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Power method: 2,5,1 matrix

Short answer

Using the power method from initial vector [1,1,1] and scaling by the largest component, after 12 iterations the dominant eigenvalue is 11.289 and the corresponding eigenvector is [0.287, 0.334, 1.000] (any multiple).

Find the largest Eigen value and corresponding vector of the following matrix using power method. [2 5 1; 5 -2 3; 1 3 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 =
  2    5    1
  5   -2    3
  1    3   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][8.0000, 6.0000, 14.0000]14.0000[0.5714, 0.4286, 1.0000]
2[0.5714, 0.4286, 1.0000][4.2857, 5.0000, 11.8571]11.8571[0.3614, 0.4217, 1.0000]
3[0.3614, 0.4217, 1.0000][3.8313, 3.9639, 11.6265]11.6265[0.3295, 0.3409, 1.0000]
4[0.3295, 0.3409, 1.0000][3.3637, 3.9658, 11.3523]11.3523[0.2963, 0.3493, 1.0000]
5[0.2963, 0.3493, 1.0000][3.3393, 3.7828, 11.3443]11.3443[0.2944, 0.3335, 1.0000]
6[0.2944, 0.3335, 1.0000][3.2560, 3.8049, 11.2947]11.2947[0.2883, 0.3369, 1.0000]
7[0.2883, 0.3369, 1.0000][3.2609, 3.7676, 11.2989]11.2989[0.2886, 0.3335, 1.0000]
8[0.2886, 0.3335, 1.0000][3.2445, 3.7761, 11.2890]11.2890[0.2874, 0.3345, 1.0000]
9[0.2874, 0.3345, 1.0000][3.2473, 3.7680, 11.2909]11.2909[0.2876, 0.3337, 1.0000]
10[0.2876, 0.3337, 1.0000][3.2438, 3.7706, 11.2888]11.2888[0.2873, 0.3340, 1.0000]
11[0.2873, 0.3340, 1.0000][3.2448, 3.7687, 11.2894]11.2894[0.2874, 0.3338, 1.0000]
12[0.2874, 0.3338, 1.0000][3.2440, 3.7694, 11.2889]11.2889[0.2874, 0.3339, 1.0000]

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

Answer: dominant eigenvalue λ ≈ 11.289; corresponding eigenvector ≈ [0.287, 0.334, 1.000]ᵀ (any multiple of it is also an eigenvector).

✓ Answer checked by Claude Opus 5.5 (high effort)

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