Answer to Question #246646 in Quantitative Methods for tabia

Question

use the power method to estimatedominant eigenvalue and the corresponding eigenvector for the matrix A =[1 2 1,0 2 5 ,1 0 -1.Start with the vector[1 1 1] and do two iterations.

Answer 1

$A=\begin{pmatrix} 1 & 2&1 \\ 0 & 2&5\\ 1&0&-1 \end{pmatrix}$

$\begin{vmatrix} 1-\lambda & 2&1 \\ 0 & 2-\lambda&5\\ 1&0&-1-\lambda \end{vmatrix}=(1-\lambda)(2-\lambda)(-1-\lambda)+10-2+\lambda=$

$=(2-3\lambda+\lambda^2)(-1-\lambda)+\lambda+8=-\lambda^3+2\lambda^2+\lambda-2+\lambda+8=$

$=-\lambda^3+2\lambda^2+2\lambda+6=0$

$\lambda_1=3.207$

$x_1=Ax_0=\begin{pmatrix} 1 & 2&1 \\ 0 & 2&5\\ 1&0&-1 \end{pmatrix}\begin{pmatrix} 1 \\ 1\\ 1 \end{pmatrix}=\begin{pmatrix} 4 \\ 7 \\ 0 \end{pmatrix}$

$x_1=x_1/||x_1||=\begin{pmatrix} 0.4961 \\ 0.8682 \\ 0 \end{pmatrix}$

$x_2=Ax_1=\begin{pmatrix} 1 & 2&1 \\ 0 & 2&5\\ 1&0&-1 \end{pmatrix}\begin{pmatrix} 0.4961 \\ 0.8682 \\ 0 \end{pmatrix}=\begin{pmatrix} 2.2325 \\ 1.7364 \\ 0.4961 \end{pmatrix}$

$x_2=x_2/||x_2||=\begin{pmatrix} 0.7775 \\ 0.6047 \\ 0.1728 \end{pmatrix}$

$Ax_2=\begin{pmatrix} 1 & 2&1 \\ 0 & 2&5\\ 1&0&-1 \end{pmatrix}\begin{pmatrix} 0.7775 \\ 0.6047 \\ 0.1728 \end{pmatrix}=\begin{pmatrix} 2.1597 \\ 2.0734 \\ 0.6047 \end{pmatrix}$

Approximation to the dominant eigenvalue $\lambda_1=3.207$ :

$\lambda_1=\frac{(Ax_2)\cdot x_2}{x_2\cdot x_2}=\frac{3.0374}{1.0019}=3.032$

Find corresponding eigenvector:

$\begin{cases} (1-\lambda)x+2y+z=0\\ (2-\lambda)y+5z=0\\ x- (1+\lambda)z=0 \end{cases}$

$\begin{cases} -2.032x+2y+z=0\\ -1.032y+5z=0\\ x-4.032z=0 \end{cases}$

$10.16x-11.032y=0$

$2y-7.193z=0$

$x=1.086y$

$y=3.6z$

$v_1=\begin{pmatrix} 1 \\ 0.9 \\ 0.25 \end{pmatrix}$