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Power method: 1,2,1 matrix

Short answer

Start with initial vector [1,1,1] and iterate, scaling by the largest component. After 13 iterations, the dominant eigenvalue converges to 3.93 and the corresponding eigenvector is [0.88, 1.00, 0.58] (any scalar multiple).

Find the largest Eigen value and the corresponding Eigen vector of the following matrix using the power method with an accuracy of 2 decimal points. [1 2 1; 2 1 2; 1 2 -1]

Power method: to find the dominant (largest in magnitude) eigenvalue λ and its eigenvector x of A, start with a trial vector x(0), compute A·x(k) repeatedly and scale it by its largest component. The scale factor tends to λ and the scaled vector tends to the eigenvector.

Step 1: Matrix and starting vector

A =
  1    2    1
  2    1    2
  1    2   -1

Take x(0) = [1, 1, 1]ᵀ. Iterate: A·x(k) = λ(k)·x(k+1), where λ(k) is the component of largest magnitude in A·x(k).

Step 2: Iterations

kx(k)A·x(k)λ(k) (largest component)x(k+1) = A·x(k)/λ(k)
1[1.0000, 1.0000, 1.0000][4.0000, 5.0000, 2.0000]5.0000[0.8000, 1.0000, 0.4000]
2[0.8000, 1.0000, 0.4000][3.2000, 3.4000, 2.4000]3.4000[0.9412, 1.0000, 0.7059]
3[0.9412, 1.0000, 0.7059][3.6471, 4.2941, 2.2353]4.2941[0.8493, 1.0000, 0.5205]
4[0.8493, 1.0000, 0.5205][3.3699, 3.7397, 2.3288]3.7397[0.9011, 1.0000, 0.6227]
5[0.9011, 1.0000, 0.6227][3.5238, 4.0476, 2.2784]4.0476[0.8706, 1.0000, 0.5629]
6[0.8706, 1.0000, 0.5629][3.4335, 3.8670, 2.3077]3.8670[0.8879, 1.0000, 0.5968]
7[0.8879, 1.0000, 0.5968][3.4847, 3.9693, 2.2911]3.9693[0.8779, 1.0000, 0.5772]
8[0.8779, 1.0000, 0.5772][3.4551, 3.9102, 2.3007]3.9102[0.8836, 1.0000, 0.5884]
9[0.8836, 1.0000, 0.5884][3.4720, 3.9440, 2.2952]3.9440[0.8803, 1.0000, 0.5820]
10[0.8803, 1.0000, 0.5820][3.4623, 3.9246, 2.2984]3.9246[0.8822, 1.0000, 0.5856]
11[0.8822, 1.0000, 0.5856][3.4678, 3.9357, 2.2966]3.9357[0.8811, 1.0000, 0.5835]
12[0.8811, 1.0000, 0.5835][3.4647, 3.9293, 2.2976]3.9293[0.8817, 1.0000, 0.5847]
13[0.8817, 1.0000, 0.5847][3.4665, 3.9330, 2.2970]3.9330[0.8814, 1.0000, 0.5840]

From the last two rows, λ and the vector are unchanged to 2 decimal places.

Answer: dominant eigenvalue λ ≈ 3.93; corresponding eigenvector ≈ [0.88, 1.00, 0.58]ᵀ (any multiple of it is also an eigenvector).

✓ Answer checked by Claude Opus 5.5 (high effort)

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