Linear Algebra and Its Applications · Problem 7.3.2

Computations with Matrices — Gilbert Strang

Mathematics · inverse-power-method

Problem

For the same

A

and the initial guess

u_0 = \begin{bmatrix} 3 \\ 4 \end{bmatrix}

, compare three inverse power steps to one shifted step with

\alpha = u_0^T A u_0 / u_0^T u_0

u_{k+1} = A^{-1} u_k = \frac{1}{3} \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix} u_k \qquad \text{or} \qquad u = (A - \alpha I)^{-1} u_0.

The limiting vector

u_\infty

is now a multiple of the other eigenvector

(1,1)