or diagonal matrix is a product of the entries alongthe leading diagonal.

Note that we have a n by n diagonal matrix therefore the determinant of the matrix is

copies

n

n

We have 0n which means 0 . Hence the only eigenvalues of the zero square

matrix is 0.■

8. We have the characteristic equation given by

2

1 2 1

det 2 1 1

1 1 2

1 1 2 1 2 11 det 2det det

1 2 1 2 1 1

1 1 2 1 2 2 2 1 2 1

1 1 2 1 2 4 2 1 1

1 3 2 1 2 3 2 1 Simplifying Brackets

A I

2

2

2

2

2

1 3 1 6 4 1 Expanding Last 2 Brackets

1 3 1 5 5 Simplifying

1 3 1 5 1 Because 5 5 5 1

1 3 1 5 Taking Out Factor 1

1 3 4

1 4 1 Factor

izing Square Brackets

Equating all this to zero gives

1 2 3

1 4 1 0

1, 4 and 1

What else do we need to find?

Eigenvectors. Let

x

y

z

u be an eigenvector belonging to 1 1 :

1 1 2 1

2 1 1 1

1 1 2 1

0 2 1 0

2 0 1 0

1 1 1 0

x

y

z

A I u u


Multiplying out the matrix gives:

0 2 0 *

2 0 0 **

0 ***

y z

x z

x y z

From the middle equation we have 2z x . Let x s where 0s then 2z s . How canwe find y?Substituting 2z s into the top equation (*):

0 2 2 0 givesy s y s

Thus our eigenvector is

1

1

2 2

x s

y s s

z s

u belonging to 1 1 .

Similarly we can find the other 2 eigenvectors. Substitute 2 4 and v be the

corresponding eigenvector:

1 4 2 1

2 1 4 1

1 1 2 4

3 2 1 0

2 3 1 0

1 1 2 0

x

y

z

A I v v


3 2 0 *

2 3 0 **

2 0 ***

x y z

x y z

x y z

Subtracting the top two equations, (*) and (**), gives

3 2 0 *

2 3 0 **

5 5 0 0

x y z

x y z

x y

From the last line we have x y . Let y s where 0s then x s and from the lastequation (***) we have

2 0 givess s z z s

Our eigenvector

1

1

1

s

s s

s

v where 0s and is the eigenvector belonging to 2 4 .

Call the last eigenvector w which belongs to 3 1 . We have


1 1 2 1

2 1 1 1

1 1 2 1

2 2 1 0

2 2 1 0

1 1 3 0

x

y

z

A I w w


2 2 0 *

2 2 0 **

3 0 ***

x y z

x y z

x y z

Multiply the last equation (***) by 2 and subtract the top equation (*):

2 2 6 0

2 2 0

0 0 5 0

x y z

x y z

z

From the last line we have 0z . Substituting this into the bottom equation (***) gives

0x y x y Let y s then x s and we have

1

1

0 0

s

s s

w where 0s

This w is the eigenvector belonging to 3 1 .

Our eigenspace for 1 is 1

1

1

2

E s

and in 3 is the line shown below:

A basis vector for 1

1

1

2

E s

is

1

1

2

.

1

1

1

2

E s


Our eigenspace for 4 is 4

1

1

1

E s

plotted in 3 is the line shown below:

A basis for 4

1

1

1

E s

is

1

1

1

. For 1 we have the eigenspace

1

1

1

0

E s

plotted in 3 below and a basis vector for 1E is

1

1

0

1

1

1

0

E s

4

1

1

1

E s

Complete Solutions 7(b) 1

Complete Solutions to Exercise 7(b)MATLAB.To find the characteristic polynomial of a matrix A enter the command poly(A).

1.The eigenvalues are determined by the characteristic equation

2

2

5 2det

2 1

5 1 4

5 5 4

6 9 0

A I

How do we solve this quadratic equation 2 6 9 0 ?

By factorizing 221, 26 9 3 3 3 0 gives 3 . How do we find

the corresponding eigenvectors?

Substituting5 2

2 1

A , 1, 2 3 into A I u O wherex

y

u and

0

0

O :

5 3 2 0

2 1 3 0

2 2 0

2 2 0

x

y

x

y

Expanding this out gives2 2 0

2 2 0

x y

x y

which yields y x

Let x s where 0s then y s which gives the eigenvector1

1

ss

s

u .

The eigenspace 3E is shown below:

x-10 -5 5 10

y

-10

-5

5

10

1-1

3E



1

.

2. The eigenvalues are determined by the characteristic equation:

2

2

2

12 7det

7 2

12 2 49

12 2 49

10 24 49

10 25

5 0

A I

This gives 1, 2 5 . How do we find the corresponding eigenvectors?

Substituting12 7

7 2

A , 1, 2 5 into A I u O wherex

y

u and0

0

O :

12 5 7 0

7 2 5 0

7 7 0

7 7 0

x

y

x

y

Expanding this out and solving gives y x . Let x s where 0s then y s which

gives the eigenvector of the form1

1u

ss

s

.

The eigenspace 5E is shown below:

x-10 -5 5 10

y

-10

-5

5

10

11


1

.

3. Proof.

5E


Leta b

c d

A then the characteristic polynomial, p , is given by

2

2

Determinant of Trace of

2

det det

Removing 2 minus signs

det

a b

c d

a d bc

a d bc

d a ad bc

a d ad bc

tr p

AA

A I

A A

Thus we have our required result.■

4. We use the result established in question 3 above and equate this to zero, which is

2 det 0tr A A

Substituting 2tr aA and 2det aA into this result gives

2

22 21, 2

det

2 0 gives

p tr

a a a a

A A

Thus the given matrix A has an eigenvalue a with multiplicity of 2.■

5. What type of matrix is

1 0 0

3 1 0

7 9 1

A ?

Triangular (lower) matrix. How do we find the eigenvalues of a triangular matrix A?

Proposition (7.6) which says that if a matrix A is a diagonal or triangular matrix then theeigenvalues of A are the entries along the leading diagonal.

Using this we have the eigenvalues 1, 2, 3 1 that is the eigenvalue is 1 with multiplicity

3. How do we find the eigenvector u?By substituting 1 into A I u O :

1 1 0 0 0 0

3 1 1 0 0 and = 0

7 9 1 1 0 0

A I u u O

x x

y y

z z

Expanding this gives


0 0 0 0

3 0 0 0

7 9 0 0

0 0 0 0 †

3 0 0 0 ††

7 9 0 0 †††

x

y

z

x y z

x y z

x y z

From the middle equation 3 0x we have

0x Substituting 0x into the bottom equation we have

7 0 9 0 0 gives 0y y What is the value of z?z can have any non-zero value, z s where 0s . Thus our eigenvector u is of the form

0 0

0 0 where 0

1

s s

s

u

A basis vector for 1E is

0

0

1

.

6. (a) We are given the matrix

5 0 0

0 5 0

0 0 2

A . We have a diagonal matrix therefore by

Proposition (7.6) we have the eigenvalues 1, 2 5 and 3 2 .

Let u be an eigenvector belonging to 1, 2 5 . Substituting 1, 2 5 into the

A I u O :

1E is this vertical axis.


5 5 0 0 0 0

0 5 5 0 0 Remember and 0

0 0 2 5 0 0

u O

x x

y y

z z

Expanding and simplifying gives

0 0 0 0

0 0 0 0

0 0 3 0

x

y

z

From the bottom equation we have 3 0 which gives 0z z . We have 1 non-zeroequation but 3 unknowns therefore there are 3 1 2 free variables, which are x and y.These can take any values, so let

andx s y t

Our eigenvector u is of the form

1 0

0 1

0 0 0

s

t s t

u where both s and t are not zero.

A basis vectors for 5E is

1 0

0 , 1

0 0

.

We also have an eigenvector v associated with the other eigenvalue 3 2 :

5 2 0 0 0 3 0 0 0

0 5 2 0 0 0 3 0 0

0 0 2 2 0 0 0 0 0

x x

y y

z z

Writing out the equations gives3 0, 3 0 and 0 0x y

From the first 2 equations we have 0, 0x y . Note that z can be any non-zero real

number, that is z s where 0s . Our eigenvector v belonging to 3 2 is of the form

0 0

0 0

1

s

s

v where 0s . A basis vector for 2

0

is 0

1

E

.

(b) The eigenvalues of the given matrix

1 2 3

0 5 1

0 0 9

B are the entries on the leading

diagonal because we have a (upper) triangular matrix. Thus

1 2 31, 5 and 9 Let u be the eigenvector belonging to 1 1 . We can find u by substituting 1 1

into B I u O :


1 1 2 3 0

0 5 1 1 0

0 0 9 1 0

0 2 3 0

0 4 1 0

0 0 8 0

x

y

z

x

y

z

B I u

Expanding this gives the simultaneous equations

0 2 3 0 †

0 4 0 ††

0 0 8 0 †††

x y z

x y z

x y z

From the bottom equation ††† we have 0z . Substituting this, 0z , into the middle

equation †† gives

4 0 which yields 0y y From the first equation † we have x can be any non-zero number, that is x s where

0s . Our eigenvector u belonging to 1 1 is of the form

1

0 0 where 0

0 0

s

s s

u

Let v be the eigenvector belonging to 2 5 . We can find v by substituting 2 5

into B I v O :

1 5 2 3 0

0 5 5 1 0

0 0 9 5 0

4 2 3 0

0 0 1 0

0 0 4 0

x

y

z

x

y

z

B I v

From the bottom row we have 0z . Substituting this, 0z , into the first row gives4 2 0 which gives 2x y y x

Let x s where 0s then 2y s and our eigenvector v belonging to 2 5 is of the

form:1

2 2 where 0

0 0

s

s s s

v

Let w be the eigenvector belonging to 3 9 . We can find w by substituting 3 9

into B I w O :


1 9 2 3 0

0 5 9 1 0

0 0 9 9 0

8 2 3 0

0 4 1 0

0 0 0 0

x

y

z

x

y

z

B I w

From the middle row we have4 0 which gives 4y z z y

Let y s then 4z s . Substituting these into the top row gives

8 2 3 0

14 78 2 3 4 0 8 14

8 4

x y z

x s s x s x s s

Our eigenvector w belonging to 3 9 is

7 / 4 7

4 where 04

4 16

x ss

y s s

z s

w

Basis vectors for eigenspaces 1 5 9, andE E E are

1

0

0

,

1

2

0

and

7

4

16

respectively.

(c) The eigenvalues of the matrix

2 0 0

2 2 0

4 10 2

C are the entries on the leading

diagonal because we have a (lower) triangular matrix. Thus

1, 2, 3 2

Let u be the eigenvector belonging to 1, 2, 3 2 . We can find u by substituting

1, 2, 3 2 into C I u O :

2 2 0 0 0

2 2 2 0 0

4 10 2 2 0

0 0 0 0

2 0 0 0

4 10 0 0

x

y

z

x

y

z

C I u

From the middle row we have2 0 0x x

Substituting 0x into the last row

4 0 10 0 0 0y y z can be any non-zero real number, let z s where 0s . Therefore the eigenvector is ofthe form


0 0

0 0 where 0

1

s s

s

u

A basis vector for 2E is

0

0

1

.

7. (a) Since the given matrix is a (upper) triangular matrix therefore the eigenvalues arethe diagonal entries which are 7, 7, 5 and 5. Thus we have

1, 2 3, 47 and 5 . Let u be the eigenvector for 1, 2 7 :

7 7 0 2 3 0

0 7 7 4 6 0

0 0 5 7 3 0

0 0 0 5 7 0

A I u

x

y

z

w

0 0 2 3 0

0 0 4 6 0

0 0 2 3 0

0 0 0 2 0

x

y

z

w

By expanding the bottom row we have 0w . Substituting this into the third row gives

0z .Thus we have 0z and 0w . Note that x and y can be any real numbers provided bothare not zero. Let x s and y t therefore the eigenvector u belonging to 1, 2 7 is of

the form1 0

0 1 where and are not both zero

0 0 0

0 0 0

u

x s

y ts t s t

z

w

Let v be the eigenvector belonging to 3, 4 5 :

7 5 0 2 3 0

0 7 5 4 6 0

0 0 5 5 3 0

0 0 0 5 5 0

A I v v

x x

y y

z z

w w

2 0 2 3 0

0 2 4 6 0

0 0 0 3 0

0 0 0 0 0

x

y

z

w

From the penultimate row we have 0w . By the second row we have2 0 2y z y z

Let z s where 0s then 2y s and from the first row we have


Because 0

2 2 3 0w

x z w x z x s

Our eigenvector v

x

y

z

w

belonging to 3, 4 5 is of the form

1

2 2

1

0 0

s

ss

s

v

because , 2 ,x s y s z s and 0w .(b) The eigenvalues are the leading diagonal entries because we have a diagonal matrix.Thus 1, 2, 3 41 and 3 . Let u be the eigenvector corresponding to 1, 2, 3 1 :

1 1 0 0 0 0

0 1 1 0 0 0

0 0 1 1 0 0

0 0 0 3 1 0

0 0 0 0 0

0 0 0 0 0

0 0 0 0 0

0 0 0 2 0

x

y

z

w

x

y

z

w

We have 4 unknowns- x, y, z and w but only 1 non-zero equation therefore we have4 1 3 free variables. From the last row we have 0w . Note that other 3 are the freevariables which means they can take any values provided all 3 are not zero.Let , andx s y t z r . Our eigenvector u is given by

1 0 0

0 1 0

0 0 1

0 0 0 0

s

ts t r

r

u provided all , andr s t are not zero

Let v be an eigenvector for 4 3 . Substituting this into the above we have

1 3 0 0 0 0

0 1 3 0 0 0

0 0 1 3 0 0

0 0 0 3 3 0

2 0 0 0 0

0 2 0 0 0

0 0 2 0 0

0 0 0 0 0

x

y

z

w

x

y

z

w

From the first 3 rows we have 0, 0 and 0x y z . The variable w is free therefore itcan take any value, s, but not zero. Thus we have

0 0

0 0where 0

0 0

1

v

x

ys s

z

w s


(c) Again we have a diagonal matrix with eigenvalues given by the entries along theleading diagonal. Thus 1, 2, 3, 4 3 . Let u be the eigenvector belonging to this

eigenvalue. We find u be substituting this 1, 2, 3, 4 3 into C I u O :

3 3 0 0 0 0

0 3 3 0 0 03

0 0 3 3 0 0

0 0 0 3 3 0

0 0 0 0 0

0 0 0 0 0

0 0 0 0 0

0 0 0 0 0

x

y

z

w

x

y

z

w

C I u

Since all the rows are zero therefore the number of free variables are 4 0 4 . Let, , andx s y t z r w q where all are not zero

Our eigenvector u belonging to 1, 2, 3, 4 3 is of the form

1 0 0 0

0 1 0 0

0 0 1 0

0 0 0 1

s

ts t r q

r

q

u

8. We need to prove Proposition (7.7) which says a matrix is invertible (non-singular) 0 is not an eigenvalue.Proof.Suppose 0 is an eigenvalue of a matrix A. Then substituting this 0 into thecharacteristic equation det 0 A I we have

det 0A

By Proposition (6.26) this is true matrix A is not invertibe (singular). Thus 0 cannot be an eigenvalue of an invertible matrix.

9. We need to prove the eigenvalue is unique.Proof.Suppose the eignvector u belongs to 2 eigenvalues and t . This means we have

and t Au u Au uSubtracting these gives

t t Au Au O u u u

Since u is an eigenvector therefore it cannot be a zero vector. Thus u Ot yields

0 which givest t Thus the eigenvalue is unique.

■

10. We need to prove that A has distinct eigenvalues provided 24dettr A A .

Proof.


The characteristic equation is given by 2 det 0tr A A .

What sort of equation is this?Quadratic equation 2 0ax bx c which has distinct roots if

2 24 0 or 4b ac b ac For our equation 2 det 0tr A A we have

1, and deta b tr c A A

Substituting these into 2 4b ac yields

2 24det 4dettr tr A A A A

gives distinct roots, which means we have distinct eigenvalues.■

We have equal eigenvalues (roots) if 2 4b ac which is 24dettr A A .

We have complex eigenvalues (roots) if 2 4b ac which is 24dettr A A .

11. (a) (i) We have a triangular matrix

1 0 0 0

3 2 0 0

5 2 3 0

9 8 1 4

A . The eigenvalues are the

entries on the leading diagonal, 1 2 3 41, 2, 3 and 4 .

