Answer to Question #264229 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

1st Iteration;

$Dx_0=$ $\begin{pmatrix} 1&1&2 \\ 2&1& 1\\ 1&1&3 \end{pmatrix}\begin{pmatrix} 1 \\ 0\\ 0 \end{pmatrix}=\begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}$

$x_1=\frac{1}{2}\begin{pmatrix} 1 \\ 2\\ 1 \end{pmatrix}=\begin{pmatrix} 0.5\\ 1\\ 0.5 \end{pmatrix}$

2nd Iteration

$Dx_1=\begin{pmatrix} 1&1&2 \\ 2&1& 1\\ 1&1&3 \end{pmatrix}\begin{pmatrix} 0.5 \\ 1 \\ 0.5 \end{pmatrix}=\begin{pmatrix} 2.5 \\ 2.5 \\ 3 \end{pmatrix}$

$x_2=\frac{1}{3}\begin{pmatrix} 2.5 \\ 2.5 \\ 3 \end{pmatrix}=\begin{pmatrix} 0.83 \\ 0.83\\ 1 \end{pmatrix}$

3rd Iteration

$Dx_2=\begin{pmatrix} 1&1&2 \\ 2&1 & 1\\ 1&1&3 \end{pmatrix}\begin{pmatrix} 0.83 \\ 0.83 \\ 1 \end{pmatrix}=\begin{pmatrix} 3.67 \\ 3.5 \\ 4.67 \end{pmatrix}$

$x_3=\frac{1}{4.67}\begin{pmatrix} 3.67 \\ 3.5\\ 4.67 \end{pmatrix}=\begin{pmatrix} 0.79 \\ 0.75 \\ 1 \end{pmatrix}$

4th Iteration

$Dx_3=\begin{pmatrix} 1&1 & 2 \\ 2&1 & 1\\ 1&1&3 \end{pmatrix}\begin{pmatrix} 0.79 \\ 0.75\\ 1 \end{pmatrix}=\begin{pmatrix} 3.54 \\ 3.32 \\ 4.54 \end{pmatrix}$

$x_4=\frac{1}{4.54}\begin{pmatrix} 3.54 \\ 3.32 \\ 4.54 \end{pmatrix}=\begin{pmatrix} 0.78 \\ 0.73\\ 1 \end{pmatrix}$

5th Iteration

$Dx_4=\begin{pmatrix} 1&1 & 2 \\ 2&1 & 1\\ 1&1&3 \end{pmatrix}\begin{pmatrix} 0.78\\ 0.73 \\ 1 \end{pmatrix}=\begin{pmatrix} 3.51 \\ 3.29\\ 4.51 \end{pmatrix}$

$x_5=\frac{1}{4.51}\begin{pmatrix} 3.51 \\ 3.29 \\ 4.51 \end{pmatrix}=\begin{pmatrix} 0.78 \\ 0.73 \\ 1 \end{pmatrix}$

6th Iteration

$Dx_5=\begin{pmatrix} 1&1 & 2\\ 2&1& 1\\ 1&1&3 \end{pmatrix}\begin{pmatrix} 0.78 \\ 0.73 \\ 1 \end{pmatrix}=\begin{pmatrix} 3.51 \\ 3.29\\ 4.51 \end{pmatrix}$

$x_6=\frac{1}{4.51}\begin{pmatrix} 3.51 \\ 3.29 \\ 4.51 \end{pmatrix}=\begin{pmatrix} 0.78 \\ 0.73\\ 1 \end{pmatrix}$

$\therefore$ The dominant eigenvalue $\lambda=4.51$

and the dominant eigenvector is

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

D^-1 by Gaus - Jordan Elimination

Augmenting $D$ with a $3×3$ identity matrix

$\begin{bmatrix} 1&1&2| 1&0&0 \\ 2&1&1|0&1&0 \\ 1&1&3|0&0&1 \end{bmatrix}$

$\frac{1}{2}R_2-R_1\to\>R_2$

$\begin{bmatrix} 1&1 & 2| 1&0&0\\ 0&\frac{-1}{2}& \frac{-3}{2}\>\>\>|-1&\frac{1}{2}&0\\ -1&1&3|0&0&1 \end{bmatrix}$

$R_3-R_1\to\>R_3$

$\begin{bmatrix} 1&1&2\>\>\>| 2&0&0\\ 0&\frac{-1}{2} &\>\> \frac{-3}{2}\>\>\>\>\>|-1&\frac{1}{2}&0\\ 0&0&\>\>1\>\>\>\>\>\>\>\>|-1&0&1 \end{bmatrix}$

$-2R_2-3R_3\to\>R_2$

$\begin{bmatrix} 1&1&2 |1&0&0 \\ 0&1&0 |5&-1&-3\\ 0&0&\>1\>\>\>\>\>|-1&0&1 \end{bmatrix}$

$R_1-2R_3\to\>R_1$

$\begin{bmatrix} 1&1 & 0|3&0&-2\\ 0&1&0| 5&-1&-3\\ 0&0&\>\>-1|-1&0&1 \end{bmatrix}$

