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Power method: 25,1,2 matrix

Short answer

Starting from [1,1,1] and iterating with the power method, scaling by the largest component gives the dominant eigenvalue 25.1821 and eigenvector [1.0000, 0.0451, 0.0685] (any scalar multiple).

Find the largest eigen-value and the corresponding eigen-vector of the following matrix. [25 1 2; 1 3 0; 2 0 -4]

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 =
 25    1    2
  1    3    0
  2    0   -4

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][28.0000, 4.0000, -2.0000]28.0000[1.0000, 0.1429, -0.0714]
2[1.0000, 0.1429, -0.0714][25.0000, 1.4286, 2.2857]25.0000[1.0000, 0.0571, 0.0914]
3[1.0000, 0.0571, 0.0914][25.2400, 1.1714, 1.6343]25.2400[1.0000, 0.0464, 0.0647]
4[1.0000, 0.0464, 0.0647][25.1759, 1.1392, 1.7410]25.1759[1.0000, 0.0453, 0.0692]
5[1.0000, 0.0453, 0.0692][25.1836, 1.1358, 1.7234]25.1836[1.0000, 0.0451, 0.0684]
6[1.0000, 0.0451, 0.0684][25.1820, 1.1353, 1.7263]25.1820[1.0000, 0.0451, 0.0686]
7[1.0000, 0.0451, 0.0686][25.1822, 1.1353, 1.7258]25.1822[1.0000, 0.0451, 0.0685]
8[1.0000, 0.0451, 0.0685][25.1821, 1.1352, 1.7259]25.1821[1.0000, 0.0451, 0.0685]

From the last two rows, λ and the vector are unchanged to 4 decimal places.

Answer: dominant eigenvalue λ ≈ 25.1821; corresponding eigenvector ≈ [1.0000, 0.0451, 0.0685]ᵀ (any multiple of it is also an eigenvector).

✓ Answer checked by Claude Opus 5.5 (high effort)

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