(ii) Eigenvalues of 5A are given by 5 which is

5 5 5 55 5 5 51 2 3 41 1, 2 32, 3 243 and 4 1024

(iii) Eigenvalues of 1A are 1 :

1 1 1 1

1 2 3 4

1 1 11, , and

2 3 4

(iv) The determinant det A is the multiplication of all the eigenvalues

det 1 2 3 4 24 A

(v) The trace is the addition of all the eigenvalues

1 2 3 4 10tr A

(b) (i) We have a triangular matrix

1 3 4 7

0 6 3 5

0 0 8 9

0 0 0 3




5 5 5 5 5 55 51 2 3 41 1, 6 7776, 8 32768 and 3 243

(iii) Eigenvalues of 1A are 1 :

1 1 1 1

1 2 3 4

1 1 11, , and

6 8 3

(iv) The determinant det A is the multiplication of all the eigenvalues


det 1 6 8 3 144 A


1 6 8 3 0tr A

(c) (i) We have a triangular matrix

2 0 0 0

0 4 0 0

0 0 7 0

0 0 0 0




5 5 5 5 5 55 51 2 3 42 32, 4 1024, 7 16807 and 0 0

(iii) 1A does not exist for the matrix A because 4 0 is an eigenvalue which means

that the matrix A is not invertible.(iv) The determinant det A is the multiplication of all the eigenvalues

det 2 4 7 0 0 A


2 4 7 0 9tr A

12. The characteristic polynomial for the given matrix is

2 1det

2 1

2 1 2

p

Expanding this and simplifying gives

2 22 3 2 3 4p

The characteristic polynomial is 2 3 4p . Thus

2 3 4p A A A I O

Rearranging this gives

1

2 3 4

13 4 3

4

A

A A I

A A I I A A I I

The inverse matrix

1 2 1 3 01 13

2 1 0 34 4

1 11

2 24

1 1 Taking in the1

2 2 minus sign4

A A I


13. The characteristic polynomial for6 5

3 4

A is 2 10 9p .

Applying the Cayley Hamilton theorem we have

2 10 9p A A A I O

Rearranging

2 6 5 1 010 9 10 9

3 4 0 1

60 50 9 0 51 50

30 40 0 9 30 31

A A I

How can we find 3A ?

3 2 210 9 10 9 A AA A A I A A

Substituting the above results into this, 3 210 9 A A A , gives3 210 9

51 50 6 510 9

30 31 3 4

510 500 54 45 510 54 500 45 456 455

300 310 27 36 300 27 310 36 273 274

A A A

14. We are given 3 24 4p . By the Cayley Hamilton theorem we have

3 24 4p A A A A I O

Rearranging this gives

1

3 2

2

2

4 4

4 4

14 Dividing both sides by 4

4

A

A A A I

A A A I I

A A A I I

Thus 1 214

4 A A A I . How do we find 4A ?

By making 3A the subject of 3 24 4 A A A I :

3 2

4 3 2

3 2

2 2 3

2 2

2

4 4

4 4

4 4 Expanding

4 4 4 4 Replacing

16 4 16 4

17 16

A A A I

A AA A A A I

A A A

A A I A A A

A A I A A

A I

Thus 4 217 16A A I .


15. We need to prove the following: If A is a n by n matrix with distinct eigenvalues

1 2 3, , , and m and then the corresponding eigenvectors 1 2 3, , ,uu u and um

are linearly independent.How do we prove this proposition?We use mathematical induction. We prove the result for 1m , assume it is true form k and then prove it for 1m k .

Proof.Clearly 1u is linearly independent.√Assume the result is true for m k , that is the eigenvectors 1 2 3, , , and kuu u u are

linearly independent.Required to prove the result for 1m k :Suppose the eigenvectors 1 2 3 1, , , , andk kuu u u u are linearly dependent.

Then we can write 1ku as a linear combination of the other vectors:

1 1 2 2 3 3 1k k kcc c c uu u u u (*)where the c’s are scalars. Multiplying this by matrix A gives

1 1 2 2 3 3 1

1 1 2 2 3 3 1

1 1 1 2 2 2 3 3 3 1 1

Because

Because

k k k

k k k

k k k k k

c

c

c

c c c

c c c c c

c c c

u

u

u

A u u u Au

Au A Au Au Au A u Au

u u u u Au u

The last line follows from the definition of eigenvalue and eigenvector. Referencing the

last line with † :

1 1 1 2 2 2 3 3 3 1 1 †k k k k kcc c c uu u u u

Multiplying the above labelled (*) by 1k gives

1 1 1 1 2 2 1 3 3 1 1 1k k k k k k k kcc c c uu u u u (**)

Subtracting (**) and † gives

1 1 1 2 1 2 1 1 1 1 2 2 2

(**) †

1 1 1 1 2 1 2 2 1 Collecting Like terms

u uu u u u O

u u u O

k k k k k k k k

k k k k k k

c cc c c c

c c c

Since 1 2 3, , , and kuu u u are linearly independent therefore

1 1 1 2 1 2 1 0k k k k kc c c None of the bracket terms are zero. Why not?Because all the eigenvalues are distinct

1 0k j for 1, 2, 3, ,j k Therefore all the c’s must be zero, that is

1 2 30, 0, 0, , 0kc c c c Substituting these into (*) gives

1 2 3 1

1

0 0 0 0 k k

k

uu u u u

O u

This is impossible. Why?


Because 1ku is an eigenvector which means it cannot be zero. Since we have a

contradiction therefore the supposition 1 2 3 1, , , , andk kuu u u u are linearly

dependent is false.Hence by mathematical induction we have 1 2 3, , , , anduu u um are linearlyindependent.

■

16. Let A be a square matrix with eigenvalues, , then the characteristic polynomial is

given by detp A I . Consider the matrix TA I :

By (1.19) (c)

Because is a Diagonal Matrix

T T TT T T

T

A I A I A B A B

A I I

Consider the characteristic polynomial q for TA :

det

det

det

det Because by (6.14) we have det det

T

TT

T

T

q

p

A I

A I

A I

A I B B

We have the same characteristic polynomial q p . Therefore the eigenvalues

(roots) of 0q are exactly the eigenvalues of 0p . Thus the matrix A and TA

have the same eigenvalues.■

Complete Solutions 7(c) 1

Complete Solutions to Exercise 7(c)

1. (a) (i) We are given the matrix1 0

0 2

A for which we need to find the eigenvalues and

corresponding eigenvectors. Since we have a diagonal matrix therefore by Proposition (7.6)the eigenvalues are the entries along the leading diagonal, 1 21 and 2 .

For the eigenvalue 1 1 :

1 1 0 0

0 2 1 0

0 0 0 gives 1 and 0

0 1 0

x

y

xx y

y

Our eigenvector belonging to 1 1 is1

0

u . The eigenvector for the other eigenvalue

2 2 is

1 2 0 0

0 2 2 0

1 0 0 gives 0 and 1

0 0 0

x

y

xx y

y

The eigenvector corresponding to 2 2 is0

1

v .

(ii) To find the invertible matrix P we need to follow the procedure outlined in the main text.Step 1:

Need to check that the eigenvectors are linearly independent. Since1

0

u and0

1

v are

not multiplies of each other therefore eigenvectors u and v are linearly independent.Step 2:The matrix P is given by

1 0

0 1

P u v I

Note that the matrix P is the identity matrix I.Step 3:The diagonal matrix D is given by 1D P AP :

1 1 0

0 2

D I AI A

Note that D is a diagonal matrix with eigenvalues 1 1 and 2 2 as the entries along the

leading diagonal.

(b) (i) We need to find the eigenvalues and corresponding eigenvectors of1 1

1 1

A :

Proposition (7.6). If matrix A is a diagonal or triangular matrix then the eigenvalues of A arethe entries along the leading diagonal.


2

21 2

1 1det det

1 1

1 1 1

1 2 1

2 2 0 gives 0 and 2

A I

For the eigenvalue 1 0 we can find the eigenvector by:

1 0 1 1 1 0 gives 0

1 1 0 1 1 0

x xx y

y y

Thus x y and let 1y then 1x . The corresponding eigenvector is1

1

u .

Similarly we can find the eigenvector v belonging to the other eigenvalue 2 2 :

1 2 1 1 1 0 gives 1 and 1

1 1 2 1 1 0

x xx y

y y

Thus1

1

v belongs to the eigenvalue 2 2 .

(ii) Step 1:

Need to check that1

1

u and1

1

v are linearly independent. Since one vector is not a

multiple of the other therefore they are linearly independent.Step 2:

Let 1 1

1 1

P u v .

Step 3:

Then taking the inverse of this matrix P we get 1 1 11

1 12

P . Thus 1D P AP is

1 1 1 1 1 1 11

1 1 1 1 1 12

0 0 1 1 Multiplying the1

2 2 1 1 two Left Hand matrices2

0 0 0 01 1Multiplying by scalar

0 4 0 22 2

D P AP

Again the leading diagonal entries are 0 and 2 which are the eigenvalues of the given matrixA.

(c) (i) The eigenvalues of3 0

4 4

A are 1 23 and 4 because by Proposition (7.6)

the entries along the leading diagonal are the eigenvalues of a triangular matrix and A is atriangular matrix.



The eigenvector u belonging to 1 3 is given by

3 3 0 0 0 0 gives 1 and 4

4 4 3 4 1 0

x xx y

y y

Hence1

4

u . The eigenvector v corresponding to 2 4 is given by

3 4 0 1 0 0 gives 0 and 1

4 4 4 4 0 0

x xx y

y y

Thus0

1

v for 2 4 .

(ii) Step 1:

The eigenvectors1

4

u and0

1

v are linearly independent.

Step 2:

Let 1 0

4 1

P u v .

Step 3:


4 1

P . Thus 1D P AP is

1 1 0 3 0 1 0

4 1 4 4 4 1

3 0 1 0

16 4 4 1

3 0

0 4

D P AP

Again the leading diagonal entries are 3 and 4 which are the eigenvalues of matrix A.

(d) (i) We are given the matrix2 2

1 3

A . The eigenvalues are

2

1 2

2 2det det

1 3

2 3 2

5 4 1 4 0 gives 1 and 4

A I

The eigenvector u belonging to 1 1 is evaluated by

2 1 2 1 2 0 gives 2 and 1

1 3 1 1 2 0

x xx y

y y

Hence2

1

u . The eigenvector v corresponding to 2 4 is given by

2 4 2 2 2 0 gives 1 and 1

1 3 4 1 1 0

x xx y

y y


Thus the eigenvector1

1

v belongs to the eigenvalue 2 4 .

(ii) Step 1:

The eigenvectors2

1

u and1

1

v are linearly independent because they are not

multiples of each other.Step 2:

Let 2 1

1 1

P u v .

Step 3:


1 23

P . Substituting these into

1P AP gives

1 1 1 2 2 2 11

1 2 1 3 1 13

1 1 2 11

4 8 1 13

3 0 1 01

0 12 0 43

P AP

Again the leading diagonal entries are 1 and 4 which are the eigenvalues of matrix A.

2. (i) By Proposition (7.19) we have 1m m A PD P and we use this to find 5A .

(a) We are given1 0

0 2

A and from question 1 part (a) we have 1 1 0

0 1

P P I and

1 0

0 2

D A . Therefore

555 5 1

5 5

1 0 Because 1 11 0

0 320 2 and 2 32

A PD P I I

(b) We are given1 1

1 1

A and from question 1 part (b) we have1 1

1 1

P ,

1 1 11

1 12

P and0 0

0 2

D . Thus

5 5 1

51 1 0 0 1 11

1 1 0 2 1 12

1 1 0 0 1 1 Taking the scalar1

1 1 0 32 1 1 1/2 to the front2

0 32 1 1 32 32 16 161 1

0 32 1 1 32 32 16 162 2

A PD P


(c) We need to find 5A for3 0

4 4

A . From question 1 part (c) we have

1 0

4 1

P , 1 1 0

4 1

P and3 0

0 4

D . Thus

5 5 1

5

5

5

1 0 3 0 1 0

4 1 0 4 4 1

1 0 243 0 1 0 Because 3 243

4 1 0 1024 4 1 and 4 1024

243 0 1 0 243 0

972 1024 4 1 3124 1024

A PD P

(d) We need to find 5A given that2 2

1 3

A . We use 1m m A PD P with 5m . What is

P, D and 1P equal to?

By question 1 part (d) we have2 1

1 1

P ,1 0

0 4

D and 1 1 11

1 23

P . Thus

substituting these into 5 5 1A PD P gives5

5

5

5

2 1 1 0 1 11

1 1 0 4 1 23

2 1 1 0 1 1 Because 1 11

1 1 0 1024 1 23 and 4 1024

2 1024 1 11

1 1024 1 23

1026 2046 342 6821

1023 2049 341 6833

A

(ii) By the above part (c) we have1/2 1/2 1

1/2

1/2

1/2

1 0 3 0 1 0

4 1 0 4 4 1

1 0 1 0 Because 3 1/ 31/ 3 04 1 4 10 1/ 2 and 4 1/ 2

1/ 3 0 1 0 1/ 3 0

4 14 / 3 1/ 2 2 4 / 3 1/ 2

A PD P

3. (a) (i) What are the eigenvalues of the diagonal matrix

1 0 0

0 2 0

0 0 3

A ?


Since we have a diagonal matrix therefore the eigenvalues are the entries on the leadingdiagonal which is 1 2 31, 2 and 3 . For 1 1 .

Let u be the eigenvector then

1 1 0 0

0 2 1 0

0 0 3 1

0 0 0 0

0 1 0 0 gives 1, 0 and 0

0 0 2 0

x

y

z

x

y x y z

z

A I u

Thus

1

0

0

u is the eigenvector belonging to 1 1 . Let v be the eigenvector belonging to

the second eigenvalue 2 2 :

1 2 0 0

2 0 2 2 0

0 0 3 2

1 0 0 0

0 0 0 0 gives 0, 1 and 0

0 0 1 0

x

y

z

x

y x y z

z

A I u

Hence the eigenvector

0

1

0

v corresponds to the eigenvalue 2 2 . Let w be the

eigenvector belonging to 3 3 :

1 3 0 0

3 0 2 3 0

0 0 3 3

2 0 0 0

0 1 0 0 gives 0, 0 and 1

0 0 0 0

x

y

z

x

y x y z

z

A I u

The eigenvector

0

0

1

w belongs to the eigenvalue 3 3 .

Step 1:

The eigenvectors

1

0

0

u ,

0

1

0

v and

0

0

1

w are linearly independent.

Step 2:


The invertible (nonsingular) matrix is 1 0 0

0 1 0

0 0 1

P u v w I .

Step 3:Hence 1 1 P I I . We have

1

1 0 0

0 2 0

0 0 3

D I AI A

Again the eigenvalues 1 1 , 2 2 and 3 3 of the matrix A are along the leading

diagonal.(iii) To find 4A we need to use 1m m A PD P with 4m :

4

4 4 1 4 4

4

1 0 0 1 0 0

0 2 0 0 16 0

0 0 3 0 0 81

A ID I D

(b) (i) What are the eigenvalues of

1 4 0

0 4 3

0 0 5

A ?

Since we have an upper triangular matrix therefore by using Proposition (7.6) we have

1 2 31, 4 and 5 . What else do we need to find?

The eigenvector for each eigenvalue. Let u be the eigenvector belonging to 1 1 :

1 1 4 0

1 0 4 1 3

0 0 5 1

0 4 0 0

0 5 3 0 gives 1, 0 and 0

0 0 6 0

x

y

z

x

y x y z

z

A I u

Thus

1

0

0

u is the eigenvector belonging to 1 1 . Let v be the eigenvector belonging to

the second eigenvalue 2 4 . We have

1 4 4 0

4 0 4 4 3

0 0 5 4

5 4 0 0

0 0 3 0 gives 4, 5 and 0

0 0 1 0

x

y

z

x

y x y z

z

A I v


Thus

4

5

0

v . Let w be the eigenvector belonging to 3 5 . We have

1 5 4 0

5 0 4 5 3

0 0 5 5

6 4 0 0

0 1 3 0 gives 2, 3 and 1

0 0 0 0

x

y

z

x

y x y z

z

A I w

Thus

2

3

1

w . We have the eigenvectors

1

0

0

u ,

4

5

0

v and

2

3

1

w corresponding to

the eigenvalues 1 2 31, 4 and 5 respectively.

