This document discusses the eigenvalue-eigenvector problem, which is important for solving differential equations, modeling population growth, and calculating matrix powers. It defines eigenvalues and eigenvectors and provides examples of solving eigenvalue problems. The power method, an iterative approach, is described for finding the dominant or lowest eigenvalue of a matrix. Solving eigenvalue problems has applications in many fields including physics, biology, economics and statistics.

1. Eigenvalue – Eigenvector Problem

2. EIGENVALUES & EIGENVECTORS
The eigenvalue problem is a problem of considerable
theoretical interest and wide-ranging application.
For example, this problem is crucial in solving systems of
differential equations, analyzing population growth models,
and calculating powers of matrices (in order to define the
exponential matrix (A100)).
Other areas such as physics, sociology, biology, economics
and statistics have focused considerable attention on
“eigenvalues” and “eigenvectors”-their applications and their
computations

4. Eigenvalue Problems
(Mathematical Background)
a11  λx1  a12x2  a1nxn  0
a21x1  a22  x2  a1nxn  0
   
an1x1  an2x2  ann  xn  0
D Determinant AI
AX0
The roots of polynomial D(λ) are the eigenvalues of the eigen system
A solution {X} to [A]{X} = λ{X} is an eigen vector
(homogeneous system)
AIX0 (eigen system)
1 2 3
4 5 6
7 8 9
1-c 2 3
4 5-c 6
7 8 9-c

6. Example 1

7. 4x1+4x2=0 4x1=-4x2
X1+x2=0 x1=-x2

8. CW

9. Example 2

11. C2+2c+2c+4=0  c(c+2)+2(c+2)=0
C3+4c2+4c-4c2-16c-16
𝑎3 − 12𝑎 − 16
𝑎 − 4
= 𝑎2
+ 4𝑎 + 4

12. R1+1/3R3
1 5 -3
3 -9 3
6 -6 0
1 5 -3
0 1 -1/2
6 -6 0

17. 15
The Power Method
(An iterative approach for determining the largest eigenvalue)
Example (3):
Iteration 1: initialization [x1, x2, x3]T = [1 1 1]T
Iteration 2: A [10 1]T
Iteration 3: A [1 -11]T
Iteration 4: A [-0.751 -0.75]T
Iteration 4: A [-0.7141 -0.714]T
(Exact solution = 6.070)

18. Class Work: Use the power method to find the dominant eigenvalue and
eigenvector for the matrix

21. 21
Power Method for Lowest Eigenvalue
(An iterative approach for determining the lowesteigenvalue)

0.141
0.141
0.281
0.562 0.281

0.281 0.422
0.422
A1
 0.281



 

   
  
0.751
0.751
0.884
0.1411
0.281
0.562 0.2811  1.124  1.124 1
0.281 0.4221 0.884
0.281
0.141
0.422

Iteration 1:
Iteration 3:
Exact solution is 0.955 which is the reciprocal of the smallest eigenvalue,
1.0472 of [A].
Iteration 2:






 
  



0.715
0.715
  0.984 1
0.984
0.281 0.1410.751 0.704
0.562 0.281 1
0.281 0.4220.751 0.704

0.141
0.281
0.422



 


     
0.709
0.709

 0.964

1
0.964
0.1410.715 0.684
0.281
0.562 0.281
1
0.281 0.4220.715 0.684

0.141
0.281
0.422
Idea: The largest eigenvalue of [A]-1 is the reciprocal of the lowest
eigenvalue of [A]


1.7783.556

 0 1.778
 3.556 1.778 0 
Example (5): A 1.778 0.281 3.556
0.422

22. Home Work
1.
2.
3.