Page 1
Lab 5 Elementary Matrix and Linear Algebra
Name ____________________
Due at Final Exam
Spring 2011
Score = ____/25
1. (3 points) Go to AKiTi Miscellaneous Mathematical Utilities page at http://www.akiti.ca/Mathfxns.html and
numerically determine the eigenvalues and eigenvectors of the following matrices. Fill in the table and printout and
attach the results from the program.
a)
 2 3 1
A   5 7 9 
7 4 8 
det  A 
1 
2 
3 
1  2  3 
b)
1 2 
4 3
3 8
0 1 
A
1 0
6 2.5


 2 1 2.5 2 
det  A 
1 
2 
3 
4 
1  2  3 4 
2. (3 points)
a) Prove that the transpose of a matrix has the same eigenvalues as the original matrix.
Is the same true of the eigenvectors? Explain your answer.
b) For real symmetric matrices the eigenvectors can always be expressed using only real numbers. Is this true of any
real square matrix? Explain your answer.
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Lab 5 Elementary Matrix and Linear Algebra
Spring 2011
3. (11 points) Complex Inner Product Spaces, Unitary and Hermitian matrices.
The concept of an inner product can be extended to vectors consisting of complex entries. In analogy with R n , the
space of n dimensional vectors with complex components is called n . The inner or Hermitian product of two
vectors in
is defined as u | v   u, v   uT v 
n
n
u v
j 1
j
j
. Here z designates the complex conjugate of z. Recall
that for a complex number z, z  a  bi with a and b real, the complex conjugate is defined as z  a  bi .
Furthermore, z  zz   a  bi  a  bi   a 2  b 2  0 . The convention of complex conjugation “on the right” in
2
the definition of u | v is arbitrary and some authors define the Hermitian product with complex conjugation on
5  2i 


the left. What’s important is to be consistent. As an example of the Hermitian product, let u  3 and


 4i 
 3  7i 
v   2i  , then u | v   5  2i  3  7i    3 2i   4i 1  i   5  31i .
 1  i 
a) Show that the Hermitian inner product satisfies the following properties. Here u, v, and w represent vectors and c
is a complex scalar.
_______
u|v  v|u
uv| w  u| w  v| w
w| u v  w| u  w| v
cu | v  c u | v
u | cv  c u | v
u | u  0 with equality if and only if u  0 .
The Cauchy Schwartz inequality is still true. For any complex scalar  and non-zero vectors u and v ,
_______
u   v | u   v  0 . So, u   v | u   v  u | u   v | u   v | u  
then u | u 
2 v|u

v|v
u|u v|v  v|u
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2
v|u
v|v
2
2
2
v|v  u |u 
v|u
v|v
2
v | v  0 . Let   
2
 0 . Rearranging this inequality gives
with equality if and only if u  cv for some complex scalar c.
Madison Area Technical College
u|v
v|v
,
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Lab 5 Elementary Matrix and Linear Algebra
Spring 2011
The adjoint or Hermitian conjugate of an n x n complex matrix is defined as A†  AT , i.e., the transpose of the
matrix of complex conjugates. Note: when moving a matrix multiplier from left to right in an inner product one
transforms to the Hermitian conjugate. Au | v   Au  v  u A v  u
T
T
T
T
_______
T
 A v 
u | A†v .
An n x n complex matrix U is called unitary if and only if U †  U 1 , i.e., U †U  UU †  I n . An n x n complex
matrix A is called Hermitian if and only if A†  A .
b) Which of the following matrices are unitary?
 1

