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Power method: tridiagonal 2,-1

Short answer

Using power method with initial vector [1,1,1], after iterations the dominant eigenvalue converges to 3.4142 and corresponding eigenvector to [-0.7071, 1.0000, -0.7071].

Find the dominant eigenvalue and eigenvector of the matrix: [2 -1 0; -1 2 -1; 0 -1 2]

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   -1    0
 -1    2   -1
  0   -1    2

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][1.0000, 0.0000, 1.0000]1.0000[1.0000, 0.0000, 1.0000]
2[1.0000, 0.0000, 1.0000][2.0000, -2.0000, 2.0000]2.0000[1.0000, -1.0000, 1.0000]
3[1.0000, -1.0000, 1.0000][3.0000, -4.0000, 3.0000]-4.0000[-0.7500, 1.0000, -0.7500]
4[-0.7500, 1.0000, -0.7500][-2.5000, 3.5000, -2.5000]3.5000[-0.7143, 1.0000, -0.7143]
5[-0.7143, 1.0000, -0.7143][-2.4286, 3.4286, -2.4286]3.4286[-0.7083, 1.0000, -0.7083]
6[-0.7083, 1.0000, -0.7083][-2.4167, 3.4167, -2.4167]3.4167[-0.7073, 1.0000, -0.7073]
7[-0.7073, 1.0000, -0.7073][-2.4146, 3.4146, -2.4146]3.4146[-0.7071, 1.0000, -0.7071]
8[-0.7071, 1.0000, -0.7071][-2.4143, 3.4143, -2.4143]3.4143[-0.7071, 1.0000, -0.7071]
9[-0.7071, 1.0000, -0.7071][-2.4142, 3.4142, -2.4142]3.4142[-0.7071, 1.0000, -0.7071]
10[-0.7071, 1.0000, -0.7071][-2.4142, 3.4142, -2.4142]3.4142[-0.7071, 1.0000, -0.7071]

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

Answer: dominant eigenvalue λ ≈ 3.4142; corresponding eigenvector ≈ [-0.7071, 1.0000, -0.7071]ᵀ (any multiple of it is also an eigenvector).

✓ Answer checked by Claude Opus 5.5 (high effort)

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