(ii) Step 1:

The eigenvectors

1

0

0

u ,

4

5

0

v and

2

3

1


Step 2:


0 5 3

0 0 1

P u v w .

Step 3:We could check PD AP to ensure that we have the correct matrices P and D. Since weneed to find 4A so we need to find 1P because 4 4 1A PD P so we could just check that wehave 1D P AP .The diagonal matrix 1D P AP . We need to find the inverse of P. Using MATLAB or our

early theory on matrices we have 1

5 4 21

0 1 35

0 0 5

P .

Substituting 1

5 4 21

0 1 35

0 0 5

P ,

1 4 0

0 4 3

0 0 5

A and

1 4 2

0 5 3

0 0 1

P into 1D P AP

gives

1

5 4 2 1 4 0 1 4 21

0 1 3 0 4 3 0 5 35

0 0 5 0 0 5 0 0 1

5 4 2 1 4 2 5 0 0 1 0 01 1

0 4 12 0 5 3 0 20 0 0 4 05 5

0 0 25 0 0 1 0 0 25 0 0 5

D P AP


(iii) To find 4A we need to use 1m m A PD P with 4m :

4

4 4 1

4

4 4

1 4 2 1 0 0 5 4 21

0 5 3 0 4 0 0 1 35

0 0 1 0 0 5 0 0 5

1 4 2 1 0 0 5 4 2Because 1 11

0 5 3 0 256 0 0 1 35 4 256 and 5 6250 0 1 0 0 625 0 0 5

1 1024 12501

0 1280 18755

A PD P

5 4 2

0 1 3

0 0 625 0 0 5

5 1020 3180 1 204 636 Multiplying by the1

0 1280 5535 0 256 1107 15 scalar

0 0 3125 0 0 625 5

(c) (i) What are the eigenvalues of

2 0 0

1 5 0

1 2 6

A ?

Since we have a lower triangular matrix therefore by using Proposition (7.6) the eigenvaluesare 1 2 32, 5 and 6 . What else do we need to find?

The eigenvector for each eigenvalue. Let u be the eigenvector belonging to 1 2 :

1

2 2 0 0

1 5 2 0

1 2 6 2

0 0 0 0

1 3 0 0 gives 12, 4 and 1

1 2 4 0

x

y

z

x

y x y z

z

A I u

Thus

12

4

1

u is the eigenvector belonging to 1 2 . Let v be the eigenvector belonging

to the second eigenvalue 2 5 . We have

2

2 5 0 0

1 5 5 0

1 2 6 5

3 0 0 0

1 0 0 0 gives 0, 1 and 2

1 2 1 0

x

y

z

x

y x y z

z

A I v



Thus

0

1

2

v . Let w be the eigenvector belonging to 3 6 . We have

3

2 6 0 0

1 5 6 0

1 2 6 6

4 0 0 0

1 1 0 0 gives 0, 0 and 1

1 2 0 0

x

y

z

x

y x y z

z

A I w

Thus

0

0

1

w . We have the eigenvectors

12

4

1

u ,

0

1

2

v and

0

0

1

w corresponding

to the eigenvalues 1 2 32, 5 and 6 respectively.

(ii) Step 1:

The eigenvectors

12

4

1

u ,

0

1

2

v and

0

0

1


Step 2:


4 1 0

1 2 1

P u v w .

Step 3:The diagonal matrix 1D P AP . We need to find the inverse of P. Using MATLAB or ourearly theory on matrices we have

1

1 0 0 1 0 0Taking in1 1

4 12 0 4 12 0negative sign12 12

9 24 12 9 24 12

P

Substituting 1

1 0 01

4 12 012

9 24 12

P ,

2 0 0

1 5 0

1 2 6

A and

12 0 0

4 1 0

1 2 1

P into

1D P AP gives


1

1 0 0 2 0 0 12 0 01

4 12 0 1 5 0 4 1 012

9 24 12 1 2 6 1 2 1

2 0 0 12 0 0 Multiplying the1

20 60 0 4 1 0 two Left Hand12

54 144 72 1 2 1 matrices

24 0 01

0 60 012

0 0 72

D P AP

2 0 0 Multiplying by the0 5 0 1

scalar0 0 6 2

Note the entries on the leading diagonal are the eigenvalues of the given matrix A.(iii) To find 4A we need to use 1m m A PD P with 4m :

4

4 4 1

12 0 0 2 0 0 1 0 01

4 1 0 0 5 0 4 12 012

1 2 1 0 0 6 9 24 12

12 0 0 16 0 0 1 0 01

4 1 0 0 625 0 4 12 012

1 2 1 0 0 1296 9 24 12

192 0 0 1 0 01

64 625 0 4 12 012

16 1250 1296 9 24

A PD P

12

192 0 0 16 0 01

2436 7500 0 203 625 012

6648 16104 15552 554 1342 1296

(d)

4. (a) We are given

1 0 0

0 1 0

0 0 1

A which is the identity matrix I. Is the identity matrix

diagonalizable?Yes because the invertible matrix P I gives the result 1 P AP III I D where D is thediagonal matrix with 1’s as the entries along the leading diagonal.

(b) Is

1 2 3

0 2 5

0 0 8

A diagonalizable?

Yes because the entries along the leading diagonal are the eigenvalues of this matrix. Thus

1 2 31, 2 and 8 are the eigenvalues and by Proposition (7.18) we conclude that

the given matrix is diagonalizable because we have 3 distinct eigenvalues.

Proposition (7.18). If a n by n matrix A has n distinct eigenvalues then the matrix A isdiagonalizable.


(c) Similar to part (b). Since

2 0 0 0

1 3 0 0

6 7 1/ 2 0

2 9 7 5

A has distinct eigenvalues

1 2 3 42, 3, 1/ 2 and 5 therefore by Proposition (7.18) we conclude that

the matrix A is diagonalizable.

5. From Example 17 we have2

1

u and1

1

v . Therefore1 2

1 1

P and

1

1 1 2 1 21

1 1 1 13

P

Substituting 1 1 21

1 13

P ,1 4

2 3

A and1 2

1 1

P into 1P AP :

1 1 2 1 4 1 21

1 1 2 3 1 13

5 10 1 2 Multiplying the two1

1 1 1 1 Left Hand Matrices3

15 0 5 0 Multiplying by the1

0 3 0 1 scalar 1/33

P AP

Note that the eigenvalues have swapped around from the diagonal matrix of Example 17.

6. Since we have distinct eigenvalues 1 22, 5 and 3 1 therefore by Proposition

(7.18) we conclude that the matrix A is diagonalizable. The diagonal matrix will have theeigenvalues along the leading diagonal of the matrix. The eigenvalues on the leadingdiagonal occur depending on the order of eigenvectors u, v and w in the matrix P.

2 0 0

0 5 0

0 0 1

D

What is the invertible matrix P going to be?It will be the eigenvectors u, v and w as the columns of the matrix P provided they are

linearly independent. Since

1

2

0

u ,

5

4

0

v and

0

0

1

w are linearly independent

therefore

1 5 0

2 4 0

0 0 1

u v w

P

Proposition (7.18). If a n by n matrix A has n distinct eigenvalues then the matrix A isdiagonalizable.


By MATLAB we have 1

4 5 01

2 1 06

0 0 6

P . How do we find 3A ?

Substitute 3m into 1m m A PD P which gives 3 3 1A PD P .

Substituting

1 5 0

2 4 0

0 0 1

P ,

2 0 0

0 5 0

0 0 1

D and 1

4 5 01

2 1 06

0 0 6

P into

3 3 1A PD P gives3

3 3 1

1 5 0 2 0 0 4 5 01

2 4 0 0 5 0 2 1 06

0 0 1 0 0 1 0 0 6

1 5 0 8 0 0 4 5 01

2 4 0 0 125 0 2 1 06

0 0 1 0 0 1 0 0 6

8 625 0 4 5 01

16 500 0 2 1 06

0 0 1 0 0 6

A PD P

1218 585 0 203 97.5 01

936 420 0 156 70 06

0 0 6 0 0 1

7. (a) We are given the matrix2 1

1 4

A . The eigenvalues are evaluated by

2

22

2 1det det

1 4

2 4 1

8 6 1

6 9 3 0 gives 3

A I

The eigenvector corresponding to 3 is given by substituting 3 into A I u O

where u is the eigenvector:

2 3 1 0

31 4 3 0

1 1 0

1 1 0

x

y

x

y

A I u

Placing the 2 by 2 matrix into row echelon form gives1 1 0

yields 1 and 10 0 0

xx y

y


Thus the eigenvector1

1

u corresponds to the eigenvalue 3 . Since there is only one

non-zero equation and 2 unknowns therefore there are 2 1 1 free variable.Thus this is the only independent eigenvector of 3 because all the other eigenvectors aremultiples of u.Since a 2 by 2 matrix has only one independent eigenvector therefore by Theorem (7.16):

Theorem (7.16). A n by n matrix A is diagonalizable it has n linearly ind e.vectors.We conclude that the given matrix is not diagonalizable.

(b) We are given the matrix2 4

1 6

A . The eigenvalues of this matrix are

2

22

2 4det det

1 6

2 6 4

12 8 4

8 16 4 0 gives 4

A I

Let u be the corresponding eigenvector. We have

2 4 4 2 4 04

1 6 4 1 2 0

x x

y y

A I u

Putting the 2 by 2 matrix into row echelon form gives2 4 0

yields 2 and 10 0 0

xx y

y

The eigenvector2

1

u belongs to the eigenvalue 4 . Since there is 1 non-zero

equation with 2 unknowns therefore the number of free variables is 2 1 1 . Hence u is theonly independent eigenvector.For the given 2 by 2 matrix we have 1 independent eigenvector which means that the matrixis not diagonalizable because the number of independent eigenvectors (1) does not equal n(2) of the square matrix.

(c) We are given

1 2 3

0 1 3

0 0 1

A which is a triangular matrix. Thus by Proposition (7.6):

Proposition (7.6). If matrix A is a diagonal or triangular then the e.values of A are the entriesalong the leading diagonal.The eigenvalue 1 . To find the corresponding eigenvector u we substitute 1 into

A I u O :

1 1 2 3

0 1 1 3

0 0 1 1

0 2 3 0

0 0 3 0 gives 1, 0 and 0

0 0 0 0

x

y

z

x

y x y z

z

A I u


The last matrix is already in row echelon form and there are 2 non-zero equations and 3unknowns which means there are 3 2 1 free variable. This means that u is the onlyeigenvector of the given matrix A.Since we are given a 3 by 3 matrix therefore 3n but we have only 1 linearly independenteigenvector therefore the given matrix is not diagonalizable.

8. (i) How do we find 11A given that4 2

9 5

A ?

We need to use Proposition (7.19), that is 1m m A PD P with 11m . To find the matrix Pwe need to find the eigenvectors which means we need the eigenvalues.Using det 0 A I to find we have

2

2

1 2

4 2det det

9 5

4 5 18

20 18

2

1 2 0 gives 1 and 2

A I

Since we have two distinct eigenvalues, 1 21 and 2 , for a 2 by 2 matrix therefore the

given matrix A can be diagonalised.Let u be the eigenvector belonging to 1 1 then substituting this into A I u gives

4 1 2

9 5 1

3 2 0 gives 2 and 3

9 6 0

x

y

xx y

y

A I u

Thus2

3

u is the eigenvector belonging to 1 1 . Letx

y

v be the eigenvector

belonging to the other eigenvalue 2 2 :

4 2 2

29 5 2

6 2 0 gives 1 and 3

9 3 0

x

y

xx y

y

A I v

Hence1

3

v is the eigenvector belonging to the other eigenvalue 2 2 . What is our

matrix P equal to?

2 1

3 3

P u v

2 1Because and

3 3

u v

What else do we need to find?The inverse of P. Thus

1

1 2 1 3 1 3 11 1

3 3 3 2 3 26 3 3

P


We also need to determine the diagonal matrix D to use 1m m A PD P . What is the diagonalmatrix D equal to in this case?It is the diagonal matrix D with the eigenvalues, 1 21 and 2 , as the entries along the

leading diagonal. (You may check that 1 P AP D ).1 0

0 2

D

Now we have all the ingredients to use the formula 1m m A PD P with 11m , that is11 11 1A PD P .

Substituting2 1

3 3

P ,1 0

0 2

D and 1 3 11

3 23

P into 11 11 1A PD P gives

11 11 1

11

11

11

2 1 1 0 3 11

3 3 0 2 3 23

2 1 1 0 3 1 Because 1 11

3 3 0 2048 3 23 and 2 2048

2 2048 3 1 Multiplying the two1

3 6144 3 2 Left Hand Matrices3

A PD P

6150 4098 2050 1366 Multiplying by the1

18441 12291 6147 4097 scalar 1/ 33

Hence 11 2050 1366

6147 4097

A .

(ii) Similarly 1 1 1 A PD P . Evaluating this gives

1 1 1

1

1

1

2 1 1 0 3 11

3 3 0 2 3 23

2 1 1 0 3 1 Because 1 11

3 3 0 1/ 2 3 23 and 2 1/ 2

2 1 2 0 3 11 1 1Taking out

3 3 0 1 3 23 2 2

4 1 3 11

6 3 3 26

A PD P

15 6 5 21 1 1Taking in

27 12 9 46 2 3

9. We prove this result by induction.Proof.

Let1 0

0 n

d

d

D be a diagonal matrix. Then 1 D D which means all the entries on the

leading diagonal are to the index 1.


Assume the result is true for m k that is

1 0

0

k

k

kn

d

d

D (*)

Required to prove the result for 1m k , that is we need to show1

11

1

0

0

k

k

kn

d

d

D

We have1

1 1 11 1

1

By (*)

0 0 0

0 0 0

k k

k k k

k kn n n

d d d

d d d

D D D D

Hence we have our result.■

10. We need to prove proposition (7.13) which is the following:Let A, B and C be square matrices. Then we have the following:I. Matrix A is similar to matrix A.II. If matrix B is similar to matrix A then the other way round is also true, that is matrix A

is similar to matrix B.III. If matrix A is similar to B and B is similar to matrix C then matrix A is similar to

matrix C.How do we prove these results?Use the definition of similar matrices:Definition (7.12). A square matrix B is similar to a matrix A if there exists an invertiblematrix P such that 1 P AP B .

Proof of I.Matrix A is similar to matrix A because we have

1 I AI Awhere I is the identity matrix.

■Proof of II.Assume matrix B is similar to matrix A then there is an invertible (nonsingular) matrix Psuch that 1B P AP . Pre-multiply this by P and post multiply this by 1P :

1 1 1

1 1

PBP P P AP P

PP A PP

IAI A

We have 11 1 1 A PBP P BP because 11 P P . Thus we have

11 1 A P BP

By Definition (7.12):


(7.12). B is similar to A if there exists an invertible matrix P such that 1 P AP B .We conclude that matrix A is similar to matrix B. This is our required result.

■Proof of III.Assume matrix A is similar to matrix B. This means that there is an invertible (nonsingular)matrix P such that

1A P BP (*)We are also given that matrix B is similar to matrix C. This means that there is an invertible(nonsingular) matrix Q such that

1B Q CQSubstituting this 1B Q CQ into (*) gives

1 1

1 1

1 11 1

Using rules of matrices

Because

A P Q CQ P

P Q C QP A BC AB C

QP C QP B A AB

By Definition (7.12) we conclude that matrix A is similar to matrix C.■

11. Need to prove that is A is diagonalizable then TA is diagonalizable.Proof.We assume that A is diagonalizable. Thus there exists an invertible matrix P such that

1 P AP D where D is a diagonal matrix.Taking the transpose of both sides of 1 P AP D gives

1 Because is a diagonal matrixT T P AP D D D

What is the Left Hand Side 1 TP AP equal to?

Using our rules on matrix transpose T T T TABC C B A we have

1 1

1 11Because

T TT T

TT T T T

P AP P A P

P A P P P

Since 11T T

P P therefore

11 1 1 1Because

TT T T T T T

P A P P A P D P AP D

By Definition (7.15):(7.15). A matrix A is diagonalizable if it is similar to a diagonal matrix.

