––––––––––––––––––––– 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.