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Power method: 15,-4,-3 matrix

Short answer

Power method with initial [1,1,1], scaling by third component after sign stabilization. Dominant eigenvalue converges to 20.0000, eigenvector to [-1.0000, 0.5000, 1.0000].

Find the dominant Eigen value and the corresponding Eigen vector of the given matrix using power method. [15 -4 -3; -10 12 -6; -20 4 -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 one of its components. The scale factor tends to λ and the scaled vector tends to the eigenvector.

Step 1: Matrix and starting vector

A =
 15   -4   -3
-10   12   -6
-20    4   -2

Take x(0) = [1, 1, 1]ᵀ. Iterate: A·x(k) = λ(k)·x(k+1), where λ(k) is the scale factor.

For k = 1 to 3, λ(k) is the component of largest magnitude in A·x(k). From k = 4 on, the first and third components of A·x(k) are almost equal in size and opposite in sign, so we always scale by the third component. This keeps the vector from flipping sign at every step. (Scaling by whichever of the two happens to be slightly larger would make the vector change sign each time and give the misleading estimate −20.)

Step 2: Iterations

kx(k)A·x(k)λ(k) (scale factor)x(k+1) = A·x(k)/λ(k)
1[1.0000, 1.0000, 1.0000][8.0000, -4.0000, -18.0000]-18.0000[-0.4444, 0.2222, 1.0000]
2[-0.4444, 0.2222, 1.0000][-10.5556, 1.1111, 7.7778]-10.5556[1.0000, -0.1053, -0.7368]
3[1.0000, -0.1053, -0.7368][17.6316, -6.8421, -18.9474]-18.9474[-0.9306, 0.3611, 1.0000]
4[-0.9306, 0.3611, 1.0000][-18.4028, 7.6389, 18.0556]18.0556[-1.0192, 0.4231, 1.0000]
5[-1.0192, 0.4231, 1.0000][-19.9808, 9.2692, 20.0769]20.0769[-0.9952, 0.4617, 1.0000]
6[-0.9952, 0.4617, 1.0000][-19.7749, 9.4923, 19.7510]19.7510[-1.0012, 0.4806, 1.0000]
7[-1.0012, 0.4806, 1.0000][-19.9406, 9.7793, 19.9467]19.9467[-0.9997, 0.4903, 1.0000]
8[-0.9997, 0.4903, 1.0000][-19.9565, 9.8803, 19.9550]19.9550[-1.0001, 0.4951, 1.0000]
9[-1.0001, 0.4951, 1.0000][-19.9816, 9.9423, 19.9820]19.9820[-1.0000, 0.4976, 1.0000]
10[-1.0000, 0.4976, 1.0000][-19.9900, 9.9705, 19.9899]19.9899[-1.0000, 0.4988, 1.0000]
11[-1.0000, 0.4988, 1.0000][-19.9952, 9.9854, 19.9952]19.9952[-1.0000, 0.4994, 1.0000]
12[-1.0000, 0.4994, 1.0000][-19.9975, 9.9927, 19.9975]19.9975[-1.0000, 0.4997, 1.0000]
13[-1.0000, 0.4997, 1.0000][-19.9988, 9.9963, 19.9988]19.9988[-1.0000, 0.4998, 1.0000]
14[-1.0000, 0.4998, 1.0000][-19.9994, 9.9982, 19.9994]19.9994[-1.0000, 0.4999, 1.0000]
15[-1.0000, 0.4999, 1.0000][-19.9997, 9.9991, 19.9997]19.9997[-1.0000, 0.5000, 1.0000]
16[-1.0000, 0.5000, 1.0000][-19.9998, 9.9995, 19.9998]19.9998[-1.0000, 0.5000, 1.0000]
17[-1.0000, 0.5000, 1.0000][-19.9999, 9.9998, 19.9999]19.9999[-1.0000, 0.5000, 1.0000]
18[-1.0000, 0.5000, 1.0000][-20.0000, 9.9999, 20.0000]20.0000[-1.0000, 0.5000, 1.0000]

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

Check: A·[−1, 0.5, 1]ᵀ = [−20, 10, 20]ᵀ = 20·[−1, 0.5, 1]ᵀ.

Answer: dominant eigenvalue λ ≈ 20.0000; corresponding eigenvector ≈ [-1.0000, 0.5000, 1.0000]ᵀ (any multiple of it is also an eigenvector).

✓ Answer checked by Claude Opus 5.5 (high effort)

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