The use of Inverse power method to find the least eigenvalue with the corresponding eigenvector.

1. Isaac Amornortey Yowetu
Inverse Power
Method May 5, 2022

2. Theorem (Inverse Eigenvalues)
Suppose that λ, x is an eigenpair of A. If λ 6= 0, then 1
λ
, x is
an eigenpair of the matrix A−1
Suppose A is an invertible matrix or non-singular matrix.
Ax = λx (1)
A−1
Ax
λ
=
λA−1
x
λ
(2)
1
λ
x = A−1
x (3)
2 / 9

3. Finding least Eigenvalue by the Inverse Power M
Considering matrices A and B
A =
−1 3
2
2
3
−1
!
Det(A)=0. Hence, it a singular or non-invertible matrix.
B =
−1 3
2
1 −1
!
Det(B) = −1
2
. Hence, it a non-singular or invertible matrix.
3 / 9

4. Problem
Find the smallest eigenvalue of the matrix using the inverse
power method
B =
−1 3
2
1 −1
!
Take the initail approximation x0 = (1, 1),
4 / 9

5. Finding Eigenvalues Using a Characteristic Equ
B =
−1 3
2
1 −1
!
Using the characteristic equation:
ρ(A) = λ2
+ 2λ − 0.5 = 0
λ =
−b ±
q
b2 − 4(ac)
2a
λ =
−2 ±
q
4 − 4(−0.5)
2
λ1 = −2.2247, λ2 = 0.2247
5 / 9

6. Solution
A = B−1
(4)
=
2 3
2 2
!
(5)
First Iteration yields
Ax0 =
2 3
2 2
!
1
1
!
=
5
4
!
(6)
By applying scaling we have the first approx.
x1 =
1
5
5
4
!
=
1.00
0.80
!
(7)
6 / 9

7. Solution
Second Iteration yields
Ax1 =
2 3
2 2
!
1
0.80
!
=
4.4
3.6
!
(8)
By applying scaling we have the second approx.
x2 =
1
4.4
4.4
3.6
!
=
1.00
0.82
!
(9)
7 / 9

8. Solution
Third Iteration yields
Ax2 =
2 3
2 2
!
1.00
0.82
!
=
4.46
3.64
!
(10)
By applying scaling we have the third approx.
x3 =
1
4.46
4.46
3.64
!
=
1.00
0.82
!
(11)
8 / 9

9. Solution
Fourth Iteration yields
Ax3 =
2 3
2 2
!
1.00
0.82
!
=
4.46
3.64
!
(12)
By applying scaling we have the fourth approx.
x4 =
1
4.46
4.46
3.64
!
=
1.00
0.82
!
(13)
The smallest eigenvalue is: x4 =
1
4.46
= 0.225
9 / 9