Eigenvalues and Eigenvectors
• A(nxn) has precisely n, not necessarily distinct
eigenvalues.
• That are the roots of p(λ) = det(λI-A) = 0
• In practice, p(λ) is difficult to obtain
Localisation of the eigenvalues
• Theorem (Hadamard-Gershgorin) : Let A an nxn
matrix and Di denote the circle in the complex plane
n
with center aii and radius
a
j 1
ij
; that is
j i


n


Di   z  C; Z  aii   aij 
i 1


j i


 4

A   0
 2

Let
D3
2
0
Z  C ; Z  4
 Z  C ; Z  2
 Z  C ; Z  9
D1 
D2
1
1

1
9

 2,
 1,
 2.
Two eigenvalue s within D1  D2 and one within D3
 ( A)  max 1i 3 i , 
7   ( A)  11
• Theorem: The eigenvalues of a real symmetric matrix A are real.
• Theorem: If
1 , 2 , ..., n are the eigenvalue s of the matrix A we have
1  2  ...  n  trace( A)
1. 2. ... . n  det( A)
• Definition: Two nxn matrices A and B are said to be similar if a
non-singular matrix S exists with A = S-1.B.S
• Theorem: Suppose A and B are similar nxn matrices
and λ is an eigenvalue of A, with associated
eigenvector X. Then λ is also an eigenvalue of B and
if A =S-1BS, then SX is an eigenvector associated with
λ and the matrix B.
• Definition:


A set of vectors V (1) , V (2) , ..., V (n) is called orthogonal
if (V (i) ) t V (j)  0 for all i  j. If, in addition, (V (i) ) t V (i)  1
for all i  1, 2, ..., n, then the set is called orthonorma l.
• Theorem: If A is a matrix and λ1, λ2, …, λk are distinct
eigenvalues of A with associated eigenvectors x(1),
x(2),…,x(k), then {x(1), x(2),…,x(k)}, is linearly
independent.
• Theorem: An orthogonal set of vectors that does not
contain the zero vector is linearly independent.
• Definition: An nxn matrix P is said to be an
orthogonal matrix if P-1 = Pt.
• Corollary: If A is a symmetric nxn matrix, then there
exist n eigenvectors of A that form an orthonormal
set.
• If A is an nxn symmetric matrix, then there exists an
orthogonal matrix P such that D = P-1AP where D is a
diagonal matrix whose diagonal entries are the
eigenvalues of A.
The algebraic eigenvalue problem
• Let A be a nxn matrix. Find A.x = λ.x where λ:scalar
and x a vector.
• det(A- λ.I) X = 0
2 0 0
A  2 2 1 , det( A   .I )  3  6.2  11.  6  0
1 1 2
1  1, 2  2, 3  3
Eigenvectors:
1 0
0
1 0 0 
P   1  1 1 and D  0 2 0
 1  2 1
0 0 3
with D  P 1. A.P
Eigenvalues of symmetric tridiagonal matrices
Sturm sequences
• The matrix A=(aij) is tridiagonal if aij = 0 if |i-j|>1
• Example:
 a1
b
 2
0
A
.
.

 0
b2
0
.
.
a2
b3
.
.
b3
a3
b4
.
.
.
.
.
.
.
bn 1
an 1
.
.
0
bn
and aij=aji
0

.
.

0
bn 

an 
Sturm sequences
P0 ( )  1  P1 ( )  (a1   )
P2 ( )  (a2   ).P1 ( )  b22 .P0 ( )
for r  1, 2,..., (n  1).
Pr 1 ( )  (ar 1   ).Pr ( )  br21.Pr 1 ( )
Example :
1
2
A
0

0
2 0 0
1 2 0

2 1 2

0 2 1
where a1  a2  a3  a4  1 and b 2  b 3  b 4  2
two roots  [-1, 3]
two roots  [-3,5]
1  2.236 and 2  -2.236
3  4.236 and 4  -0.236
Jacobi Method (Eigenvalues and Eigenvectors)
• Suppose that A = (aij) is a symmetric matrix.
• Let Ω denote the first orthogonal matrix:
.
. 0
1 0 .
0 .
.


.
c s
.


.
s
c
.