 2
 1

A 2

 0


 0

3 i
2 2
0
3 i
2 2
0
0
0
1 i
2
i
2

 1
0 


 2

 1
0 

 B 2

1 i 

 0
2 


i 

 0
2

3 i
2 2
0
3 i
2 2
0
0
0
1 i
2
i
2


0 






0 
 C

1 i 


2 


i 


2

1
2
1
2
0
0
1 i
2
1  i
2
0
1 3 i
2 2
0
0
1 3 i
2 2
0

0 


0 

i 

2
i 

2
sin   cos   sin   i cos   cos   


D   sin   sin    cos   i cos   sin   
 i cos  

0
sin  
c) What is true about the determinant of a unitary matrix?
d) Is every orthogonal matrix a unitary matrix? Explain your answer.
e) Look up the definition of a group in algebra. Show that the set of n x n unitary matrices form a group with
respect to the operation of matrix multiplication.
f) Let SU(n) be the set of all n x n unitary matrices with a determinant of one. Show that SU(n) is a group with
respect to the operation of matrix multiplication. Find an element of SU(2) that is not the identity matrix.
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Lab 5 Elementary Matrix and Linear Algebra
Spring 2011
g) Suppose A is an n x n matrix and U is an n x n unitary matrix.
What is the det U † AU  ?
What is the trace U † AU  ?
h) Can a non-identity matrix be both Hermitian and unitary? Explain your answer.
i) For any matrix below which is Hermitian determine its eigenvalues.
 3 5 
5 11 


3 5 
5 11


3i 5 
 5 11


3i 5 
 5 11i 


 3 5i 
5i 11


 3 5i 
 5i 11


__________
j) Suppose A is n x n Hermitian matrix and u is a vector in
n
. What is Au | u  Au | u ?
k) Suppose A is n x n Hermitian matrix and for a non-zero  , A   . What is    ?
Choose one of the following six problems. Each problem is worth 8 points. You may do extra problems to earn up
to 40 extra lab points.
4. The Spectral or Fourier Representation of Hermitian Matrices
a) If A is an n x n Hermitian matrix then it has n linearly independent eigenvectors which span
 
n
j
j1
n
. Let
designate a set of n normalized linearly independent eigenvectors of A with A j   j  j and
 j |  j  1.
Determine  j |  k .
b) Let B be an n x n matrix defined as Bi j  i ' th component of  j , i.e., B  1
2
n  .
†
Determine BB .
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Madison Area Technical College
Page 5
Lab 5 Elementary Matrix and Linear Algebra
Spring 2011
Wolfram Alpha can be used to determine the eigenvalues and a set of eigenvectors for a matrix as illustrated below.
c) Determine the eigenvalues and a set of orthonormal eigenvectors of the matrix A.
 5
A
 3  i
3  i

2 
1  ________________ 1  ________________ 2  ________________  2  ________________
2
 . Compute the following:
 i 
Let v  
2

j 1
j
v |  j  j  ________________
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Av  ________________
Madison Area Technical College
Page 6
Lab 5 Elementary Matrix and Linear Algebra
Spring 2011
d) Determine the eigenvalues and a set of orthonormal eigenvectors of the matrix A.
 11

 8
 3
A
 8
 3i

 4
3
8
9
8

3i
4
3i 

4 
3 i

4 
5 

2 
1  ________________ 1  ________________ 2  ________________  2  ________________
 3  ________________  3  ________________
 2i 
 
Let v  i . Compute the following:
 
 1 
3

j 1
j
v |  j  j  ________________
Av  ________________
e) Let A be an n x n Hermitian matrix with an orthonormal set of eigenvectors
 j |  k   jk . Let v be any vector in
n
What does the sum

j 1
j
n
n
j
j1
so that A j   j  j and
.
v |  j  j represent?
n
If A is non-singular, what does the sum
1

j 1
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 
v |  j  j represent?
j
Madison Area Technical College
Page 7
Lab 5 Elementary Matrix and Linear Algebra
Spring 2011
Problems 5 and 6 are dedicated to the memory of Professor C. H. Blanchard (1923 – 2009) of the University of
Wisconsin-Madison Physics Department. He was the best and most inspiring teacher I have ever encountered.
5. The Moment of Inertia Tensor
In the analysis of the rotation of a rigid body one defines a second rank tensor, called the moment of inertia tensor
which can be represented by the following 3 x 3 real symmetric matrix.
I inertia
 I xx

  I xy
 I xz

I xy
I yy
I yz
I xz 

I yz 
I zz 
For a mass density m  x, y, z  , the matrix elements of the moment of inertia tensor are defined by the following
triple integrals over the region of three dimensional space S bounded by the surface of the rigid body.
I xx 
   x, y, z   y
2
m

 z 2 dV
S
   m  x, y, z  xydV
I xy 
S
   m  x, y, z  xzdV
I xz 
S
I yy 
   x, y, z   x
2
m

 z 2 dV
S
   m  x, y, z  yzdV
I yz 
S
I zz 
   x, y, z   x
m
2

 y 2 dV
S
 x 
 
For an angular velocity vector    y , the angular momentum of the rigid body is given by
 
z 
1
I inertia and the kinetic energy of rotation is  T I inertia . For a uniform solid right circular cylinder of radius a
2
and height h inclined as shown in Figure 1, the moment of inertia tensor is
I inertia
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 h 2 5a 2


