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Question 3 by Power Method

The document details the application of the Power Method to a specific matrix to estimate its dominant eigenvalue and eigenvector. It includes an initial guess vector and outlines the iterative process, providing approximate eigenvalues and normalized eigenvectors for each iteration. The final results indicate that the maximum eigenvalue is approximately 7.0000 with a corresponding eigenvector of approximately [0.0667, 1.0000].

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Muhammad Waheed
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52 views3 pages

Question 3 by Power Method

The document details the application of the Power Method to a specific matrix to estimate its dominant eigenvalue and eigenvector. It includes an initial guess vector and outlines the iterative process, providing approximate eigenvalues and normalized eigenvectors for each iteration. The final results indicate that the maximum eigenvalue is approximately 7.0000 with a corresponding eigenvector of approximately [0.0667, 1.0000].

Uploaded by

Muhammad Waheed
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Power Method Iterations

May 20, 2025

We apply the Power Method to the matrix:


 
1 −3 2
A = 4 4 −1
6 3 5
Initial guess vector:
   
1 1
1  
x(0) = 1 , normalized: √ 1
1 3 1

Ax(k)
We iterate using x(k+1) = ∥Ax(k) ∥∞
, and estimate eigenvalue using Rayleigh quotient.

Iterations:

1
Iteration Approx. Eigenvalue Normalized
 Eigenvector
 x(k)
0.0000
1 8.4999 0.5000
1.0000
0.0769
2 6.7500 0.1538
 1.0000 
0.2727
3 6.1744 −0.0126
1.0000
0.3503
4 6.5826 0.0061
1.0000
0.2996
5 6.8414 0.0656
1.0000
0.3086
6 6.9606 0.0690
1.0000
0.2982
7 6.9909 0.0664
1.0000
0.3003
8 6.9979 0.0668
1.0000
0.2998
9 6.9995 0.0667
1.0000
0.3000
10 6.9999 0.0667
1.0000
0.2999
11 7.0000 0.0667
1.0000
0.3000
12 7.0000 0.0667
1.0000
0.3000
13 7.0000 0.0667
1.0000
0.3000
14 7.0000 0.0667
1.0000
0.3000
15 7.0000 0.0667
1.0000
0.3000
16 7.0000 0.0667
1.0000
0.3000
17 7.0000 0.0667
1.0000
0.3000
18 7.0000 0.0667
1.0000
2
Final Result:
 
0.3000
λmax ≈ 7.0000 , x ≈ 0.0667
1.0000

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