Answer to Question #256126 in Linear Algebra for kofi

Question

Consider the linear eigenproblem Ax=λx for the matrix

D=[1 1 2

2 1 1

1 1 3 ]

1. Solve for the largest (in magnitude) eigenvalue of the matrix and the corresponding eigenvector by the power method with 𝑥^(0)T=[1 0 0]

2. Solve for the smallest eigenvalue of the matrix and the corresponding eigenvector by the inverse power method using the matrix inverse. Use Gauss-Jordan elimination to find the matrix inverse.

Answer 1

1.

$x_1=Ax_0=\begin{pmatrix} 1 & 1&2 \\ 2 & 1&1\\ 1&1&3 \end{pmatrix}\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}=\begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}$

$x_2=Ax_1=\begin{pmatrix} 1 & 1&2 \\ 2 & 1&1\\ 1&1&3 \end{pmatrix}\begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}=\begin{pmatrix} 5 \\ 5 \\ 6 \end{pmatrix}$

$x_3=Ax_2=\begin{pmatrix} 1 & 1&2 \\ 2 & 1&1\\ 1&1&3 \end{pmatrix}\begin{pmatrix} 5 \\ 5 \\ 6 \end{pmatrix}=\begin{pmatrix} 22 \\ 21 \\ 28 \end{pmatrix}$

$x_4=Ax_3=\begin{pmatrix} 1 & 1&2 \\ 2 & 1&1\\ 1&1&3 \end{pmatrix}\begin{pmatrix} 22 \\ 21 \\ 28 \end{pmatrix}=\begin{pmatrix} 99 \\ 93\\ 127 \end{pmatrix}$

$x_5=Ax_4=\begin{pmatrix} 1 & 1&2 \\ 2 & 1&1\\ 1&1&3 \end{pmatrix}\begin{pmatrix} 99 \\ 93 \\ 127 \end{pmatrix}=\begin{pmatrix} 446 \\ 418\\ 573 \end{pmatrix}=573\begin{pmatrix} 0.78 \\ 0.73\\ 1 \end{pmatrix}$

dominant eigenvector:

$x=\begin{pmatrix} 0.78 \\ 0.73\\ 1 \end{pmatrix}$

$Ax=\begin{pmatrix} 1 & 1&2 \\ 2 & 1&1\\ 1&1&3 \end{pmatrix}\begin{pmatrix} 0.78 \\ 0.73\\ 1 \end{pmatrix}=\begin{pmatrix} 2.51 \\ 4.75\\ 10.26 \end{pmatrix}$

$Ax\cdot x=15.69$

$x\cdot x=2.14$

dominant eigenvalue:

$\lambda=\frac{Ax\cdot x}{x\cdot x}=\frac{15.69}{2.14}=7.33$

2.

$\begin{pmatrix} 1 & 1&2 &|1&0&0\\ 2 & 1&1&|0&1&0\\ 1&1&3&|0&0&1 \end{pmatrix}\to\begin{pmatrix} 1 & 0&0 &|-2&1&1\\ 0 & 1&0&|5&-1&-3\\ 0&0&1&|-1&0&1 \end{pmatrix}$

$A^{-1}=\begin{pmatrix} -2& 1&1 \\ 5 & -1&-3\\ 1&0&1 \end{pmatrix}$

$x_1=A^{-1}x_0=\begin{pmatrix} -2& 1&1 \\ 5 & -1&-3\\ 1&0&1 \end{pmatrix}\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}=\begin{pmatrix} -2 \\ 5 \\ 1 \end{pmatrix}$

$x_2=A^{-1}x_1=\begin{pmatrix} -2& 1&1 \\ 5 & -1&-3\\ 1&0&1 \end{pmatrix}\begin{pmatrix} -2 \\ 5 \\ 1 \end{pmatrix}=\begin{pmatrix} 2 \\ 2 \\ -1 \end{pmatrix}$

$x_3=A^{-1}x_2=\begin{pmatrix} -2& 1&1 \\ 5 & -1&-3\\ 1&0&1 \end{pmatrix}\begin{pmatrix} 2 \\ 2 \\ -1 \end{pmatrix}=\begin{pmatrix} -3 \\ 11 \\ 1 \end{pmatrix}$

$x_4=A^{-1}x_3=\begin{pmatrix} -2& 1&1 \\ 5 & -1&-3\\ 1&0&1 \end{pmatrix}\begin{pmatrix} -3 \\ 11 \\ 1 \end{pmatrix}=\begin{pmatrix} 18 \\ -29 \\ -2 \end{pmatrix}$

$x_5=A^{-1}x_4=\begin{pmatrix} -2& 1&1 \\ 5 & -1&-3\\ 1&0&1 \end{pmatrix}\begin{pmatrix} 18 \\ -29 \\ -2 \end{pmatrix}=\begin{pmatrix} -67 \\ 125 \\ 16 \end{pmatrix}=16\begin{pmatrix} -4.19 \\ 7.81 \\ 1 \end{pmatrix}$

the corresponding eigenvector the smallest eigenvalue:

$x=\begin{pmatrix} -4.19 \\ 7.81 \\ 1 \end{pmatrix}$

$A^{-1}x=\begin{pmatrix} -2& 1&1 \\ 5 & -1&-3\\ 1&0&1 \end{pmatrix}\begin{pmatrix} -4.19 \\ 7.81 \\ 1 \end{pmatrix}=\begin{pmatrix} 17.19 \\ -31.76 \\ -3.19 \end{pmatrix}$

the smallest eigenvalue:

$\lambda=\frac{x\cdot x}{A^{-1}x\cdot x}=\frac{26.37}{-323.26}=-0.08$