4
16



0

 3  h2 a 2 

  
 4  3 4 
0
h2 a 2

3
4
0
3  h2 a 2 
  
4  3 4 


0

2
2

h 7a


12 16 

Madison Area Technical College
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Lab 5 Elementary Matrix and Linear Algebra
Spring 2011
Figure 1
a) Determine the eigenvalues and an orthonormal set of eigenvectors for this I inertia .
b) The orthonormal eigenvectors are called the principal axes of the rigid body. Explain this terminology.
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Madison Area Technical College
Page 9
Lab 5 Elementary Matrix and Linear Algebra
Spring 2011
6. Coupled Pendulums
A simple pendulum consists of an object of mass m suspended by an essentially weightless and non stretchable
string of length that is allowed to swing freely. As in figure 2 let the angle denote the displacement from the
downwards vertical.  is related to the standard polar angle  by  
3
  , so that the position vector of the mass
2
  3

 3
 
   iˆ  sin 
   ˆj   sin   iˆ  cos   ˆj  . The acceleration vector is thus
is given by R  cos 
2
2


 
 
2
 d 
ˆ
ˆ d 
   sin   iˆ  cos   ˆj  
  cos   i  sin   j  2
2
 dt 
dt
dt
2
2

d 2  ˆ 
d 2
 d 
 d 


   sin   

cos

i

cos


sin








 dt 
 dt 




dt 2 
dt 2
a
d 2R
2

 ˆj

Figure 2
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Madison Area Technical College
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Lab 5 Elementary Matrix and Linear Algebra
Spring 2011
The forces acting on the mass are the tension in the string, T , which points to the origin and gravity which points
straight down. F  T   sin   iˆ  cos   ˆj   mg ˆj  T sin   iˆ  T cos    mg  ˆj . From Newton’s second law

 d
of motion, F  ma  m   sin   
 dt

2
d 2  ˆ

i  m

cos





dt 2 

 d
cos   
 dt

2
d 2  ˆ

 j , so that we have

sin





dt 2 
the coupled differential equations

 d
T sin    m   sin   
 dt


 d
T cos    mg  m cos   
 dt

 d
From the first equation T  m 
 dt
2
cos   d 2



sin   dt 2

2
d 

  cos   2

dt
2
2
d 

  sin   2

dt
2







 . Substituting this expression into the second equation

gives the following.
 d
m 
 dt
cos   d 

 cos    sin 
  dt 2

2


 d
  mg  m cos   
 dt


 cos 2   d 2 

d 2 

  g  sin  

 sin   dt 2 
dt 2 

2
2

d 2
d 2
2
2

sin


cos

  2
  2

dt
dt

2




  g sin  

d 2
dt
2
d 

  sin   2

dt
2

g
sin  
We seek solutions that satisfy the initial values of   0  and    0  . This non linear second order initial value
problem has a solution which can be expressed as an elliptic function. A simpler result is obtained if we restrict our
attention to small angles where sin     . The differential equation then is linear and is given by
3/9/2016
Madison Area Technical College
d 2
dt
2
g
  .
Page 11
Lab 5 Elementary Matrix and Linear Algebra
Spring 2011
Looking for exponential solutions of the form   t   et gives the auxiliary equation  2  
  i
g
g
which has solutions
 i0 . Thus,   t   b1ei0t  b2ei0t  b1  b2  cos 0t   i b1  b2  sin 0t  . Requiring that the
_____
solution be real means that   t     t  , so that
 b1  b2  cos 0t   i  b1  b2  sin 0t    b1  b2  cos 0t   i b1  b2  sin 0t  .
Since sine and cosine are linearly independent functions it must be true that
b1  b2  b1  b2
.
b1  b2  b2  b1
c  ic2
d  id2
Taking b1  1
for real c1 , c2 , d1, d 2 yields d 2  c2 and d1  c1 so that b1  b2  c1 and
, b2  1
2
2
b1  b2  ic2 . So the general solution in the small angle approximation is   t   c1 cos 0t   c2 sin 0t  . Fitting
the initial conditions gives  t     0  cos 0t  
T
2
0

2
g
and a frequency of f 
   0
sin 0t  . The solution is periodic with a period given by
0
g
1