We conclude that the matrix TA is diagonalizable.■

12. We are given the matrix3 5

0 2

A . The eigenvalues and eigenvectors are

1 2

1 53, and 2,

0 1

u v


The diagonalizing matrix is the eigenvector matrix1 5

0 1

P and3 0

0 2

D . The

formula for finding higher powers of matrices is:1m m A PD P

We need to find inverse matrix 1P :

1 1 5

0 1

P

Substituting 2, 3 and 4m and the matrices into this formula 1m m A PD P gives2

2 2 1

2

1 5 1 5 1 5 9 0 1 5 9 253 0

0 1 0 1 0 1 0 4 0 1 0 40 2

A PD P

Similarly we have

3 1 5 27 0 1 5 27 95

0 1 0 8 0 1 0 8

A

4 1 5 81 0 1 5 81 325

0 1 0 16 0 1 0 16

A

Substituting3 5

0 2

A , 2 9 25

0 4

A , 3 27 95

0 8

A and 4 81 325

0 16

A into

2 3 4

2 3 4

2 3 4

exp2! 3! 4!

1 0 3 5 9 25 27 95 81 325

0 1 0 2 0 4 0 8 0 162! 3! 4!

t t tt t

t t tt

A I A A A A

13. We are given that1 1

1 0

F . The eigenvalues are given by

2

1 1det det

1 0

1 0 1

1 0

F I

Solving the quadratic by using the formula2 4

2

b b ac

a with 1, 1a b and

1c gives

21 1 4 1 1 1 5

2 2

The eigevalues are 1

1 5

2

and 2

1 5

2

. What are the eigenvectors?

Let u be the eigenvector belonging to 1

1 5

2

.


22 1

1

1 51 1

1 5 22 1 5

1 02

1 51

2

1 51

2

1 0 1 5 1 5Because and

1 0 2 2

x

y

x

y

x

y

F I u

Expanding matrices we have

21

1

0gives and 1

0

x yx y

x y

Note that this is the solution because 1 2

1 5 1 51

2 2

. Thus 1

1

u is the

eigenvector belonging to 1

1 5

2

. Let v be the eigenvector belonging to 2

1 5

2

1

2

1 51 1

1 5 22 1 5

1 02

1 51

2

1 51

2

1 0

1 0

x

y

x

y

x

y

F I v

Expanding the matrices we have

12

2

0gives and 1

0

x yx y

x y

Thus 2

1

v is the eigenvector belonging to 2

1 5

2

.

Since the eigenvectors u and v are linearly independent therefore the eigenvector matrix P is

given by 1 2

1 1

P u v .

The eigenvalue matrix 1

2

0

0

D .

To ensure that we have the correct matrices P and D we check that PD FP :


2 211 2 1 2

2 1 2

0

01 1

PD

1 21 2

1 2

1 11 1

1 0 1 1

FP

The bottom rows of PD and FP are equal but the top rows seem to be different. Can we showthat these are equivalent?

2

21

1

1 5 1 2 5 5

2 4

6 2 5 3 5 1 51 1

4 2 2

Similarly we have 22 21 .

PD FP so we have the correct eigenvector and eigenvalues matrices P and D respectively.14. By Question 3 of the last Exercise 7(b) the characteristic polynomial p is given by

2 detp tr A A

We find by equating this to zero and solving the resulting quadratic equation by using theformula with 1, and deta b tr c A A

2

2

2

det 0

4

2

4det

2

tr

b b ac

a

tr tr

A A

A A A

Since we are given that 24dettr A A therefore we have 2 distinct roots of the

quadratic, say 2

1

4det

2

tr tr

A A A

and 2

2

4det

2

tr tr

A A A

.

Since we have a 2 by 2 matrix and 2 distinct eigenvalues therefore by Proposition (7.18):Proposition (7.18). If a square n by n matrix A has n distinct eigenvalues then the matrix A isdiagonalizable.We conclude that the matrix A is diagonalizable.

■

15. Required to prove that if A is an invertible and diagonalizable matrix then 1A is alsodiagonalizable.Proof.Assuming A is diagonalizable means that there exists matrices P and D where D is a diagonalmatrix such that

1 P AP DUsing our rules of matrices we have


1

1 1

11 1

1 11 1 1 1 1 1

Pre-multiplying by and Post multiplying by

Applying

P AP D

A PDP P P

A PDP

P D P ABC C B A

This last result 11 1 1 1 A P D P means that 1D and 1A are similar matrices because:

Definition (7.12). A square matrix B is similar to a matrix A if there exists an invertiblematrix P such that 1 P AP B .

1D is a diagonal matrix therefore 1A is diagonalizable.■

16. We need to prove that if A is diagonalizable then mA (where m ) is diagonalizable.Proof.A is diagonalizable therefore by Proposition (7.19):

Proposition (7.19). If a n by n matrix A is diagonalizable with 1 P AP D where D is adiagonal matrix then 1m m A PD PWe have

1 11 1 1 1Rememberm m m A PD P P D P P P

Thus mA is similar to mD and the result of question 9 we have mD is a diagonal matrix. ByDefinition (7.15):Definition (7.15). A n by n matrix A is diagonalizable if it is similar to a diagonal matrix.We conclude that the matrix A is diagonalizable.

■

17. Required to prove that AB is similar to BA provided A and B are invertible matrices.Proof.Let the invertible matrix P A then 1 1 P A and

1 1

1 Because

P AB P A AB A

A A BA A BC AB C

IBA BA

Since we have 1 P AB P BA therefore by Definition (7.12):

Definition (7.12). A square matrix B is similar to a matrix A if there exists an invertiblematrix P such that 1 P AP B .We conclude that AB is similar to BA.

■

18. (i) Need to prove that if matrices A and B are similar then tr trA B where tr is

trace.Proof.Assume matrices A and B are similar. Then by Proposition (7.14):

(7.14). Let A and B be similar matrices. The eigenvalues of these matrices are identical.


We have matrices A and B have the same eigenvalues, call them 1 2 3, , , and n .

Then by Proposition (7.9) part (b):

(7.9) (b) Trace of a matrix A is given by sum of eigenvalues, 1 2 3 ntr A

We have

1 2 3 ntr tr A BHence we have our required result, that is tr trA B .

■(ii) Required to prove that if matrices A and B are similar then det detA B .

Proof.Suppose matrices A and B are similar. By Proposition (7.14) matrices A and B have the sameeigenvalues and by (7.9) part (a):

(7.9) (a) The determinant of the matrix A is given by 1 2 3det n A .

Hence det detA B .

■19. Let 1 2 3, , , and n be the eigenvalues of matrix A. We are given that

1 2 31, 1, 1, and 1n We are also told that the matrix A is diagonalizable which means that by Proposition (7.19)we have

1m m A PD PWhat is mD equal to?

1 0

0 0

0

0

0

m

m

m

n

D

By using the hint in the question we know that if 1x then 0mx as m . Applying

this to 1 2 3, , , ,m m m m

n we have 1 2lim lim , , lim 0m m m

nm m m

.

1 0 0 0

lim 0 0 0 0

0 00

0 0

00

m

m

mm

n

D O

Thus as m the matrix mD is the zero matrix. Hence as m we have1 1m m A PD P POP O

This means that as m the matrix mA is the zero matrix.

20. We need to prove that the eigevalues of mA are 1 2, , ,m m m

n providing

A is diagonalizable with eigenvalues 1 2, , , n .

Proof.We are told that A is diagonalizable so by Proposition (7.19):

Proposition (7.19). If A is diagonalizable then 1m m A PD PWe have

1m m A PD P (*)


where D is the diagonal matrix with eigenvalues along the leading diagonal:

1 0

0 n

D

What is mD equal to?

1 0

0

m

m

m

n

D

From (*) we have mA is similar to mD and by Proposition (7.13) part (b) the diagonal matrixmD is similar to mA . Therefore

1 m m P A P DThe eigenvalues of mA are the entries on the leading diagonal in mD . Thus

1 2, , ,m m m

n are the eigenvalues of mA .

21. Theorem (7.16) is the following:A n by n matrix A is diagonalizable it has n linearly independent eigenvectors.How do we prove this theorem?We have the arrows pointing in both directions therefore we need to prove

1) A is diagonalizable it has n linearly independent eigenvectors.2) A has n linearly independent eigenvectors A is diagonalizable.

Proof.

. Assume the matrix A is diagonalizable and the eigenvectors of A are

1 2 3, , , , nv v v vThis means there is an invertible matrix P such that 1 P AP D where D is a diagonalmatrix. Remember eigenvector matrix P contains the eigenvectors of matrix A:

1 2 3 nP v v v vMatrix P is invertible and from chapter 2 result (2.23):

Proposition (2.23) . Let A be the n by n matrix whose columns are given by the vectors

1 2 3, , , and nvv v v :

1 2 nA v v vThen vectors 1 2, , , nv v v are linearly independent matrix A is invertible.

Hence the eigenvectors are linearly independent.

. Assume matrix A has n linearly independent eigenvectors belonging to the eigenvalues

1 2 3, , , , n respectively and

1 2 3 nP v v v vConsider the matrix multiplication 1P AP . We can carry out this matrix multiplication as

1 1 P AP P AP :

1 2 3

1 2 3

n

n

AP A v v v v

Av Av Av Av

Since 1 2, , , nv v v are eigenvectors of matrix A so by the definition (7.1):


(7.1) Au u

We have 1 1 1 2 2 2 3 3 3, , , , n n n Av v Av v Av v Av v . Therefore

1 1 2 2 3 3 n n AP v v v v

We can also evaluate PD where D is a diagonal matrix. Let1 0

0 0

0

0

0 n

k

k

D then

1

1 2 3 1 1 2 2 3 3

0

0 0

0

0

0n n n

n

k

k k k k

k

PD v v v v v v v v

Let 1 1 2 2, , , n nk k k then we have

1 1 2 2 3 3 n nk k k k PD v v v v APLeft multiplying both sides of PD AP by 1P gives

1 1 1 P PD P P D ID D P AP

Hence 1 P AP D means the eigenvector matrix P diagonalizes the given matrix A. Hencematrix A is diagonalizable.

■

Complete Solutions 7(d) 1

Complete Solutions to Exercise 7(d)Provide solutions for questions 10 and 11.

1. (a) We are given the matrix1 0

0 2

A and we need to find an orthogonal matrix Q

which diagonlizes this matrix. The eigenvalues and eigenvectors of matrix A are:

1

11,

0

u and 2

02,

1

v

The eigenvectors u and v are orthogonal and are also have a norm of 1 so we do not needto normalize them. Thus

1 0

0 1

Q u v I

Check that T Q AQ D where1 0

0 2

D A :

T Q AQ IAI ANote that matrix A is already a diagonal matrix so no surprise that the orthogonaldiagonalizing matrix is the identity matrix.

(b) We are given the matrix1 1

1 1

A . Need to find the eigenvalues and associated

eigenvectors:

1

10,

1

u and 2

12,

1

v

Since A is a symmetric matrix therefore eigenvectors u and v are orthogonal. Is theorthogonal matrix Q u v ?

No because u and v are not normalized. We need to normalize these eigenvectors. How?By dividing by the norm (length). What is the norm of u equal to?

2

2 221 1 11 1 2

1 1 1

u

This is the norm squared2

2u , so taking the square root of both sides gives 2u .

Similarly we have

2 2 21 11 1 2

1 1

v

Thus 2v . What are our perpendicular unit eigenvectors u and v equal to?

11 1

12

u uu

and 11 1

12

v v

v

Hence our orthogonal matrix 1 11

1 12

Q u v . Checking that AQ QD :

1 1 1 1 0 21 1

1 1 1 1 0 22 2

1 1 0 0 0 21 1

1 1 0 2 0 22 2

AQ

QD


(c) Similarly we are given2 1

1 2

A and we need to find the eigenvalues and

corresponding eigenvectors:

1

11,

1

u and 2

13,

1

v

Again we need to normalize the eigenvectors:

2 221 11 1 2

1 1

u

Thus 2u and similarly we have 2v . Therefore our orthonormal eigenvectors

are

11 1

12

u uu

and 11 1

12

v v

v

Hence our orthogonal matrix 1 11

1 12

Q u v . Checking that AQ QD with

2 1 1 11 and

1 2 1 12

A Q :

2 1 1 1 1 31 1

1 2 1 1 1 32 2

1 1 1 0 1 31 1

1 1 0 3 1 32 2

AQ

QD

(d) We need to find an orthogonal matrix Q that diagonalizes5 12

12 5

A . First we

determine the eigenvalues and eigenvectors:

1

313,

2

u and 2

213,

3

v

Normalising these eigenvectors gives

2 2 23 33 2 13

2 2

u

Hence 13u . Similarly we have 13v . Our orthonormal eigenvectors are

31

213

u and

21

313

v

Our orthogonal vector is 3 21

2 313

Q u v . Checking that AQ QD :

5 12 3 2 39 261 1

12 5 2 3 26 3913 13

3 2 13 0 39 261 1

2 3 0 13 26 3913 13

AQ

QD

2. The working out is very similar to question 1.


(a) We are given matrix9 3

3 1

A . Determining the eigenvalues and eigenvectors gives

1

10,

3

u and 2

310,

1

v

Normalising the eigenvector u and v yields:

2 2 21 11 3 10

3 3

u

2 2 23 33 1 10

1 1

v

Taking the square root of both the norm squares gives 10u and 10v . What are

the orthonormal (perpendicular unit) eigenvectors?

11 1

310

u u

uand

31 1

110

v v

v

Thus 1 31

3 110

Q u v . We can check AQ QD :

9 3 1 3 0 301 1

3 1 3 1 0 1010 10

1 3 0 0 0 301 1

3 1 0 10 0 1010 10

AQ

QD

(b) Similarly for3 2

2 2

A we have

1 4 ,2

2

u and 2

21,

2

v

What else do we need to do?Normalize the eigenvectors u and v:

22 22 2

2 2 62 2

u

22 22 2

2 2 62 2

v

Taking the square root of both sides gives 6 u v . What are the perpendicular unit

eigenvectors?

21 1

6 2

u u

uand 1 1 2

6 2

v v

v

Our orthogonal matrix is given by 2 21

6 2 2

Q u v . Checking AQ QD :


3 2 2 2 8 21 1

6 62 2 2 2 4 2 2

2 2 4 0 8 21 1

0 16 62 2 4 2 2

AQ

QD

(c) We need to find an orthogonal matrix which diagonalizes5 3

3 3

A . The

eigenvectors and eigenvalues of this matrix are:

1

36,

3

u and 2

32,

3

v

We need to normalize the eigenvectors. How?By dividing each vector by its norm:

22 23 33 3 12

3 3

u

22 23 33 3 12

3 3

v

By taking the square root we have 12 u v . What are the normalized eigenvectors

u and v equal to?

31 1

12 3

u u

uand 1 1 3

12 3

v v

v

Hence the orthogonal vector is 3 31

12 3 3

Q u v . We can check AQ QD :

5 3 3 3 18 2 31 1

12 123 3 3 3 6 3 6

3 3 6 0 18 2 31 1

0 212 123 3 6 3 6

AQ

QD

(d) For the given matrix5 12

12 1

A we need to find an orthogonal matrix which

diagonalizes A. First we find the eigenvalues and eigenvectors:

1

21,

12

u and 2

127,

2

v

Normalising the eigenvectors gives:

22 22 22 12 16

12 12

u

We have2

16u . Taking the square root gives 4u . Similarly we have 4v .

Normalized eigenvectors are


21

4 12

u and 1 12

4 2

v

What is the orthogonal matrix Q equal to?

2 121

4 12 2

2 2 31Because 12 4 3 2 3

4 2 3 2

1 3 Taking out 2 and1

writing 2/4 1/ 22 3 1

Q u v

Checking that AQ QD with5 12 5 2 3

12 1 2 3 1

A :

5 2 3 1 3 1 7 31 1

2 22 3 1 3 1 3 7

1 3 1 0 1 7 31 1

0 72 23 1 3 7

AQ

QD

3. (a) We are given the matrix

1 0 0

0 2 0

0 0 3

A . Since A is a diagonal matrix so the

orthogonal matrix is the identity matrix I. The matrix has eigenvalues 1, 2 and 3.Thus the orthogonal matrix is given by

1 0 0

0 1 0

0 0 1

Q I

Checking that T T Q AQ I AI A . The diagonal matrix is the given matrix A.

