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Power method: 1,4,-1 matrix

Short answer

Starting from [1,1,1]ᵀ, repeatedly multiply by the matrix and normalize by the largest component. After 10 iterations the dominant eigenvalue converges to approximately 12.471, with corresponding eigenvector approximately [0.092, 0.512, 1.000]ᵀ.

Find the dominant Eigen value and the corresponding vector of the following matrix using power method. [1 4 -1; 4 2 5; -1 5 10]

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    4   -1
  4    2    5
 -1    5   10

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, 11.0000, 14.0000]14.0000[0.2857, 0.7857, 1.0000]
2[0.2857, 0.7857, 1.0000][2.4286, 7.7143, 13.6429]13.6429[0.1780, 0.5654, 1.0000]
3[0.1780, 0.5654, 1.0000][1.4398, 6.8429, 12.6492]12.6492[0.1138, 0.5410, 1.0000]
4[0.1138, 0.5410, 1.0000][1.2777, 6.5373, 12.5911]12.5911[0.1015, 0.5192, 1.0000]
5[0.1015, 0.5192, 1.0000][1.1783, 6.4443, 12.4945]12.4945[0.0943, 0.5158, 1.0000]
6[0.0943, 0.5158, 1.0000][1.1574, 6.4088, 12.4846]12.4846[0.0927, 0.5133, 1.0000]
7[0.0927, 0.5133, 1.0000][1.1460, 6.3975, 12.4740]12.4740[0.0919, 0.5129, 1.0000]
8[0.0919, 0.5129, 1.0000][1.1433, 6.3932, 12.4725]12.4725[0.0917, 0.5126, 1.0000]
9[0.0917, 0.5126, 1.0000][1.1420, 6.3919, 12.4713]12.4713[0.0916, 0.5125, 1.0000]
10[0.0916, 0.5125, 1.0000][1.1417, 6.3913, 12.4711]12.4711[0.0915, 0.5125, 1.0000]

From the last two rows, λ and the vector are unchanged to 3 decimal places.

Answer: dominant eigenvalue λ ≈ 12.471; corresponding eigenvector ≈ [0.092, 0.512, 1.000]ᵀ (any multiple of it is also an eigenvector).

✓ Answer checked by Claude Opus 5.5 (high effort)

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