. If the pendulum is “released from rest”,    0   0 and
T 2
  t     0  cos 0t  , so that   0  is the amplitude of the angle oscillations.
Two pendulums each of length are coupled by a cross string as shown on Figure 3. The effect of the cross string
can be modeled by a force on pendulum 1 by pendulum 2 equal to k 1  2  where k is a positive proportionality
constant. By Newton’s third law of motion the force on pendulum 2 by pendulum 1 is k  2 1  . In the small
angle approximation the two angles satisfy the following linear differential equation.
 g
 k
d 2  1  
 
dt 2  2   k

3/9/2016

  1 
 

g
  k 2

k
Madison Area Technical College
Page 12
Lab 5 Elementary Matrix and Linear Algebra
Spring 2011
Figure 3
Graphic from page 17 of Nine Experiments for a Third Grade Hour, 2009 Memorial Edition by C. H. Blanchard.
Figure 4
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Madison Area Technical College
Page 13
 g
  k
Let A  
 k

Lab 5 Elementary Matrix and Linear Algebra
Spring 2011


2
 . A is a real symmetric matrix. It has orthonormal eigenvectors 1 ,  2 that span R .
g
  k

k
2
d 2c 2
 1  t  
d 2 1  d c1

c
t


c
t

So for any value of t, 
 2  A c11  c 2 2  .
 1   1 2   2 . Thus, 2    2 1 
2

t



dt
dt
dt
2
2




a) Display the second order differential equations for both c1  t  and c2  t  .
b) Solve for both c1  t  and c2  t  with initial conditions for each defined by the values of 1  0  ,  2  0  , 1  0  ,
and 2  0  .
c) Use the solutions for c1  t  and c2  t  to display the solutions for 1  t  and 2  t  .
d) Is the motion of the coupled pendulums always periodic, i.e. is there a period T with both 1  t  T   1  t  and
 2  t  T    2  t  for all values of t? Explain your answer. What conditions on the constants k , , g is required for
periodicity?
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Madison Area Technical College
Page 14
Lab 5 Elementary Matrix and Linear Algebra
Spring 2011
e) Display the solutions for 1  t  and 2  t  if both masses are released from rest, i.e. 1  0    2  0   0 .
f) If 1  0    2  0   0 and  2  0   0 describe the motion. In particular, does the second pendulum oscillate?
g) Display the solutions for 1  t  and 2  t  if1  0    2  0   0 and 2  0   1  0    0 .
h) Display the solutions for 1  t  and 2  t  if1  0    2  0   0 and 2  0   1  0    0 .
i) The solutions of g) and h) are periodic solutions which consist of oscillations at a single frequency. These are
called the normal modes of the system. One of these modes is called a “walking mode”. Which one is it?
j) One of the normal modes is faster, i.e. has a higher frequency. Which normal mode is it?
k) What is the connection between the eigenvectors of A and the normal modes?
l) What is the connection between the eigenvalues of A and the normal modes?
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Madison Area Technical College
Page 15
Lab 5 Elementary Matrix and Linear Algebra
Spring 2011
7. Rayleigh-Ritz Variational Calculations of Extreme Eigenvalues
Suppose A is an n x n real symmetric matrix. In many applications such as bound states in Quantum Mechanics and
economic optimizations the extreme eigenvalues max and min are of particular interest. Consider an n x 1 vector
 called a “trial function”. The eigenvectors of A are linearly independent and an orthonormal basis for
 
R n consisting of eigenvectors of A is always possible. Designate this orthonormal basis by  j
A j   j  j ,   j ,  k    jk , and max  1  2
n
j1
with
 n  min . The trial function can be expanded as
    ,  j   j .
n
j1

n
n

n
j 1
k 1

j 1 k 1
 A,     A  ,  j   j ,   ,  k   k      ,  j    ,  k   A j ,  k 

n
   j   ,  j    ,  k    j ,  k     j   ,  j    ,  k   jk    j   ,  j 
n
n
n
j 1 k 1
n
n
j 1 k 1
j 1
  ,  
A,   


For   0 , the Rayleigh Quotient is defined as
n
 ,  
 min  ,  j 
n
j 1
 ,  
  ,  
j
2
j
 ,  
. It can be bracketed as follows:
  ,  
A ,   



n
 ,  
n
min
2
j 1
j 1
max
2
j
 ,  
  ,  
n
2
2
2
 A,     j 1
 ,  
 ,   max  ,  
 A,    
min 
 ,   max
j 1
j
j