(b) Steps 1 and 2: We are given the matrix

2 2 2

2 2 2

2 2 2

A . The eigenvalues and

corresponding eigenvectors are

1 2 3

1 1 1

0, 1 , 0, 1 and 6, 1

2 0 1

u v w

Step 3: We need to check that the eigenvectors u and v are orthogonal since they belong tothe same eigenvalue 1 20 and 0 . We have

1 1

1 1 1 1 1 1 2 0 0

2 0

u v


Because 0 u v therefore u and v are orthogonal.Step 4: We need to normalize these eigenvectors:

22 2

22 2

2 2 2

1 11 1

1 161 1 2 2 2

1 11 1

1 121 1 0 0 0

1 11 1

1 131 1 1 1 1

u

v

w

Step 5: Since the eigenvectors , andu v w are orthonormal vectors therefore

1/ 6 1/ 2 1/ 3

1/ 6 1/ 2 1/ 3

2 / 6 0 1/ 3

Q u v w

Step 6: By using MATLAB or otherwise you can check that we have0 0 0

0 0 0

0 0 6

T

Q AQ D

(c) Steps 1 and 2: For the given matrix

0 0 0

0 1 1

0 1 1

A we have

1 2 3

1 0 0

0, 0 , 0, 1 and 2, 1

0 1 1

u v w

Step 3: We need to check that the eigenvectors u and v are orthogonal since they belong tothe same eigenvalue 1 20 and 0 . We have

1 0

0 1 1 0 0 1 0 1 0

0 1

u v

Because 0 u v therefore the eigenvectors u and v are orthogonal.Step 4: The eigenvector u is already normalized but we need to normalize v and w:

2 22 2

2 2 2 2

0 0

1 1 0 +1 1 2

1 1

0 0

1 1 0 1 1 2

1 1

v

w


Taking the square root gives 2 and 2 v w . Normalized eigenvectors are

0 0

1 1 1 11 and 1

2 21 1

v v w wv w

Step 5:The orthogonal matrix Q is given by

1 0 0

0 1/ 2 1/ 2

0 1/ 2 1/ 2

12 0 0 Taking out and

1 20 1 1

12 rewriting 1 20 1 12

Q u v w

Step 6:You may like to check that T Q AQ D where D is the diagonal matrix given by

0 0 0

0 0 0

0 0 2

D

4. (a) Steps1 and 2:

We are given the symmetric matrix

1 2 2

2 1 2

2 2 1

A and we first find the eigenvalues and

the corresponding eigenvectors:

1 2 3

2 2 1

1, 1 , 1, 3 and 5, 1

1 1 1

u v w

Step 3: This time the eigenvectors u and v are not orthogonal because

2 2

1 3 2 2 1 3 1 1 6

1 1

u v (*)

We need to convert u and v, eigenvectors belonging to the same eigenvalue 1 2 1 ,

into an orthogonal set of vectors. How?By applying the Gram Schmidt Process (4.16) which is

1

2

1

1

q u

12 12

1

v qq v q

q(**)

What is2

1 1 andv q q equal to?


Since 1 q u therefore 1By (*)

6 v q v u u v . What is2

1q equal to?

2 2 2 21 1 1

2 2

1 1 2 1 1 6

1 1

q q q u u †

Substituting 1 6 v q and2

1 6q into 12 12

1

v qq v q

qgives

1

2 12

1

2 2 2 2 0 06

3 1 3 1 2 2 16

1 1 1 1 2 1

v qq v q

q

We can ignore the scalar 2 in 2q and normalize *2

0

1

1

q .

Now *1 2, andq q w are orthogonal. (Check by finding the dot product.)

What else do we need to do in order to find the orthogonal matrix Q?Step 4: Need to normalize *

1 2, andq q w :

2

1

1 11 1

2By † 61 1

16 and 61

qq q

q q

* *2 2 2 2

2

0 01 1 1

1 121 1 1 1

q qq

2 2 2

1 11 1 1

1 131 1 1 1 1

w ww

Step 5: Our orthonormal vectors are *1 2, andq q w . Thus

*1 2

2 / 6 0 1/ 3

1/ 6 1/ 2 1/ 3

1/ 6 1/ 2 1/ 3

Q q q w

Step 6: You can check that T Q AQ D where D is a diagonal matrix by using MATLAB.MATLAB instructions are as follows:A=[1 2 2; 2 1 2; 2 2 1]

A =

1 2 2

2 1 2

2 2 1


Q=[-2/sqrt(6) 0 1/sqrt(3); 1/sqrt(6) -1/sqrt(2) 1/sqrt(3); 1/sqrt(6) 1/sqrt(2) 1/sqrt(3)]

Q =

-0.8165 0 0.5774

0.4082 -0.7071 0.5774

0.4082 0.7071 0.5774

>> Q'*A*Q

ans =

-1.0000 -0.0000 0.0000

0 -1.0000 0

0 0 5.0000Thus we have the diagonal matrix D with the entries on the leading diagonal as theeigenvalues of the matrix A.

(b) Steps1 and 2:

We are given the symmetric matrix

2 1 1

1 2 1

1 1 2

A which is matrix in part (a) but with the

1’s and 2’s interchanged. The eigenvalues and eigenvectors of this matrix are

1 2 3

1 1 1

1, 1 , 1, 0 and 4, 1

0 1 1

u v w

Step 3: Again the eigenvectors u and v belonging to the same eigenvalue are notorthogonal because

21 1

1 0 1 0 0 1

0 1

u v (†)

Need to use the Gram Schmidt Process to convert these u and v into orthogonal vectors:

1

1

1

0

q u

12 12

1

v qq v q

q

Since 1 q u therefore 1By (†)

1 v q v u u v . Also

2 2 2 21 1 1

1 1

1 1 1 1 0 2

0 0

q q q u u


1 2q into 12 12

1

v qq v q

qgives


12 12

1

1 1 1 1/ 2 1/ 2 11 1

0 1 0 1/ 2 1/ 2 12 2

1 0 1 1 2

v qq v q

q

Now 1 2, andq q w are orthogonal. (You may check by showing the dot product is zero.)

Step 4: Need to normalize 1 2, andq q w :

1 1

1

11 1

12

0

q qq

2

1Because 2 q

Remember for normalising we can ignore the fraction. For normalising 2q we ignore the

½:

* *2 2* 2 2 2

2

1 11 1 1

1 161 1 2 2 2

q qq

2 2 2

1 11 1 1

1 131 1 1 1 1

w ww

Step 5: Our orthonormal vectors are *1 2, andq q w . Thus

*1 2

1/ 2 1/ 6 1/ 3

1/ 2 1/ 6 1/ 3

0 2 / 6 1/ 3

Q q q w

Step 6: You can check that T Q AQ D where D is a diagonal matrix by using MATLAB.MATLAB instructions are as follows:A=[2 1 1; 1 2 1; 1 1 2]

A =

2 1 1

1 2 1

1 1 2

>> Q=[-1/sqrt(2) -1/sqrt(6) 1/sqrt(3); 1/sqrt(2) -1/sqrt(6) 1/sqrt(3); 0 2/sqrt(6) 1/sqrt(3)]

Q =

-0.7071 -0.4082 0.5774

0.7071 -0.4082 0.5774

0 0.8165 0.5774

>> Q'*A*Q

ans =

1.0000 0 0

0 1.0000 0.0000

0 0 4.0000


(c) Steps1 and 2:

We are given the matrix

5 4 2

4 5 2

2 2 8

A . The eigenvalues and corresponding

eigenvectors are

1 2 3

1 1 2

9, 1 , 9, 0 and 0, 2

0 2 1

u v w

Step 3:Check the eigenvectors u and v belonging to the same eigenvalue 1 2 9 are

orthogonal:

1 1

1 0 1 1 1 0 0 2 1

0 2

u v (☺)

Thus u and v are not orthogonal so we need to use the Gram Schmidt Process to placethese into an orthogonal set 1q and 2q say:

1u q

12 12

1

v qq v q

q(*)

Similar to solutions to parts (a) and (b) above we have

1 1 v q v u u v [By (☺)]

2 2 2 21

1 1

1 1 1 1 0 2

0 0

q


1 2q into (*) gives

12 12

1

1 11

0 12

2 0

1 1 1 1/ 2 1/ 2 11 1

0 1 0 1/ 2 1/ 2 12 2

2 0 2 0 2 4

v qq v q

q

Note that the vectors 1q and 2q are a basis for the eigenspace 9E .

Step 4:We need to normalize our orthogonal set of vectors 1q , 2q and w.

1 1

1

11 1

12

0

q qq

Remember to normalize 2q we can remove the fraction and normalize *2q where *

2q is the

vector 2q but without the ½. What is the norm of *2q ?


2 2* 2 22

1 1

1 1 1 1 4 18

4 4

q

Taking the square root of both sides gives *2 18 3 2 q . We have

* *2 2*

2

11 1

13 2

4

q qq

The norm squared of w is

2 2 2 2

2 2

2 2 2 2 1 9

1 1

w

Taking the square root gives 9 3 w . The normalized vector is

2

1 12

31

w ww

Step 5: Form the matrix Q whose columns are the orthonormal vectors 1

11

12

0

q ,

*2

11

13 2

4

q and 2

12

31

w :

*1 2

1/ 2 1/ 3 2 2 / 3

1/ 2 1/ 3 2 2 / 3

0 4 / 3 2 1/ 3

Q q q w

Step 6: Check that 1 T Q AQ Q AQ D by using MATLAB.>> A=[-5 4 2; 4 -5 2; 2 2 -8]

A =

-5 4 2

4 -5 2

2 2 -8>> Q=[-1/sqrt(2) 1/(3*sqrt(2)) 2/3; 1/sqrt(2) 1/(3*sqrt(2)) 2/3; 0 -4/(3*sqrt(2)) 1/3]

Q =

-0.7071 0.2357 0.6667

0.7071 0.2357 0.6667

0 -0.9428 0.3333

>> Q'*A*Q

ans =

-9.0000 0 0

0 -9.0000 0

0 0.0000 0.0000


5. The matrix1 1

1 1

A is identical to the matrix of question 1(b).

Our orthogonal matrix Q was given by1 11

1 12

Q .

To find the powers of matrix A we use the formula m m TA QD Q where D is the diagonal

matrix whose leading diagonal entries are the eigenvalues of A, that is0 0

0 2

D .

10

10 10

10

10

9

9 9

1 1 0 0 1 11 1

1 1 0 2 1 12 2

1 1 0 0 1 11

1 1 0 2 1 12

1 1 0 0 1 12

1 1 0 1 1 12

0 1 1 12

0 1 1 1

1 12 2

1 1

T

A QD Q

A

To prove that 12m mA A we use the above with m in place of 10.

1 1

1 1 0 0 1 11

1 1 0 2 1 12

1 1 0 0 1 12

1 1 0 2 1 12

1 12 2

1 1

m m Tm

m

m

m m

A QD Q

A

6. We need to prove that if A is a diagonal matrix then orthogonal diagonalizing matrixQ I .

Proof.Let A be a diagonal matrix and if Q I then

1 1 Q AQ I AI IAI ATherefore the identity matrix I orthogonally diagonalizes A. Thus our orthogonal matrixQ I .

■

7 (a) We need to prove that the zero matrix O is orthogonally diagonalizable.Proof.The eigenvalues of the zero matrix O is given by

det 0 O I which gives 1 2 3 0n The corresponding eigenvectors for these 1 2 3 0n are


1 2

1 0 0

0 1, , ,

0 0

0 1

n

v v v

The orthogonal matrix is 1 2 n Q v v v I . Thus

1 1 Q OQ I OI IOI OThus the zero matrix O is diagonalizable.We could also use the result of question 6 above since O is a diagonal matrix.

■(b) We need to prove that the identity matrix I is orthogonally diagonalizable.Proof.Since the identity matrix is a diagonal matrix therefore we can use the result of question 6to deduce that the identity matrix is diagonalizable and the orthogonal matrix Q I .

■

8. We need to show thata b

b c

A is diagonalizable and find the orthogonal matrix Q.

Since A is a symmetric matrix therefore by Theorem (7.27) we conclude that the matrix Ais orthogonally diagonalizable. The eigenvalues of matrix A are given by

2

2 2

det det

0 (*)

a b

b c

a c b

a c ac b

A I

We have a quadratic equation which we can solve by using the quadratic equation formula:

2 2

2 2 2

2 2 2

2 2

4

2

2 4 4

2

2 4

2

4

2

a c a c ac b

a c a c ac ac b

a c a c ac b

a c a c b

Thus we have 2 22 2

1 2

4 4 and

2 2

a c a c b a c a c b

.

What are the eigenvectors u and v equal to?Let u be the eigenvector for 1 then:

11

1

0

0

a b x

b c y

A I u

We have the simultaneous equations:


1

1

0

0

a x by

bx c y

Solving these for x and y gives 1,x b y a because substituting these

1,x b y a into the above equations gives

1 1 1

1 1 1

2 21 1 1

2 21 1

By (*)

0

0

a x by a b b a

bx c y bb c a

b c ca a

c a ca b

Thus1

b

a

u . Similarly let v be the eigenvector for 2 then

22

2

0

0

a b x

b c y

A I u

This gives the simultaneous equations

2

2

0

0

a x by

bx c y

Solving these gives 2 ,x c y b because substituting these 2 ,x c y b into the

above equations gives

2 2 2

2 22 2 2

2 22 2

By (*)

2 2

0

0

a x by a c bb

a ac c b

a c ac b

b c c b

Hence the eigenvector is 2 c

b

v .

We need to show that the eigenvectors are orthogonal because we may have a repeatedeigenvalue such as 1 2 :

22 1

1

2 1 **

b cb c a b

a b

b a c

u v

How do we show this is zero?By using the hint which was

2 0x px q has roots a and b then a b p and (*) from above

2 2 0 (*)a c ac b

has the roots 1 2 and . Therefore 1 2 a c and substituting this into (**) we have

2 1 0 0b a c b u v

Hence u and v are orthogonal set of vectors. We need to normalize these eigenvectors:


2

2

1 1 1

221

b b b

a a a

b a

u

Taking the square root gives the norm 221b a u . The normalized vector is

22

11

1 1 b

ab a

u uu

Similarly we normalize the eigenvector v:

2 22 2 22

c cc b

b b

v

Taking the square root of both sides gives the norm 2 22 c b v .

2

2 22

1 1 c

bc b

v v

v

What is our orthogonal vector which diagonalizes the given matrix A?

2

2 22 21 2

1

2 22 21 2

cb

b a c b

a b

b a c b

Q u v

9. We need to prove that if Q orthogonally diagonalizes the matrix A then Q alsodiagonalizes the matrix 1A .Proof.We have T Q AQ D because Q diagonalizes the matrix A. Taking the inverse of bothsides we have

11

1 11 1 1 1 1

11 1 1 1

11 1 1

Because

Since is orthogonal so

Because

T

T

T

D Q AQ

Q A Q ABC C B A

Q A Q Q Q Q

Q A Q Q Q

We have 1 1 1 Q A Q D where 1D is the diagonal matrix with the leading diagonal

entries1 2 3

1 1 1 1, , , and

n which are the eigenvalues of 1A . Thus Q orthogonally

diagonalizes the matrix 1A .■

10. We need to prove that (7.23) which claims:

Let A be a symmetric matrix with distinct eigenvalues 1 2, , , n and

corresponding eigenvectors 1 2, ,v v and nv . Then these eigenvectors are orthogonal.


We prove this result by induction. The 3 steps of induction are:1. Check the result is correct for some base case 0n k .

2. Assume the result is true for n k .3. Prove the result for 1n k by using steps 1 and 2.Proof.Step 1:For 2n the result is true because of:

Proposition (7.22). Let A be a symmetric matrix. If 1 and 2 are distinct

eigenvalues of matrix A then their corresponding eigenvectors u and v respectivelyare orthogonal.

Step 2:Assume the result is true for n k :The eigenvectors 1 2, , , kv v v which belong to 1 2, , , k are orthogonal.

Step 3:Required to prove this result for 1n k . We need to prove:

The eigenvectors 1 2 1, , , kv v v which belong to 1 2 1, , , k are orthogonal.