.
. 0


0
.
.
.
0
1


• Select row p and column q for which apq ≠ 0
• Form the preliminary quantities

a qq - a pp
2a pq
; t
sign(  )
(    2  1)
with sign(  )  1 if   0
sign(  )  -1 if   0
and c 
1
t 1
2
; s  ct
• Start with A(0) = A, and Q(0) = I
• Construct the sequence :
A (k 1)   k A (k) Tk
 ( k 1)
(k )
Q
 k Q
• Stop with A(k) is a diagonal matrix
• D = Q.A.QT where D diagonal matrix
• Q Eigenvectors matrix.
Example
1 2 3
2 1 4
A
3 4 1

4 3 2
apq  a14  4
4
3
2

1
Eigenvalue s : - 8  i  10
p  1 and q  4   
t
aqq  a pp
2a pq
0
sign ( )
1

1

c

(    2  1) (1/ 2 )
(t 2  1) (1/ 2 )
S  ct
sc
 2

 2
0

 0

2

 2
0 0
1 0
0 1
0 0
2
2
2

2 
0 
0 
2

2 
i  1,2,3,4
A
(1)

 3

 2
  ( 0 ) . A ( 0 ) .T0   2

 3

 0

a pq  4  a 23
1

0
1  
0

0
0
2
2
2
2
0
p2 q3
0
2

2
2
2
0
0

0

0

1
2

2
2
2
1
4
4
1

0 

5 2
2 
5 2

2 
0 

5 2 5 2
2
2
a -a
  33 22  0
8
A (2)
t 1
Q (1)   (0) .Q (0)   (0)
cs
1
2
 3  1 0 0
  1  3 0 0

 1.A (1) .1T  
0
0 5 5


0 5 5
0
Q (2)  1.Q (1)










2
2
0
0
2
2
0
0
2
2
2
2
2

2
2
2
0
0
2

2 
0 


0 

2 
2 
A( 3)
A( 4 )
 3  1 0 0 
1  3 0 0 

  ( 2) . A ( 2) .T2  
3
0 0 0


0
0
0
10


 4 0
 0 2

0
0

0
0
0
0 0

0 0

0 10
0
Q (4)   3 .Q (3)
Eigenvecto rs (column of Q (4) ).
Q (3)   (2) .Q (2)
1
1
 1
 2 2 2
 1
1
1

2
2
 2
 1  1 1
 2
2 2
 1 1
1


2 2
2
1
2
1

2
1
2
1

2
Power Method
 4 1 1 
A   1 3  2
 1  2 3 
has the dominant eigenvalue 1  6 and associated
v (1)  (1,-1,1)t
The procedure outline above obtaining 2
proceeds as follows :
y0  (1,0,0)  y1  A. y0  (4,1, 1)
T
y1

4 
T
(11)  4  z1  1,0.25, 0.25T
y2  A.z1  (5,  3.5, 3.5)T
 y2

 5  (12 )  5
z 2  1,  0.7, 0.7 
T
y3  A.z 2  (5.4,  4.5, 4.5)T
 y3

(13)  5.4  z3  1,0.833, 0.8333T
y4  A.z3  (5.666,  5.1666, 5.1666)T
 y4

 5.666  (14 )  5.666
z 4  1,0.911, 0.911
T
 5.4
.......... .......... .......... .......... .......... .......... .......
y10  A.z9  (5.988,  5.978, 5.97)T
 y10
 (110 )  5.988
z10  1,  0.997, 0.997
T
Let
1  6 and v (1)  (1,  1, 1)T

 5.988
• Theorem: Suppose that λ1, λ2 … λn are eigenvalues of
A with associated eigenvectors V(1), V(2),…,V(n) , and
λ1 has multiplicity one. If x is any vector with the
property that XtV(1)=1, then
B = A - λ1V(1)Xt
Has eigenvalues 0, λ2 … λn with associated
eigenvectors V(1), W(2),…,W(n) , where V(i) and W(i) are
related the equation
V(i) = (λi- λ1 ).W(i) + λ1(Xt.W(i))V(1)
For each i = 2,3,…,n
• A particularly useful deflation technique, called the Wielandt’s
deflation procedure, results by defining
x 
1
1vi1
 ai1 
a 
 i2 
 . 