Any choice of   0 gives a lower bound to max and an upper bound to min . By allowing  to vary over all non
zero vectors of R n we have the following characterizations.
 A,  
R ,  0   ,  
 A,  
max  max
R ,  0   ,  
min 
min
n
n
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Madison Area Technical College
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Lab 5 Elementary Matrix and Linear Algebra
If  varies over just a subspace W of R n then min 
 A,  
or if  2 W
W ,  0   ,  
then max  max
 A,  
W ,  0   ,  
max
and
 A,  
W ,  0   ,  
min
 A,     max  A,  
and max
. If 1 W
W ,  0   ,  
W ,  0   ,  
min
then min 
 A,  
. How well the variational calculations
W ,  0   ,  
min
approximate the extreme eigenvalues depends on “how close” the

subspace W is to 1 and  2 .Let  
Spring 2011
 ,  
, then
 A,    A, 

 with  ,    1 . So in practice
 ,  
variational calculations are done using normalized trial functions, i.e.  ,    1. This is equivalent to varying the
trial function over a subspace of the unit n-sphere. As an example consider the 2 x 2 real symmetric matrix
a c 
A
. The eigenvalues are solutions of the characteristic polynomial  2    a  b   ab  c2  0 .

c b
max  1 
min  2 
ab
a  b
2
 4c 2
2
ab
a  b
2
 4c 2
2

ab
 a b 
2
 
 c
2
 2 

ab
 a b 
2
 
 c
2
 2 
2
2
cos   
 with 0    2 spans the unit circle (the unit 2-sphere).
 sin   
The trial function   
Thus, 1  max  A,   and 2  min
0 2
0 2
 A,   . This can be verified as follows.
 a cos    c sin   cos   
2
2
f     A ,    
,
  a cos    2c cos   sin    b sin  
c
cos


b
sin

sin









Using the trigonometric identities
cos 2   
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1  cos  2 
2
; sin 2   
1  cos  2 
2
; sin   cos   
sin  2 
2
Madison Area Technical College
Page 17
f   
Lab 5 Elementary Matrix and Linear Algebra
a  b a  b

cos  2   c sin  2  ; cos   
2
2
a b
 a b 
2
2 
 c
2


2
Spring 2011
; sin   
c
 a b 
2

 c
2


2
ab
 a b 
2
 
  c cos   cos  2   sin   sin  2  
2
 2 
2

2
ab
 a b 
2

 
  c cos  2   
2
 2 
 ab
 
 a b 
2
 
So the maximum of f   at   is
  c  max and the minimum of f   at    is
2 4
2
2
 2 
2
ab
 a b 
2
 
  c  min .
2
2


2
 cos   


For a 3 x 3 real symmetric matrix, trial functions of the form 1   sin    would in general give only
 0 
approximations to max and min since  1 does not span all of the unit 3-sphere. The trial function
sin   cos   


 2   sin   sin    with 0    2 ; 0     does span the unit 3-sphere and so
 cos   
max  max  A 2 ,  2  and min  min  A 2 ,  2  .
0  2
0 
0  2
0 
sin   sin   cos   


sin   sin   sin   

For a 4 x 4 real symmetric matrix, trial functions of the form  
with
 sin   cos   


cos  


0    2 ; 0     ; 0     would give the exact values of max and min , but this would involve finding
extreme values of a function of three independent variables.
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Madison Area Technical College
Page 18
Lab 5 Elementary Matrix and Linear Algebra
Spring 2011
Fill in the following tables for each matrix below.
a)
10 6 
A

 6 5
cos   


 sin   
min 
f min 
max 
f max 
f     A,  
b)
7 4
A

1 2 
cos   


 sin   
min 
f min 
max 
f max 
f     A,  
c)
 5
A
 3  i
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3  i

2 
cos   


 sin   
min 
f min 
max 
f max 
f     A,  
Madison Area Technical College
Page 19
Lab 5 Elementary Matrix and Linear Algebra
Spring 2011
d)
4 0 6
A   0 2 0 
 6 0 4 
cos   


   sin   
 0 
min 
f min 
max 
f max 
f     A,  
e)
4 0 6
A   0 2 0 
 6 0 4 
sin   cos   


   sin   sin    f  ,    A,  
 cos   
min 
f min 
max 
f max 
f)
11
0
A
0

5
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5
0 
0

0 0 11
0 0
4 6
6 4
cos   


0 


 sin   


 0 
min 
f min 
max 
f max 
f     A,  
Madison Area Technical College
Page 20
Lab 5 Elementary Matrix and Linear Algebra
Spring 2011
g)
11
0
A
0