Consider 1 2 1, , , ,k kv v v v . By step 2 we know that the eigenvectors

1 2, , , kv v v are orthogonal. Since the eigenvalue 1k associated with 1kv is

distinct from each of the other eigenvalues 1 2, , , k so by step 1 the eigenvector

1kv is orthogonal to each of the eigenvectors 1 2, , , kv v v . This is our required

result.■

Complete Solutions 7(e) 1

Complete Solutions 7(e)

1. (a) The eigenvalues and eigenvectors of matrix TA A are

1 1

11,

0

v and 2 2

04,

1

v

The singular values of matrix A are 1 21 1 and 4 2 . Using 1 1 1 u Av and

2 2 2 u Av we have

1

1 0 1 1

0 2 0 0

u and 2

1 0 0 02

0 2 1 2

u

Hence from this last result 2

02

2

u we have 2

0 01

2 12

u .

We have 1 2

1 0

0 1

U u u , 1

2

0 1 0

0 0 2

D and 1 2

1 0

0 1

V v v .

Note that U V I . The triple factorization of the given matrix A isT A UDV IDI D

(b) The eigenvalues and normalized eigenvectors of matrix TA A are

1 1

1181,

25

v and 2 2

211,

15

v

The singular values of matrix A are 1 281 9 and 1 . Using 1 1 1 u Av and

2 2 2 u Av we have

1

1 4 1 91 19

4 7 2 185 5

u and 2

1 4 2 21 1

4 7 1 15 5

u

We have 1

9 11 1

18 29 5 5

u . Substituting this and 2u gives

1 2

1 21

2 15

U u u , 1

2

0 9 0

0 0 1

D and

1 2

1 2 1 21 1

2 1 2 15 5

TTT

V v v .

You may like to check the factorization TA UDV .

(c) The eigenvalues and normalized eigenvectors of matrix TA A are

1 1

116,

25

v and 2 2

211,

15

v

The singular values of matrix A are 1 26 and 1 .

Using 1 11

1

u Av and 2 2

2

1

u Av we have


1 1

1 0 111 1 1

0 1 226 5 30

1 2 5

u Av and 2 2

1 0 221 1

0 1 115 5

1 2 0

u Av

Since A is a 3 by 2 matrix so U is a 3 by 3 matrix which means we need to find the vector

3u which is orthogonal to both 1u and 2u :

3

11 2 5 0

1, 2, 1 22 1 0 0

1

x

y x y z

z

u

Normalizing the vector 3u gives

3

11

26

1

u

We have 1

11

230

5

u , 2

21

15

0

u and 3

11

26

1

u .

Substituting these and the above into U, D and V gives:

1 2 3

1/ 30 2 / 5 1/ 6

2 / 30 1/ 5 2 / 6

5 / 30 0 1/ 6

U u u u ,1

2

6 00

0 10

0 00 0

D and

1 2

1 2 1 21 1

2 1 2 15 5

TTT

V v v


(d) The product

1 0 1

0 1 2

1 2 5

T

A A . The eigenvalues and normalized eigenvectors of

matrix TA A are

1 1

11

6, 230

5

v , 2 2

21

1, 15

0

v and 3 3

11

0, 26

1

v

The singular values of matrix A are 1 26 and 1 .

Using 1 11

1

u Av and 2 2

2

1

u Av we have


1 1

11 0 1 61 1 1 1

20 1 2 126 6 30 180

5

u Av and

2 2

21 0 1 21 1 1

10 1 2 11 5 5

0

u Av

Since A is a 2 by 3 matrix so U is a 2 by 2 matrix, D is a 2 by 3 matrix and V is a 3 by 3matrix:Substituting these and the above into U, D and V gives:

1 2

6 / 180 2 / 5

12 / 180 1/ 5

U u u , 1

2

0 0 6 0 00 0 0 1 0

D and

1 2 3

1/ 30 2 / 5 1/ 6

2 / 30 1/ 5 2 / 6

5 / 30 0 1/ 6

V v v v

You may like to check the factorization TA UDV by first transposing matrix V.

(e) The product1 3 1 1 10 10

1 3 3 3 10 10T

A A . The eigenvalues and normalized

eigenvectors of matrix TA A are

1 1

1120,

12

v and 2 2

110,

12

v

The positive singular value of matrix A is 1 20 .

Using 1 11

1

u Av we have

1 1

1 1 11 1 1

3 3 120 20 2

2 11 1Because 40 4 10 2 10

6 340 10

u Av

Since A is a 2 by 2 matrix so U is a 2 by 2 matrix. What is 2u equal to?

2u needs to be orthogonal to 1u which means 2 1 0 u u therefore by inspection and

normalizing we have

2

31

110

u

Substituting these and the above into U, D and V gives:

1 2

1 31

3 110

U u u , 1 0 20 00 0 0 0

D and

1 2

1 1 1 11 1

1 1 1 12 2

TTT

V v v



(f) The product

10 10 101 31 1 1

10 10 101 33 3 3

10 10 101 3

T

A A . The eigenvalues and normalized

eigenvectors of matrix TA A are

1 1

11

30, 13

1

v , 2 2

11

0, 12

0

v and 3 3

11

0, 16

2

v

The positive singular value of matrix A is 1 30 .

Using 1 11

1

u Av we have

1 1

11 1 11 1 1

13 3 320 30 3

1

3 11 1Because 90 9 10 3 10

9 390 10

u Av

Since A is a 2 by 3 matrix so U is a 2 by 2 matrix, D is a 2 by 3 matrix and V is a 3 by 3matrix. What is 2u equal to?

2u needs to be orthogonal to 1u . We need

2 1 0 u u

As in part (e) we have 2

31

110

u . We have

1 2

1 31

3 110

U u u ,30 0 0

0 0 0

D and

1 2 3

1/ 3 1/ 2 1/ 6 1/ 3 1/ 3 1/ 3

1/ 3 1/ 2 1/ 6 1/ 2 1/ 2 0

1/ 3 0 2 / 6 1/ 6 1/ 6 2 / 6

T

TT

V v v v


2. We need to prove that:

Let A be any matrix. Then the eigenvalues of TA A are positive or zero.

Proof.Let 1 2, , , n be the eigenvalues of TA A with eigenvectors 1 2, , , nv v vrespectively. For an arbitrary eigenvector jv we have

Tj j jA Av v (*)

By Proposition (7.31):

(7.31). Let A be any matrix. Then TA A is a symmetric matrix.


This means that TA A can be orthogonally diagonalized so the eigenvectors areorthonormal.

Consider the norm square2

jAv :

2

by (*)

Because 1

T

j j j j j

T Tj j

Tj j j j j j j j j j

Av Av Av Av Av

v A Av

v v v v v v v

We have2

0j j Av . Hence all the eigenvalues of TA A are positive or zero.

■

3. Required to prove:

Let matrix A have k positive singular values. Then the rank of matrix A is k.

Proof.Let A be a m by n matrix. By SVD:

(7.30). We can decompose any given matrix A of size m by n with singular values

1 2 0k where k m , into TUDV , that isTA UDV

where U is a m by m orthogonal matrix, D is a m by n matrix and V is an n by n orthogonalmatrix.

We have TA UDV where U and V are orthogonal matrices. Since U is orthogonal so it isinvertible because 1 T U U and similarly V is orthogonal so 1 T V V . Hence

1 11T V V V which means that TV is invertible. By hint we have:

T

T

rank rank

rank rank k

A UDV

DV D

This completes our proof.■

4. (a) We need to prove that 1 2, , , ku u u form an orthonormal basis for the column

space of matrix A.Proof.U is a m by m orthogonal matrix so the column vectors of matrix U are orthonormal:

1 2 nU u u uBy result of question 3 we have the rank of matrix A is k.By Proposition (3.29) of chapter 3:

(3.29). Row rankrank A of A Column rank of A

Hence the dimension of the column space is k which means we need k basis vectors for thecolumn space of matrix A.We are given that 1 2, , , k are positive and from the main theorem of the section

(7.30) for 1, 2, 3, ,j k we have


1j j

ju Av

By Proposition (3.36):

(3.36). The linear system Ax b has a solution b can be generated by the columnspace of matrix A.

Therefore ju is in the column space of matrix A. We have k orthonormal vectors

1 2, , , ku u u which are in the column space of matrix A. Since orthogonal vectors are

linearly independent so they 1 2, , , ku u u form an orthonormal basis for the column

space of matrix A.■

(b) We need to prove that 1 2, , , kv v v form an orthonormal basis for the row space of

matrix A.Proof.Since V is an n by n orthogonal matrix whose columns are the eigenvectors of TA A :

1 2 nV v v vThese eigenvectors are an orthonormal set of vectors because V is orthogonal.By result of question 3 we have the rank of matrix A is k. By Definition (3.28) of chapter 3:

(3.28). The rank of a matrix A is the row rank of A.

Hence the dimension of the row space is k which means we need k basis vectors for the rowspace. A subset of k vectors in the above, 1 2, , , kS v v v , is also orthonormal. As

these vectors are orthogonal so they are linearly independent which means they form abasis. Therefore S forms an orthonormal basis for the row space of matrix A.

■(c) Required to prove:

The set of vectors 1 2, , ,k k n v v v form an orthonormal basis for the null space of

matrix A.

Proof.By Theorem (3.34):

(3.34). If A is a matrix with n columns (number of unknowns) then

nullity rank n A A

We have the dimension of null space is n k and there are n k vectors in the set

1 2, , ,k k n v v v . Remember this set 1 2, , ,k k n v v v represent the orthonormal

eigenvectors of TA A which correspond to the zero eigenvalues 1 2 0k k k .

This means that for 1, 2, ,j k k k n we have

0Tj j j j A A v v v O

These vectors 1 2, , ,k k n v v v form an orthonormal basis for the null space of TA A .

The null space of TA A and A are identical. Hence 1 2, , ,k k n v v v form an

orthonormal basis for the null space of matrix A.■

5. Proof.The eigenvalues of TA A are unique. Why?


By Question 9 of Exercise 7(b):

Let A be a square matrix and be an eigenvalue with the corresponding eigenvector u.The eigenvalue is unique for the eigenvector u.

The singular values are given by the positive roots:

1 1 2 2, , , n n Therefore the singular values are unique.

■

6. We need to prove that the column vectors of matrix U in TA UDV are theeigenvectors of TAA .Proof.Using the singular value decomposition TA UDV we have

By using

Because

' where ' is a diagonal matrix

TT T T

T TT T T T T T T

TT T T T

T T T T

I

AA UDV UDV

UDV V D U XYZ Z Y X

UD V V D U X X

U DD U U D U D DD

Remember U is an orthogonal matrix so it inverse is given by TU . Left-multiplying theabove result 'T TAA U D U by TU and right-multiplying by U gives

' 'T T T T

I I

U AA U U U D U U D

Since 'T T U AA U D , the matrix U diagonalizes TAA and the columns of U are the

eigenvectors of TAA . This completes our proof.■

7. Required to prove that:

The singular values of A and TA are identical.

Proof.The singular values of a matrix A are given by the square roots of the eigenvalues

1 2, , , n of TAA

1 1 2 2, , , n n The singular values of a matrix TA are given by the square roots of the eigenvalues

1 2, , , nt t t of TTAA .

By Question 16 of Exercise 7(b):

The eigenvalues of the transposed matrix, TA , are exactly the eigenvalues of the matrix A.

Hence TTAA will have the same eigenvalues as TAA which means:

1 1 2 2, , , n nt t t Therefore the singular values of TA are the same. Hence the singular values of both A andthe transposed matrix TA are identical.

■


8. (a) We have to prove that:

The set of vectors 1 2, , , ku u u form an orthonormal basis for the range of T.

Proof.Let 1 2, , , ku u u be the first k column vectors of matrix U. By Proposition (7.34)

part(a):

(7.34) (a) The set of vectors 1 2, , , ku u u is an orthonormal basis for the column space

of matrix A.

Also we are given that T x Ax so by Proposition (5.13) of chapter 5:

(5.13). Let : n mT be a linear transformation given by T x Ax . Then

range T is the column space of A.

So the range of the transformation T is the column space of matrix A therefore

1 2, , , ku u u is an orthonormal basis for the range.

■(b) Required to prove that:

The set of vectors 1 2, , ,k k n v v v form an orthonormal basis for the kernel of T.

Proof.Remember the kernel of a transformation T are the vectors in the start vector space whichare mapped to the zero vector. In our case we have T x Ax so it is the vectors x which

satisfy T x Ax O . Of course this is the null space of matrix A. By Proposition (7.34):

(7.34) (c) The set of vectors 1 2, , ,k k n v v v form an orthonormal basis for the null

space of matrix A.

Hence 1 2, , ,k k n v v v is an orthonormal basis for kernel of T.

■

Complete Solutions Miscellaneous Exercise 7 1

Complete Solutions Miscellaneous Exercise 7

1. (a) What are the eigenvalues of1 1

0 3

A equal to?

Since we have a (upper) triangular matrix therefore the eigenvalues are the entries on theleading diagonal of A. We have 1 1 and 2 3 .

(b) Let u and v be the corresponding eigenvectors of 1 1 and 2 3 respectively. We

can find these by

1 1 1 0 1 0

0 3 1 0 2 0

x x

y y

A I u

This gives 1x and 0y . Thus1

0

u is the eigenvector for 1 1 . Similarly for

2 3 we have

1 3 1 2 1 0

30 3 3 0 0 0

x x

y y

A I v

This gives 1x and 2y . Thus1

2

v . What is the matrix Q equal to?

1 1

0 2

Q u v . What is the diagonal matrix D equal to?

D is the diagonal matrix with entries along the leading diagonal given by the eigenvalues.

We have 1

2

0 1 0

0 0 3

D .

(c) How do we find 5A ?

By applying (7.19) 1m m A QD Q . We need to find the inverse of1 1

0 2

Q :

1 2 11

0 12

Q

Substituting1 1

0 2

Q ,1 0

0 3

D and 1 2 11

0 12

Q into 1m m A QD Q with

5m gives5

5

5

5

1 1 1 0 2 11

0 2 0 3 0 12

1 1 2 11 01

0 2 0 12 0 3

1 1 1 0 2 11

0 2 0 243 0 12

1 243 2 11

0 486 0 12

2 242 1 1211

0 486 0 2432

A


Hence 5 1 121

0 243

A .

2. To find 1, 000, 001A of4 5

3 4A

we need to determine the characteristic equation

Ap .

2

4 5det det

3 4

4 4 15

4 4 15

1 0

Ap

A I

By the Cayley Hamilton Theorem we have2 2 which gives A I O A I

Using the rules of indices on matrices we have

500, 0001, 000, 001 2 4 5

3 4

A A A IA A

3. We need to find the eigenvalues and eigenvectors of

0 2 2

2 0 2

2 2 0

A

. Let be the

eigenvalues:

2

2

2 2

det det 2 2

2 2

2 2 2 2det 2det 2det

2 2 2 2

4 2 2 4 2 4 2

2 2 4 2 4 2

2 2 8 2

2 2 8

2 2 8

2 2 4 0

A I

We have 1, 2 2 and 3 4 . Let u be the eigenvector belonging to 1, 2 2 :

2 2 2 0

2 2 2 2 0

2 2 2 0

x

y

z

A I u

Putting the matrix into reduced row echelon form gives


1 1 1 0

0 0 0 0

0 0 0 0

x y z

Since we have 3 unknowns and only 1 non-zero equation therefore there are 3 1 2 freevariables. The expansion of the above gives 0x y z . Let andz s y t thenx t s . Thus our general eigenvector is given by

1 1

0 1

1 0

x s t

y t s t

z s

u

A basis for the eigenspace 2E is

1 1

0 , 1

1 0

. [In this case we do not need to choose

our eigenvectors so that they are orthogonal.]Let v be the eigenvector for 3 4 :

4 2 2 0

4 2 4 2 0 gives 1

2 2 4 0

x

y x y z

z

A I v


1

1

1

.

4. We have the following answers:(a) Let A be an invertible n n matrix with eigenvalue . Then 1 is an eigenvalue of

1A .(b) An n n matrix A is diagonalizable if and only if A has n linearly independenteigenvectors .

5. We are given the matrix7 5

3 7

A . The eigenvalues are given by

2 2

7 5det det

3 7

7 7 15

49 15 64 8 8 0

A I

Our two eigenvalues are 1 8 and 2 8 . Let u and v be the corresponding

eigenvectors of 1 8 and 2 8 respectively. We have

7 8 5 0

83 7 8 0

1 5 0 gives 5 and 1

3 15 0

x

y

xx y

y

A I u


Similarly we have

7 8 5 0

83 7 8 0

15 5 0 gives 1 and 3

3 1 0

x

y

xx y

y

A I v

Thus 1 2

5 18, and 8,

1 3

u v .

