Answers To Final Exam for Linear Algebra

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Answers to Final Exam for Linear Algebra

(Spring, 2003)

––––––––––––––––––––– Part 1 –-––––––––-––––––––––––

1. (10%)

Let A and B be non-singular n

 n matrices. Show that AB and BA have the same set of eigenvalues.

Answer

Let u be the eigenvector of AB belonging to the eigenvalue

, i.e.,

AB u

  u

Multiplying both sides with B gives

BA Bu

 

Bu so that v

Bu is an eigenvector of BA with eigenvalue

.

Conversely, we see that if v is an eigenvector of BA with eigenvalue

, then is the eigenvector of AB belonging to the eigenvalue

. u

 

1

B v

Since this applies to all

, AB and BA must have the same set of eigenvalues.

2. (10%)

Find a similarity transform to put the matrix A

 a b

0 a

in the Jordan form.

Answer

We need to fine S such that

 s t u v

 a

0 b a a

0

SAS

1 a

1



 s t a

 u v

 0 or

 sa sb

 ta ua ub

 va u

0 v

 sb as

 u at

 v au av

 so that

S

 s

0 t sb

1 a

. Hence

S

1 

1

2 s b

 sb

 t

0 s

where s , t are arbitrary non-zero constants.

3. (20%)

Let A

0 0 0 1

0 0 1 0

0 1 0 0

1 0 0 0

What are the eigenvalues?

For each n

2 , find scalars a and b so that A n  aI

 bA .

Answer det A

 

I

 det

0

0

0 0 1

 

1

1

 

0

0

1 0 0

 

 

4 

2

2 

1

2 

1

2

3 

 

 2 

1

Therefore, the eigenvalues are

  

1 , both being doubly degenerate. Thus, the

Cayley- Hamilton equation is

A 2 

I

2 

0

A

2 

I

Hence,

A n

I

A

for

Thus, a

1

2

1

  n n

 even odd b

1

2

1

  n

3. (20%)

Given a square matrix A with distinct eigenvalues, we define the spectral matrices as

G r

  s r a r

 s

 a s where a and r a.

b.

G r a s

are the eigenvalues of

is an eigen-matrix of

G G

0 for r s r s

.

A

A . Show that

with eigenvalue a r

.

Answer

Using the Cayley- Hamilton equation, we have

   r

   r

  s r a r

 s

 a s

0 so that

AG r

 a G r r i.e., G r

is an eigen-matrix of A with eigenvalue

G G r s

  a r

 t

 a t

 a s

 u

 a u a . Next, r

If r

 s , the second product will contain the factor

  r

. Hence,

G G

0 r s

––––––––––––––––––––– Part 2 –-––––––––-––––––––––––

Let

 

be a non-orthogonal basis with inner products s ij

 e e i j

.

Let A be a linear transformation defined by its matrix elements a ij

 e i

A e j

.

4. (10%)

Find the characteristic (secular) equation for A .

Answer

If u is an eigenvector of A with eigenvalue

, then

A u

  u

Writing u

  u e i i

, we have i i

A u e i i

   i u e i i

Taking the inner product with e j

gives

 i e j

A e i u i

   i e e j i u i i

 e j

A e i

  e e i u i

0

The secular equation is therefore

det A

 

S

0

5. (10%)

What is the criterion, in matrix form, for A to be hermitian (self adjoint)?

Answer

By definition, A is hermitian if for arbitrary vector u and v , u Av

Au v (1)

In matrix form, we have

 † u v u S v where S

 

is the overlap matrix. Hence (1) becomes

† u SAv

   †

Au Sv

 u

† †

A Sv for all u and v i.e.,

 †

SA A S or

6. (20%)

A

† 

SAS

1

Let

A

  

1

1

1

2

 and

S

  

1 1

 1 2

Find the eigenvalues and eigenvectors of A .

Are the eigenvectors orthogonal?

Answer

The secular equation is det

1

  

1

 

2 2

0

     

1

   

1

 

1

   

0

1

 

1

0

so that the eigenvalues are

  

1

Since the eigenvalues are distinct, they are orthogonal.

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