5
sin   cos   


0

 f  ,    A,  

 sin   sin   


 cos   
5
0 
0

0 0 11
0 0
4 6
6 4
min 
f min 
max 
f max 
h) For some of the above cases the interval min , max  was not identical to  f min , f max  . Explain why in each case.
i) (2 point bonus) The Rayleigh-Ritz variational method can be used to find the extreme eigenvalues of complex
Hermitian matrices. Explain how the trial functions need to be chosen to make this work,
8. Iterative Scheme to Calculate Eigenvalue of Greatest Absolute Value
Consider a real n x n matrix A with n distinct real eigenvalues. Let  be the most extreme eigenvalue (i.e., the
eigenvalue of largest absolute value) of A with eigenvector 1 and designate the remaining eigenvectors of A
as  k k  2 with A k  k  k . Let  be a vector in R n . Because the eigenvectors of A form a basis for R n ,  has
n
an expansion in the eigenspace of A:   c11 



Then Am ,    Am
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
n
n
c  , c 
k 1
k
k
j 1
j
n
c 
k 2
k
k
. Assume that ck  0 .
n
n
n

m


c
c

,


kmck c j   k ,  j  .





j
1
j
1
j
j 1
k  2 j 1

Madison Area Technical College
Page 21
Lab 5 Elementary Matrix and Linear Algebra
A
A

,  
m 1
m
, 
 m 1c1  c j  1 ,  j    km 1ck  c j   k ,  j 
n
n
n
j 1
n
k 2
n
n
j 1
k 2
j 1
j 1
 m c1  c j  1 ,  j    km ck  c j   k ,  j 
n
 n

 c1  c j  1 ,  j     k
k 2  
 j 1

n
 n

 m c1  c j  1 ,  j     k
k 2  
 j 1
m 1
 
c1  c j  1 ,  j     k 
j 1
k 2   
n

n
n
m 1
n

ck  c j   k ,  j  
j 1

m
n


 ck  c j   k ,  j  

j 1




m 1
ck  c j   k ,  j 
n
j 1
 
c1  c j  1 ,  j     k  ck  c j   k ,  j 
j 1
k 2   
j 1
n
m
n
n
 k 
 1 , all of the sums    ck  c j   k ,  j  and
Since

k 2   
j 1
k
Spring 2011
m
n
 k 

 
k 2   
n
m 1
ck  c j   k ,  j  will approach zero
n
j 1
 A ,   converges to  . Fill in the following tables.
as m increases. Thus as m increases the ratio
 A ,  
m 1
m
10 6 
A

 6 5
a)
  ______________
1
 
1
 A ,  
 A ,  
m 1
m
m
1
2
3
4
5
6
7
3/9/2016
Madison Area Technical College
Page 22
Lab 5 Elementary Matrix and Linear Algebra
7 4
A

1 2 
b)
  ______________
Spring 2011
2
 
1 
 A ,  
 A ,  
m 1
m
m
1
2
3
4
5
6
7
7 4
A

1 2 
c)
  ______________
1
 
1
 A ,  
 A ,  
m 1
m
m
1
2
3
4
5
6
7
3/9/2016
Madison Area Technical College
Page 23
Lab 5 Elementary Matrix and Linear Algebra
4 0 6
A   0 2 0 
 6 0 4 
d)
  ______________
 A ,  
 A ,  
Spring 2011
1
  1
1
m 1
m
m
1
2
3
4
5
6
7
11
0
A
0

5
e)
5
0 
0

0 0 11
0 0
4 6
6 4
  ______________
 A ,  
 A ,  
1
1
 
1

1
m 1
m
m
1
2
3
4
5
6
7
f) Why are the results for b) and c) different?
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Madison Area Technical College
Page 24
Lab 5 Elementary Matrix and Linear Algebra
Spring 2011
9. Eigenvalues of a Tridiagonal Toeplitz Matrix
Consider the n by n matrix G(n), defined by the following formula:
G n i , j  ai , j  bi  1, j  ci  1, j
1  i, j  n
a
c

