Question

Compute the largest eigenvalue in magnitude, and the corresponding eigenvector of the matrix
A =1 −1 1
2 0 3
1 4 −1
by performing four iterations of the power method.

Answer 1

$A=\begin{bmatrix} 1 & -1&1 \\ 2 & 0&3\\ 1&4&-1 \end{bmatrix}$

Begin with an initial nonzero approximation of

$x_0=\begin{bmatrix} 1 \\ 1\\ 1 \end{bmatrix}$

Then obtain the following approximations

Iteration

$x_1=Ax_0=\begin{bmatrix} 1 & -1&1 \\ 2 & 0&3\\ 1&4&-1 \end{bmatrix}\begin{bmatrix} 1 \\ 1\\ 1 \end{bmatrix} =\begin{bmatrix} 1 \\ 5\\ 4 \end{bmatrix}$

“Scaled” Approximation

$\to1\begin{bmatrix} 1 \\ 5\\ 4 \end{bmatrix}$

Iteration

$x_2=Ax_1=\begin{bmatrix} 1 & -1&1 \\ 2 & 0&3\\ 1&4&-1 \end{bmatrix}\begin{bmatrix} 1 \\ 5\\ 4 \end{bmatrix} =\begin{bmatrix} 0 \\ 14\\ 17 \end{bmatrix}$

“Scaled” Approximation

$\to14\begin{bmatrix} 0 \\ 1\\ 1.2 \end{bmatrix}$

Iteration

$x_3=Ax_2=\begin{bmatrix} 1 & -1&1 \\ 2 & 0&3\\ 1&4&-1 \end{bmatrix}\begin{bmatrix} 0 \\ 14\\ 17 \end{bmatrix} =\begin{bmatrix} 3 \\ 51\\ 39 \end{bmatrix}$

“Scaled” Approximation

$\to3\begin{bmatrix} 1 \\ 17\\ 13 \end{bmatrix}$

Iteration

$x_4=Ax_3=\begin{bmatrix} 1 & -1&1 \\ 2 & 0&3\\ 1&4&-1 \end{bmatrix}\begin{bmatrix} 3 \\ 51\\ 39 \end{bmatrix} =\begin{bmatrix} -9 \\ 123\\ 168 \end{bmatrix}$

“Scaled” Approximation

$\to-9\begin{bmatrix} 1 \\ -13.7\\ -18.7 \end{bmatrix}$

Note that the approximations appear to be approaching scalar multiples of

$\begin{bmatrix} 1 \\ -14\\ -19 \end{bmatrix}$

With $x=(1,-13.7,-18.7)$ as the approximation of a dominant eigenvector of $A$ , use the Rayleigh quotient to obtain an approximation of the dominant eigenvalue of $A$ . First compute the product $Ax$

$Ax=\begin{bmatrix} 1 & -1&1 \\ 2 & 0&3\\ 1&4&-1 \end{bmatrix}\begin{bmatrix} 1 \\ -13.7\\ -18.7 \end{bmatrix} =\begin{bmatrix} -4 \\ -54.1\\ -35.1 \end{bmatrix}$

Then, because 

$Ax\cdot x=-4\cdot 1+(-54.1)(-13.7)+(-35.1)(-18.7)=1393.54\\ x\cdot x=1\cdot1+(-13.7)(-13.7)+(-18.7)(-18.7)=538.38$

you can compute the Rayleigh quotient to be

$\lambda=\frac{Ax\cdot x}{x\cdot x}=\frac{1393.54}{538.38}=2.6$

which is a good approximation of the dominant eigenvalue

$\lambda=3$

Answer $\begin{bmatrix} 1 \\ -14\\ -19 \end{bmatrix}, \lambda=3$