(a) The matrix 5 1

1 3

S u v and the diagonal matrix is8 0

0 8

.

(b) Let D be the diagonal matrix for B. We know from part (a) that

3 1 B A S S where8 0

0 8

What is B equal to?1/3 1 B S S (†)

1/3

1/3 8 0 2 0

0 8 0 2

. What else do we need to find?

1S . Thus1

1 5 1 3 1 3 11 1

1 3 1 5 1 516 16

S .

Substituting5 1

1 3

S , 1/3 2 0

0 2

and 1 3 11

1 516

S into (†) gives

1/3 1 5 1 2 0 3 11

1 3 0 2 1 516

10 2 3 11

2 6 1 516

28 20 7 51 1

12 28 3 716 4

B S S

Note that the given matrix7 5

3 7

A so7 51 1

3 74 4

B A .

6. (a) (i) The eigenvalues and the corresponding eigenvectors of the given matrix4 3

1 0

A are

1

33,

1

u and 2

11,

1

v

(ii) The matrix P is given by 3 1

1 1

P u v .

(iii) The eigenvalues of 2008A are 20083 and 20081 1 . The determinant of 2008A is2008 20083 1 3 .


(b) We need to determine whether

1 1 0

0 1 0

0 0 0

B is diagonalizable. Since we have an

upper triangular matrix therefore the eigenvalues are the entries along the leadingdiagonal, that is 1, 2 1 and 3 0 . By theorem (7.16) we have B is diagonalizable if

and only if it has 3 linearly independent eigenvectors.Let u be the eigenvector belonging to 1, 2 1 :

1 1 1 0 0

0 1 1 0 0

0 0 0 0

0 1 0 0

0 0 0 0

0 0 0 0

x

y

z

x

y

z

B I u

Since there is only 1 non-zero equation and 3 unknowns therefore there are 3 1 2 freevariables. Hence we have 2 linearly independent eigenvectors corresponding to 1, 2 1 .

Since 3 0 is a distinct eigenvalue from 1, 2 1 therefore it has a linearly independent

eigenvector. We have 3 linearly independent eigenvectors for a 3 by 3 matrix so thematrix is diagonalizable. (Alternatively you can find the 3 linearly independenteigenvectors.)

7. (a) Since1 0

10 2

A is a triangular matrix therefore it’s eigenvalues are given by the

entries on the leading diagonal, that is 1 1 and 2 2 . We have two distinct

eigenvalues for a 2 by 2 matrix so the matrix A is diagonalizable.Let u be the eigenvector belonging to 1 1 :

0 0 0 1

gives10 1 0 10

x x

y y

A I u u

Let v be the eigenvector belonging to 2 2 :

1 0 0 0

2 gives10 0 0 1

x x

y y

A I v v

The diagonalizing matrix S is given by 1 0

10 1

S u v .

(b) The eigenvalues of the given matrix2 0

0 3

A are 1 2 and 2 3 .

Let u be the eigenvector belonging to 1 2 :

0 0 0 1

2 gives0 1 0 0

x x

y y

A I u u



1 0 0 0

3 gives0 0 0 1

x x

y y

A I v v

The diagonalizing matrix S is given by 0 1

1 0

S v u .

Check that this S is indeed the diagonalizing matrix:

1 10 1 2 0 0 1

1 0 0 3 1 0

0 3 0 1 3 0

2 0 1 0 0 2

S AS S S

B

8. Note that the given matrix B is a symmetric matrix. The eigenvalues are given by

2

1 1det det

1 1

1 1 1

1 1 1 2 0

B I

The eigenvalues are 1 2 and 2 2 . Since the eigenvalues are distinct therefore

the matrix B is diagonalizable and the matrix P is given by P u v where u and v are

the eigenvectors belonging to 1 2 and 2 2 .

We have

1 2 1 02 gives 1 2, 1

01 1 2

xx y

y

B I u

1 2 1 02 gives 1 2, 1

01 1 2

xx y

y

B I v

Hence the eigenvalues and eigenvectors are given by

1 2 ,1 2

1

u and 2 2 ,1 2

1

v

The matrix 1 2 1 2

1 1

P u v and

2 0

0 2

D .

9. (a) For the given matrix

4 2 2

0 3 1

0 1 3

A we are told that 1 2 and 2 4 are two

eigenvalues of the matrix A. How do we find the third eigenvalue?By Proposition (7.9) (b) we know that the trace (addition of all the entries on the leadingdiagonal) of a matrix 1 2 3tr A . What is the trace of A?

4 3 3 10tr A

Substituting 1 2 , 2 4 and 10tr A into 1 2 3tr A gives

3 32 4 10 4


Let w be the eigenvector belonging to 3 4 :

0 2 2 0 0

4 0 1 1 0 gives 1

0 1 1 0 1

x x

y y

z z

A I w w

Note that w is linearly independent of the given eigenvector for 2 4 . Hence the last

pair is

0

4, 1

1

.

(b) The matrix P has entries along it’s columns given by the eigenvectors of part (a).2 1 0

1 0 1

1 0 1

P and

2 0 0

0 4 0

0 0 4

D

10. We need to prove that if 1 and 2 are distinct eigenvalues of a symmetric matrix A,

then the corresponding eigenspaces are orthogonal.Proof.Let 1 and 2 be distinct eigenvalues with the corresponding eigenvectors u and v of the

symmetric matrix A. By definition of eigenvalues and eigenvectors we have

1Au u and 2Av v (*)

Taking the transpose of 1Au u gives

1

1 Because and

TT

T TT T T T T Tk k

Au u

u A u XY Y X X X

Multiple both sides of the last line by the eigenvector v:

1T T Tu A v u v (†)

Since the given matrix A is symmetric therefore T A A . Substituting this T A A into(†) gives

1

2 1

2 1

2 1

By (*)

0

0 gives 0

T T

T T

T T

T T

u Av u v

u v u v

u v u v

u v u v

This last statement follows because we are given that 2 1 and are distinct. Hence

0 0T u v u vWe conclude that u and v are orthogonal.

■

11. (a) The eigenvalues of6 2

2 3

A are given by


2

6 2det det

2 3

6 3 4

6 3 4

9 14 2 7 0

A I

The eigenvalues are 1 2 and 2 7 . Let u be the eigenvector belonging to 1 2 .

6 2 2 0

22 3 2 0

4 2 0 gives 1 and 2

2 1 0

x

y

xx y

y

A I u


6 7 2 0

72 3 7 0

1 2 0 gives 2 and 1

2 4 0

x

y

xx y

y

A I v

We have the eigenvalues and eigenvectors given by

1 2 ,1

2

u and 2 7 ,2

1

v

Note that since the given matrix A is symmetric therefore the eigenvectors u and v areorthogonal.

(b) Let

1 1 1 1

1 1 1 1

1 1 1 1

1 1 1 1

A then

1 1 1 1 1 2 1

1 1 1 1 1 2 12 2

1 1 1 1 1 2 1

1 1 1 1 1 2 1

1 1 1 1 1 1 1

1 1 1 1 1 1 1

1 1 1 1 1 1 1

1 1 1 1 2 5 2

v v

w

A

A

Hence v

is an eigenvector belonging to an eigenvalue of 2 but w

is not an eigenvector ofA.

12. (a) The eigenvalues and eigenvectors for the matrix

1 1 1

1 1 1

1 1 1

A can be evaluated

by the usual procedure outlined above. We obtain the following:


1 2 3

1 0 1

1, 1 , 2, 1 and 0, 1

1 1 0

u v w

(b) The given matrix A is diagonalizable because the eigenvectors u, v and w are linearlyindependent. It is easier to spot that we have distinct eigenvalues 1 21, 2 and

3 0 therefore the corresponding eigenvectors are linearly independent.

13. (a) (i) We have1 1 1 1 0 1

1 1 1 1 0 0 1

1 1 1 2 0 2

1 1 1 1 0 1

1 1 1 1 0 0 1

1 1 1 0 0 0

Au

Av

1 1 1 1 3 1

1 1 1 1 3 3 1

1 1 1 1 3 1

Aw

Hence the following are the eigenvalues and corresponding eigenvectors of A:

1 2

1 1

0, 1 , 0, 1

2 0

u v and 3

1

3, 1

1

w

(ii) Since u, v and w are orthogonal therefore the orthogonal matrix Q has its columnentries given by the normalizing these eigenvectors. Normalizing a non-zero vector x is

equal toxx

:

1 1

1 11 , 1

6 22 0

u v and 1

11

31

w

Thus 1/ 6 1/ 2 1/ 3

1/ 6 1/ 2 1/ 3

2 / 6 0 1/ 3

Q u v w .

(iii) Since the eigenvalues are 1 20, 0 and 3 3 therefore the characteristic

equation is given by

2 3 2

0 0 3

3 3 0

Ap

Cayley Hamilton Theorem states that the matrix A satisfies this characteristic equation.We have

3 23Ap A A A O which gives 3 23A A

Substituting the entries for A into this 3 23A A yields


3 2

1 1 1 1 1 1

3 3 1 1 1 1 1 1

1 1 1 1 1 1

3 3 3 9 9 9

3 3 3 3 9 9 9

3 3 3 9 9 9

A A

(b) Need to prove that B and TB have the same eigenvalues.Proof.The characteristic equation TB

p of the matrix TB is given by

det

det Because

det Because

det Because det det

T

T

B

T TT

T TT T

T

B

p

p

B I

B I I I

B I X Y X Y

B I X X

Hence TB and B is given by the same characteristic equation T BBp p therefore

they have the same eigenvalues.■

14. (a) An eigenvalue of a square matrix A is a scalar such thatAu u

where u O is a column vector called the eigenvector belonging to eigenvalue .

(b) Need to show that

1

1

1

u is an eigenvector of

0 2 1

1 1 1

1 2 0

A :

0 2 1 1 3 1

1 1 1 1 3 3 1

1 2 0 1 3 1

Au

Hence the corresponding eigenvalue is 1 3 .

(c) To show that A is diagonalizable we need to find the other eigenvalues andeigenvectors. We can use the trace and determinant of A to find these other twoeigenvalues 2 and 3 :

2 33 1 [Because trace of A is 1]

2 33 3 [Because determinant of A is 3]

Solving these equations gives 2 3 1 . Hence we have 1 3 and 2, 3 1 . We are

given the eigenvector belonging to 1 3 in part (b). Let v be the eigenvector belonging

to 2, 3 1 :


1 2 1 0

1 2 1 0

1 2 1 0

x

y

z

A I v

Placing this into row echelon form gives

1 2 1 0

0 0 0 0

0 0 0 0

We have 1 non-zero equation and 3 unknowns therefore there are 3 1 2 free variables.Let z s and y t then 2x t s . Thus the general eigenvector v belonging to

2, 3 1 is given by

2 1 2

0 1

1 0

x t s

y t s t

z s

v


1 2

0 , 1

1 0

. Since we have 3 linearly independent

vectors

1 1 2

1 , 0 , 1

1 1 0

for a 3 by 3 matrix therefore the matrix A is diagonalizable.

The non-singular matrix

1 1 2

1 0 1

1 1 0

P . To find the inverse of this matrix we apply

row operations but have a look at early chapters to find the inverse. It is given by

1

1 2 11

1 2 34

1 2 1

P

Checking 1P AP is a diagonal. (You are not asked to check but if time allows carry outthe following check.)

1

1 2 1 0 2 1 1 1 21

1 2 3 1 1 1 1 0 14

1 2 1 1 2 0 1 1 0

3 6 3 1 1 21

1 2 3 1 0 14

1 2 1 1 1 0

12 0 0 3 0 01

0 4 0 0 1 04

0 0 4 0 0 1

P AP


15. (a) For the given matrix

1 2 2

8 11 8

4 4 1

A the characteristic equation p is

25 3 0p The eigenvalues and eigenvectors are given by

1 5 has eigenvector

1

4

2

u and 2, 3 3 has eigenvectors 1

1

0

1

v , 2

1

1

0

v

The eigenvector u is a basis for the eigenspace 5E and 1 2,v v is a basis for 3E .

The matrix 1 2

1 1 1

4 0 1

2 1 0

P u v v and from the theory of early chapters

1

1 1 1

2 2 3

4 3 4

P .

The diagonal matrix 1 P AP D has the entries on the leading diagonal of theeigenvalues:

5 0 0

0 3 0

0 0 3

D

(b) Cayley Hamilton theorem states that the given matrix A satisfies its characteristic

equation. This means that 25 3p A A I A I O . Verifying this result gives:

2

2

1 5 2 2 1 3 2 2

5 3 8 11 5 8 8 11 3 8

4 4 1 5 4 4 1 3

A I A I

4 2 2 2 2 2 2 2 2

8 6 8 8 8 8 8 8 8

4 4 6 4 4 4 4 4 4

4 2 2 4 4 4

8 6 8 16 16 16

4 4 6 8 8 8

0 0 0

0 0 0

0 0 0

O

Hence Cayley Hamilton theorem is satisfied.

16. (a) We need to find the characteristic equation detAP t t A I :


2 2 2

2 2

3 0 0

1 1 0 0det det

1 1 9

1 1 1 10

1 0 0 1 0 0

det 1 9 3det 1 9

1 1 10 1 1 10

1 10 9 3 10 9

10 9 3 10 9

3 10 9

t

tt

t

t

t

t t t

t t

t t t t t t

t t t t t t

t t t t

A I

Hence we have our result 2 23 10 9AP t t t t t .

(b) The eigenvalues are given by 2 23 10 9 0AP t t t t t . We need to solve

this equation. How?The first quadratic 2 3 0t t with 1, 1 and 3a b c gives the roots

2 24 1 1 12 1 11

2 2 2

b b act

a

In this case t is not real because we have 11 . (You might have come across the squareroot of negative numbers in your other modules. A number of this type is called acomplex number.)Solving the other quadratic gives

2 10 9 1 9 0t t t t

Real eigenvalues of A are 1 1t and 2 9t .

Let u be the eigenvector belonging to 1 1t :

1 3 0 0 0

1 1 1 0 0 0

1 1 1 9 0

1 1 1 10 1 0

1 3 0 0 0 0

1 2 0 0 0 0 gives

1 1 1 9 0 9

1 1 1 9 0 1

x

y

z

w

x x

y y

z z

w w

A I u

u

A basis for 1E is

0

0

9

1

.

17. You need to be very careful with this question. The eigenvalues of the givenmatrix are:


0 1 2 0 0

1 1 1 0 1

det det 0 0 1 0 0

0 0 2 3 0

1 2 3 4 0

1 0 0

1 1 0 1 Expanding along1 det

0 0 3 0 the middle row

1 2 4

1 0Expanding along

1 3 det 1 1 1the t

1 2

A I

2

hird row

Expanding along1 3 2 1 1 1

the last column

1 3 2 1 1

Taking out the negative signs and simplifying we have

2

3 2

2

2 2

det 1 3 2 1 1

1 3 1

1 3 1 1

1 1 3 0

A I

The eigenvalues are 1, 2 3, 41, 1 and 5 3 .

18. (a) The characteristic equation of the given matrix is:

1 1 0

det det 1 2 1

0 1 1

1 3 0

A I

The eigenvalues are 1 20, 1 and 3 3 .

(b) Let u, v and w be the eigenvectors of 1 20, 1 and 3 3 respectively. We have

1 1 0 0 1

1 2 1 0 gives 1

0 1 1 0 1

x x

y y

z z

Au u

0 1 0 0 1

1 1 1 0 gives 0

0 1 0 0 1

x x

y y

z z

A I v v


2 1 0 0 1

3 1 1 1 0 gives 2

0 1 2 0 1

x x

y y

z z

A I w w

(c) The basis

1 1 1

2 , 0 , 1

1 1 1

. The diagonal matrix

3 0 0

0 1 0

0 0 0

D

(d) 1 1 1

2 0 1

1 1 1

S w v u .

(e) The inverse of the above matrix S can be found using row operations. We have

1

1 2 11

3 0 36

2 2 2

S

(f) The vectors in

1 1 1

2 , 0 , 1

1 1 1

are orthogonal so we need to normalize them.