0

0
i.e., G n  


0
0

 0
b
a
c
0

0
0
0
0
b
a
c

0
0
0
0
0
b
a

0
0
0








0
0
0
0

a
c
0
,
0
0
0
0

b
a
c
0
0

0

0


0
b

a 
Matrices such as G(n) whose values along a given diagonal are constant are referred to as Toeplitz matrices.
If bc = 0 the only eigenvalue of G(n) is a. To get a more useful result we must assume that bc  0 .
Let X be an n component eigenvector of G(n) with eigenvalue  . The components of X satisfy the equation
bX k 1  a   X k  cX k 1  0
1 k  n
(1)
with the boundary conditions that
X 0  X n1  0 . (2)
Equation (1) is a linear homogeneous recurrence relation of degree 2 and it can be solved in a manner analogous
a
c

with solving a linear second order differential equation. Since X 0  0 and X k 1   X k  1 
Xk  , a
b
b


nontrivial (i.e., nonzero) eigenvector requires that X1 be nonzero. Without loss of generality we can take X 1  1 .
For a given eigenvalue this initial condition uniquely determines X. To see this, suppose Y is a second eigenvector
with Y1 = 1. Because equation (1) is linear and homogeneous, Z = Y – X also satisfies equation (1). However, Z0 =
Z1 = 0 so that the recursion requires that Z is the zero vector.
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Madison Area Technical College
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Lab 5 Elementary Matrix and Linear Algebra
Spring 2011
To construct the general solution of equation (1) we look for exponential solutions of the form X k  r k with r  0 .
2
a
c
 a
 
 and
Hence, r k 1 br 2  a   r  c   0 which gives the two solutions r1 



2b
b
 2b 
r2 
2
c
 a
 
  . If the discriminant is zero, r1  r2 and the general solution of equation (1) would be
2b
b
 2b 
a
of the form X k   r1k   kr1k for constants  and  . However, the boundary conditions of equation (2) would
force both constants to be zero. Thus, a nontrivial eigenvector requires that r1  r2 . For this case the general
solution of equation (4) is given by X k   r1 k   r2 k . The initial conditions X 0  0 and X 1  1 are satisfied when
   
1
. Using the equations for r1 and r2 to solve for  gives
r1  r2
a
c
c
 br1  a   br2 . (3)
r1
r2
 c
 c
This results in the following quadratic equation for r1 , r1 2  r1 
 r2    0 , so that
 br
 b
 2

 c

c
 r2  
 r2 
 br

br2
c
 2
 . Since r  r , it must be true that
r1 
r1 r2  . The boundary condition at
1
2
2
b
k = n + 1 can be expressed as X n1 


1
r n1  r2 n1  0 , so that for any integer j, r1 n1  r2 n1e i 2 j . This
r1  r2 1
equation has the following n distinct solutions each of which corresponds to a different nontrivial eigenvector:
r1  r2 e
i
2 j
n1
2 j
2 j
 j
i
c i n1
c i
e
for 1  j  n . Hence, r1 2  r1 r2 e n1  e n1 , so that r1 
, where the radical
b
b
designates the principal square root of a complex number. From equation (3) the n eigenvalues of G(n) are given by
3/9/2016
Madison Area Technical College
Page 26
Lab 5 Elementary Matrix and Linear Algebra
j a
c
i
e
 j
n 1
c
b
Spring 2011
 j
c i
2c
 j 
 b e n1  a 
cos
 . (4)
b
c
 n  1
b
It needs to be noted that for complex c and b it is not necessarily true that
2c
 2 bc . The result could be off by a
c
b
1
1
minus sign (recall that  i  

i
1
1
1
1

1
 i ). On the other hand,
1
 n  1  j  
2c
2c
j 
2c
 j 

 , so that allowing the index j to
  a
  a
cos
cos  
cos
 n  1

n  1
n 1 
c
c
c

b
b
b
 j 
 will indeed generate all the eigenvalues of G(n).
run from 1 to n in the expression a  2 bc cos
 n  1
j  a 
Compute and display the five eigenvalues and associated eigenvectors of the following matrix.
i
0
0
 8
 4i 8
i
0

 0 4i 8
i

0 4i 8
 0
 0
0
0 4i
3/9/2016
0
0 
0

i
8 
Madison Area Technical College