a
 in 
1
Where vi is a coordinate of V(1) which is nonzero, and the
values ai1, ai2, …, ain are the entries in the ith row of A.
With this definition, we have
n
1
1
xt v (1)  (1) ai1 , a i2 , ..., a in (v1(1) , v2(1) ,..., vn(1) )  (1)  aij v (j1)
1vi
1vi j 1
•Where the sum is the ith coordinate of the product AV(1). Since
AV(1)=V(1), this implies that
1
x v  (1) (1vi(1) )  1
1vi
t (1)
The ith row of B = A - λ1V(1)Xt consists entirely of zero entries.
The matrix B can be replace by an (n-1)x(n-1) matrix B’ that is
obtained by deleting the ith row and column from B.
B’ will have eigenvalues λ2, λ3 … λn . If | λ2 | > | λ3 |, The power
method can be reapplied to the matrix B’ to determine this new
dominant eigenvalues and an eigenvetors.
 4 1 1 
A   1 3  2
 1  2 3 
v (1)  (1,-1,1)t
The procedure outline above obtaining 2 proceeds as follows
4
1
2
x   1  
6
3
 1 
1
2


(1) t
v x   1 
3
 1 
1
6
1
6
1

6
t
 2
 3
1  2


6  3
 2
 3
1
6
1
6
1
6
1
6
 1

6
1
6 
and B  A - 1v (1) x t
0 0
  3 2
- 3 - 1
 2
 3
4

1
1

 
2


   1 3  2  6 
 3
 1  2 3   2
 3
0
- 1
2 
1
6
1
6
1
6
1
6
 1

6
1
6 
 2  1
Deleting the first row and column gives B  


1
2


'
Which has eigenvalues λ2 = 3 and λ3 = 1. For λ2 = 3 the
eigenvector, W(2)’ , can be obtained by solving the second
order linear system
(B’ – 3I)W(2)’ = 0
Resulting in
W(2)’ = (1, -1)t
Thus, W(2) = (0, 1, -1)t and, from equation
V(i) = (λi- λ1 ).W(i) + λ1(Xt.W(i))V(1)
V(2) =(3-6).(0,1,-1)t + 6((2/3,-1/6,1/6).(0, 1, -1)t)(1,-1,1)t
=(-2,-1,1)t
W(3)’ = (1, 1)t
W(3) = (0, 1, 1)t
V(3) =(1-6).(0,1,1)t + 6((2/3,-1/6,1/6).(0, 1, 1)t)(1,-1,1)t
=(0, -4, 4)t
Power and deflation method
6
2
A
1

 1
 1
4 1 0

1 4  1

0 1 3 
2
1
y0  (1,1,1,1)T  y1  A. y0  (8,7,5,1)T


y1   8    8  z1  1, 7 , 5 , 1  1, 0.875, 0.625, 0.125
8 8 8
y2  A.z1  (8.25, 6.125, 4.25,  1.25)T  y2   8.25  (12)  8.25
(1)
1
T
T
z2  1, 0.7424, 0.5151,  0.1515
T
y3  A.z2  (8.1514, 5.4847, 3.954,  0.9696)T
 y3

 8.1514
   8.1514  z3  1, 0.6728, 0.485107,  0.24163
( 3)
1
T
y4  A.z3  (8.0724, 5.1765, 3.8549,  2.21)T
 y4
1( 4)  8.072
z4  1, 0.64125, 0.4775,  0.27377
T
y5  A.z4  8.03381, 5.04254, 3.82518,  2.29885
T
y5

 8.03381
1(5)  8.03
z5  1, 0.627, 0.476,  0.2861
T
Let 1  8 and v
(1)
 1, 0.627, 0.476,  0.2861
T

 8.072
Deflation method for
2  ?
Find x t .v (1)  1
 3 
 ai1 
 6 
4 

a 
 2 
 14 
1
1
i
2
  .
  
x 
.
with v i(1)  1

(1)
1
 ai 3 
8  1 
i .v i
 8 




 1 
a

1


 i4 

8
0.470
0.357
0.214 
 0.75
 0.25
0.156
0.119
 0.071
(1)
t

v .x  
 0.125
0.078
0.059
 0.035


0.035 
  0.125  0.078  0.059
B  A  1.v (1) .x t  A  8.v (1) .x t 
0
0

0

0
 1.762
 1.856
2.746
0.048
0.373
3.524
0.627
 0.524
0.716 
0.572 

 0.714

2.714 
• Deleting the first row an column gives
2.746
B'   0.373

0.627
0.048
3.524
 0.524
0.572 
 0.714

0.714 