Normalizing a non-zero vector v is equal tovv

. Call this new basis ' :

1 1 11 1 1

' 2 , 0 , 16 2 3

1 1 1

(g) Let the new S be called S’. We have

1/ 6 1/ 2 1/ 3

' 2 / 6 0 1/ 3

1/ 6 1/ 2 1/ 3

S .

(h) Since S’ is an orthogonal matrix therefore

1

1/ 6 1/ 2 1/ 3

' ' 2 / 6 0 1/ 3

1/ 6 1/ 2 1/ 3

1/ 6 2 / 6 1/ 6

1/ 2 0 1/ 2

1/ 3 1/ 3 1/ 3

T

T

S S

19. (a) What are the eigenvalues of the given matrix A equal to?

Since0

a b

c

A is a triangular matrix therefore the eigenvalues are the entries on the

leading diagonal. We have 1 a and 2 c .


Let u be the eigenvector belonging to 1 a :

0

0 0 1 gives

0 0 0

a a b xa

c a y

b x x

c a y y

A I u

u

Let v be the eigenvector belonging to 2 c :

0

0 gives

0 0 0

a c b xc

c c y

a c b x x b

y y c a

A I v

v

The eigenvector matrix1

0

b

c a

S and the inverse of this matrix is

1 1

0 1

c a b

c a

S

The eigenvalue matrix0

0

a

c

.

(b) From part (a) we have 1 A S S . By applying (7.19) 1m m A S S with 1000m we have

1000 1000 1

1000

1000

1000

1000 1000

1000

1000 1000 1000

1000

1

1 0 1

0 0 0 1

1 01

0 0 10

1

0 10

1

0

b a c a b

c a c c a

b c a ba

c ac a c

c a ba bc

c a cc a

c a a a b bc

c a cc a

a

A S S

000 1000 1000

1000

/ 1Taking in

0

b c a c a

c ac

By using the algebraic identity

1 2 2 1n n n n n nx y x y x x y xy y we can write

1000 1000 999 998 997 2 998 999c a c a c c a c a ca a The entry

1000 1000 999 998 997 2 998 999

999 998 997 2 998 999 Cancelling

b c a b c a c c a c a ca a

c a c a

b c c a c a ca a c a


Hence

1000 999 998 997 2 998 9991000

10000

a b c c a c a ca a

c

A

20. (a) A square matrix A is diagonalizable if there exists an invertible (non-singular)matrix P such that 1 P AP D where D is a diagonal matrix.(b) An eigenvalue of a square matrix A is a scalar such that

Au uwhere u O is a column vector called the eigenvector belonging to eigenvalue .Eigenvalues and eigenvectors come in pairs like chalk and cheese because one is vectorand the other is a scalar.(c) If an eigenvalue 0 occurs m times where 1m then we say 0 is an eigenvalue with

multiplicity of m or 0 has multiplicity m. This is called algebraic multiplicity.

We do not cover geometric multiplicity of an eigenvalue in this book.However geometric multiplicity of an eigenvalue 0 of a square matrix A is the

dimension of the subspace of eigenvector u belonging to 0 .

(d) As noted in part (c) geometric multiplicity of an eigenvalue is not covered in thebook.

We are given that a 3 3 matrix A has characteristic polynomial 21 2 .

The algebraic multiplicity of 1 1 is one and 2, 3 2 is two.

A 3 3 matrix A is diagonalizable if and only if it has 3 linearly independenteigenvectors. Matrix A is diagonalizable if and only if 1 1 has geometric multiplicity

of 1 and 2, 3 2 has geometric multiplicity of 2.

(e) (i) We are given the matrix

1 2 1

1 1 2

1 0 3

A and it’s characteristic polynomial

21 2 . The eigenvalues are 1 1 and 2, 3 2 .

Let u be the eigenvector belonging to 1 1 . We have

1 1 2 1 0

1 1 1 2 0

1 0 3 1 0

0 2 1 0

1 0 2 0 gives 4, 1 and 2

1 0 2 0

x

y

z

x

y x y z

z

A I u



1 2 2 1 0

2 1 1 2 2 0

1 0 3 2 0

1 2 1 0

1 1 2 0 gives 1, 1 and 1

1 0 1 0

x

y

z

x

y x y z

z

A I v

We have 1 1 ,

4

1

2

u and 2, 3 2 ,

1

1

1

v . The eigenvectors of 2, 3 2 are all

multiples of v which means they are linearly dependent on v so we only have 2 linearlyindependent eigenvectors u and v of the given 3 3 matrix. Hence A is notdiagonalizable.

(ii) We are given the matrix

1 1 1

1 1 1

1 1 3

B and it’s characteristic polynomial

21 2 which gives the eigenvalues 1 1 and 2, 3 2 . Let u be the eigenvector

belonging to 1 1 . We have

1 1 1 1 0

1 1 1 1 0

1 1 3 1 0

0 1 1 0 1

1 0 1 0 gives 1

1 1 2 0 1

x

y

z

x x

y y

z z

B I u

u


1 2 1 1 0

2 1 1 2 1 0

1 1 3 2 0

1 1 1 0

1 1 1 0

1 1 1 0

x

y

z

x

y

z

B I v

Putting the last matrix in row echelon form gives

1 1 1 0

0 0 0 0

0 0 0 0

x y z

Since we have 1 non-zero equation with 3 unknowns therefore there are 3 1 2 freevariables. Let z s and y t then from the first row we have

0x y z x z y Substituting y t and z s we have x s t and the general eigenvector is given by


1 1

0 1

1 0

x s t

y t s t

z s

v

Hence the matrix B is diagonalizable because it has 3 linearly independent eigenvectors

given by

1 1 1

1 , 0 , 1

1 1 0

. The diagonalizing matrix P is given by

1 1 1

1 0 1

1 1 0

P .

21. (a) The vectors 1 2 3, ,w w w in 3 are said to be orthonormal if and only if

(i) 1 2 3, ,w w w are orthogonal, that is 1 2 2 3 2 3 0 w w w w w w .

(ii) 1 2 3, ,w w w are unit vectors, that is 1 2 3 1 w w w .

(b) By the Gram Schmidt process we have 1 1 w v

0

1

1

and

2 12 2 12

1

v ww v w

w(*)

We have 2 1

1 0

1 1 0 1 0 1

0 1

v w ,2 2 2 2

1

0 0

1 1 0 1 1 2

1 1

w .

Substituting these 2 1 1 v w ,2

1 2w into the above (*) gives

2

1 0 1 1 21 1

1 1 1 1/ 2 1/ 2 12 2

0 1 0 1/ 2 1/ 2 1

w

Remember to simplify our arithmetic we can ignore the fraction ½:

Let 2

2

1

1

w . How do we find 3w ?

3 1 3 23 3 1 22 2

1 2

v w v ww v w w

w w

We have 3 1

5 0

4 1 0 4 1 6 1 10

6 1

v w and

3 2

5 2

4 1 5 2 4 1 6 1 8

6 1

v w


2 22 22

2 2

1 1 2 1 1 6

1 1

w

Substituting these 3

5

4

6

v , 3 1 10 v w , 3 2 8 v w ,2

1 2w and2

2 6 w into the

above 3 1 3 23 3 1 22 2

1 2

v w v ww v w w

w wgives

3

5 0 210 8

4 1 12 6

6 1 1

5 0 2 5 0 8 / 3 7 / 3 14 7

4 5 1 1 4 5 4 / 3 7 / 3 13 3

6 1 1 6 5 4 / 3 7 / 3 1

w

Again ignore the fraction, we have 3

1

1

1

w . Our orthogonal vectors are

1

0

1

1

w , 2

2

1

1

w and 3

1

1

1

w

Since we want an orthonormal set so we need to normalise these orthogonal vectors:

1

01

12

1

w , 2

21

16

1

w and 3

11

13

1

w

(c) (i) The characteristic equation of the given symmetric matrix

3 2 4

2 0 2

4 2 3

A is

3 26 15 8 0 where is an eigenvalue. Factorising this gives

23 26 15 8 1 8 0

The eigenvalues are 1, 2 1 and 3 8 .

(ii) Let u be the eigenvector corresponding to 1, 2 1 :


3 1 2 4 0

2 0 1 2 0

4 2 3 1 0

4 2 4 0

2 1 2 0

4 2 4 0

x

y

z

x

y

z

A I u

Putting this into row echelon form gives

2 1 2 0

0 0 0 0

0 0 0 0

x y z

2 2 0x y z

We have 1 non-zero equation and 3 unknowns therefore there are 3 1 2 free variables.This means that we have 2 linearly independent eigenvectors for 1, 2 1 . We select our

vectors 1u and 2u so that they are orthogonal:

1

1

4

1

u and 2

1

0

1

u


1 1

4 , 0

1 1

.

Let v be the eigenvector for 3 8 . We have

3 8 2 4 0

8 2 0 8 2 0

4 2 3 8 0

5 2 4 0

2 8 2 0 gives 2 and 1

4 2 5 0

x

y

z

x

y x z y

z

A I v

The eigenvector corresponding to 3 8 is

2

1

2

v . The set of linearly independent

vectors are

1 1 2

4 , 0 , 1

1 1 2

. [We have actually found an orthogonal set of

eigenvectors.](d) We need to prove that if 3

1 2 3, , v v v is an orthogonal set then they are linearly

independent.We assume that none of the vectors 3

1 2 3, , v v v are the zero because if any one of

them is then the set is linearly dependent.


Let 1 2 3, andk k k be scalars such that

1 1 2 2 3 3k k k v v v O (*)

How do we show linear independence?Required to prove that 1 2 3 0k k k . Consider the inner product

1 1

1 1 1 1 2 2 3 3

1 1 1 1 1 1 2 2 1 1 3 3

21 1 1 1 2 1 2 1 3 1 3

0 because orthogonality 0 because orthogonality

221 1

0 ,

, By (*)

, , ,

, , ,

0

k

k k k k

k k k k k k

k k k k k

k

v O

v v v v

v v v v v v

v v v v v v

v

Since the vector 1v is in the set of orthonormal vectors therefore it is a unit vector so we

have 1 1v . Hence from above22

1 1 0k v we have 1 0k .

Similarly 2 3 0k k . Hence 31 2 3, , v v v are linearly independent.

■

22. (a) A set of vectors 1 2, , , nv v v are orthogonal if and only if

, 0i j v v provided i j

This set 1 2, , , nv v v is orthonormal if it is orthogonal and each of these is a unit

vector, that is for each 1, 2, 3, ,i n we have2

, 1i i i v v v

(b) A square matrix Q is orthogonal if and only if 1T Q Q .To prove that the set of columns of an orthogonal matrix forms an orthonormal set wehave to show that

0 if

1 ifT

ij

i j

i j

QQ I I

For the rest of the proof see Chapter 4??(c) (i) Need to show that 1, 0, 1, 0u and 1, 0, 1, 0 v are eigenvectors of the

matrix A:5 0 1 0 1 4 1

0 1 0 1 0 0 04 4

1 0 5 0 1 4 1

0 1 0 1 0 0 0

Au u

and5 0 1 0 1 6 1

0 1 0 1 0 0 06 6

1 0 5 0 1 6 1

0 1 0 1 0 0 0

Av v

Hence the eigenvalues are 1 4 and 2 6 .

(ii) How do we find the other two eigenvalues of A?


We can use the trace and determinant of the matrix A to find remaining eigenvalues 3and 4 . We have

5 1 5 1 12tr A

det 5[5 1 0 5 ] 1[ 1 1 1 ] 0 A

We need to solve the equations

3 44 6 12 [Because 1 4 and 2 6 ]

3 4 3 44 6 24 0 From the bottom equation one of the eigenvalues is zero. Let 3 0 . Substituting this

3 0 into the top equation gives 4 2 . Hence the other two eigenvalues are 3 0 and 4 2 . Let w be the eigenvector belonging to 3 0 :

5 0 1 0 0 0

0 1 0 1 0 1 gives

1 0 5 0 0 0

0 1 0 1 0 1

x x

y y

z z

r r

Aw w

Let x be the eigenvector belonging to 4 2 :

5 2 0 1 0 0

0 1 2 0 1 02

1 0 5 2 0 0

0 1 0 1 2 0

3 0 1 0 0 0

0 1 0 1 0 1 gives

1 0 3 0 0 0

0 1 0 1 0 1

x

y

z

r

x x

y y

z z

r r

A I x

x

(iii) Since we have distinct eigenvalues therefore the eigenvectors are linearlyindependent. Note that the given matrix A is symmetrical therefore our invertible(nonsingular) matrix P is an orthogonal matrix given by

1 1 0 0

0 0 1 11

1 1 0 02

0 0 1 1

P u v w x

where , , ,u v w x are normalised eigenvectors. Hence the matrix Q is

1

1 0 1 0Because is an

1 0 1 01orthogonal matrix

0 1 0 12so

0 1 0 1

T

T

P

Q P

P P Q

Our diagonal matrix is given by


4 0 0 0

0 6 0 0

0 0 0 0

0 0 0 2

23. We need to convert the given quadratic 2 23 4 6x xy y into diagonal form. Let

x

y

x

2 23 4 6 Tx xy y x Ax where3 2

2 6

A

The characteristic equation of A is given by 2 7 0 . The eigenvalues are

1 2 and 2 7

Let u be the eigenvector belonging to 1 2 :

1 2 0 2

2 gives2 4 0 1

x x

y y

A I u u


4 2 0 1

7 gives2 1 0 2

x x

y y

A I v v

Normalizing the eigenvectors u and v:

21

15

u and 11

25

v

The orthogonal matrix Q is given by

2 11

1 25

Q u v

LetX

Y

y be the new axes. By (7.30) we have x Qy which gives 1y Q x . Since Q

is orthogonal therefore

1 2 1 2 11 1

1 2 1 25 5

T

T

Q Q

Thus using 1y Q x we have

2 1 21 1

1 2 25 5

X x x y

Y y x y

Hence we have 1 12 and 2

5 5X x y Y x y . The diagonal form is

2 2 2 2 2 21 23 4 6 2 7x xy y X Y X Y [Because 1 2 and 2 7 ]

24. We are given the quadratic 2 2 2 22 4 4 3xy xz yz z aX bY cZ . We need towrite this in matrix form:


2

0 1 2

2 4 4 3 where 1 0 2 ,

2 2 3

T

x

xy xz yz z y

z

x Ax A x

The eigenvalues of the matrix A are 1, 2 1 and 3 5 .

Let u be the eigenvector belonging to 1, 2 1 :

1 1 2 0

1 1 2 0

2 2 4 0

x

y

z

A I u

Putting this into reduced row echelon form gives

1 1 2 0

0 0 0 0

0 0 0 0

We have two linearly independent eigenvectors:1

1

1

u and

1

1

0

v

Choose your eigenvectors so that they are orthogonal, that is 0 u v .Let w be the other eigenvector belonging to 3 5 :

5 1 2 0 1

5 1 5 2 0 gives 1

2 2 2 0 2

x x

y y

z z

A I w w

Normalizing the eigenvectors u, v and w gives

1

11

31

u , 1

11

20

v and 1

11

62

w

Our orthogonal vector is 1/ 3 1/ 2 1/ 6

1/ 3 1/ 2 1/ 6

1/ 3 0 2 / 6

Q u v w .

Let

X

Y

Z

y be the new axes. By (7.30) we have x Qy which gives 1y Q x . Since Q

is orthogonal therefore

1

1/ 3 1/ 2 1/ 6 1/ 3 1/ 3 1/ 3

1/ 3 1/ 2 1/ 6 1/ 2 1/ 2 0

1/ 3 0 2 / 6 1/ 6 1/ 6 2 / 6

T

T

Q Q

Thus using 1y Q x we have


1

31/ 3 1/ 3 1/ 31

1/ 2 1/ 2 02

1/ 6 1/ 6 2 / 6 12

6

x y zX x

x yY y

Z zx y z

Hence we have 1 1,

3 2X x y z Y y x and 1

26

Z x y z .

Since the eigenvalues are 1, 2 1 and 3 5 therefore the diagonal form is2 2 2 2 2 2 2

1 2 32 4 4 3 5xy xz yz z X Y Z X Y Z

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Learn Mathematics & Exercises Online Website | Maths For All solutions... · 2017. 3. 24. · 0 0 0 Taking Away along det det 0 0 the Leading Diagonal 0 0 0 0 0 0 0 det 0 0 0 0 0