$R_1-R_2\to\>R_1$

$\begin{bmatrix} 1&0&\>\>\>\>0 | \quad-2&1&1\\ 0&1&0|\quad 5&-1&-3\\ 0&0&1|-1&0&1 \end{bmatrix}$

D^-1 is the right side of Augmented matrix

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

Solving for the largest eigenvalue of D^-1

1st Iteration

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

$x_1=\frac{1}{5}\begin{pmatrix} -2 \\ 5 \\ -1 \end{pmatrix}=\begin{pmatrix} -0.4 \\ 1\\ -0.2 \end{pmatrix}$

2nd Iteration

$D^{-1}x_1=\begin{pmatrix} -2&1 & 1\\ 5&-1 & -3\\ -1&0&1 \end{pmatrix}\begin{pmatrix} -0.4 \\ 1\\ -0.2 \end{pmatrix}=\begin{pmatrix} 1.6 \\ -2.4 \\ 0.2 \end{pmatrix}$

$x_2=\frac{1}{2.4}\begin{pmatrix} 1.6 \\ -2.4 \\ 0.2 \end{pmatrix}=\begin{pmatrix} 0.67 \\ -1 \\ 0.08 \end{pmatrix}$

3rd Iteration

$D^-x_2=\begin{pmatrix} -2&1&1 \\ 5&-1 & -3\\ -1&0&1 \end{pmatrix}\begin{pmatrix} 0.67 \\ -1 \\ 0.08 \end{pmatrix}=\begin{pmatrix} -2.25 \\ 4.08\\ -0.58 \end{pmatrix}$

$x_3=\frac{1}{4.08}\begin{pmatrix} -2.25 \\ 4.08 \\ -0.58 \end{pmatrix}=\begin{pmatrix} -0.55 \\ 1 \\ -0.14 \end{pmatrix}$

4th Iteration

$D^-x_3=\begin{pmatrix} -2&1 & 1 \\ 5&-1& -3\\ -1&0&1 \end{pmatrix}\begin{pmatrix} -0.55 \\ 1 \\ -0.14 \end{pmatrix}=\begin{pmatrix} 1.96 \\ -3.33 \\ 0.41 \end{pmatrix}$

$x_4=\frac{1}{3.33}\begin{pmatrix} 1.96 \\ -3.33\\ 0.41 \end{pmatrix}=\begin{pmatrix} 0.59 \\ -1 \\ 0.12 \end{pmatrix}$

5th Iteration

$D^-x_4=\begin{pmatrix} -2&1 & 1 \\ 5&-1& -3\\ -1&0&1 \end{pmatrix}\begin{pmatrix} 0.59 \\ -1\\ 0.12 \end{pmatrix}=\begin{pmatrix} -2.06 \\ 3.58 \\ -0.47 \end{pmatrix}$

$x_5=\frac{1}{3.58}\begin{pmatrix} -2.06 \\ 3.58 \\ -0.47 \end{pmatrix}=\begin{pmatrix} -0.57 \\ 1 \\ -0.13 \end{pmatrix}$

6th Iteration

$D^{-1}x_5=\begin{pmatrix} -2&1 & 1 \\ 5&-1& -3\\ -1&0&1 \end{pmatrix}\begin{pmatrix} -0.57 \\ 1 \\ -0.13 \end{pmatrix}=\begin{pmatrix} 2.02 \\ -3.48 \\ 0.44 \end{pmatrix}$

$x_6=\frac{1}{3.48}\begin{pmatrix} 2.02\\ -3.48 \\ 0.44 \end{pmatrix}=\begin{pmatrix} 0.58 \\ -1 \\ 0.13 \end{pmatrix}$

8th Iteration

$D^{-1}x_7=\begin{pmatrix} -2&1 & 1 \\ 5&-1 & -3\\ -1&0&1 \end{pmatrix}\begin{pmatrix} -0.58 \\ 1\\ -0.13 \end{pmatrix}=\begin{pmatrix} 2.03 \\ -3.5 \\ 0.45 \end{pmatrix}$

$x_8=\frac{1}{3.5}\begin{pmatrix} 2.03 \\ -3.5 \\ 0.45 \end{pmatrix}=\begin{pmatrix} 0.58 \\ -1 \\ 0.13 \end{pmatrix}$

9th Iteration

$D^{-1}x_8=\begin{pmatrix} -2&1& 1\\ 5&-1 & -3\\ -1&0&1 \end{pmatrix}\begin{pmatrix} 0.58 \\ -1 \\ 0.13 \end{pmatrix}=\begin{pmatrix} -2.03 \\ 3.51 \\ -0.45 \end{pmatrix}$

$x_9=\frac{1}{3.51}\begin{pmatrix} -2.03 \\ 3.51 \\ -0.45 \end{pmatrix}=\begin{pmatrix} 0.58 \\ 1 \\ -0.13 \end{pmatrix}$

10th Iteration

D$^-x_9=\begin{pmatrix} -2&1 & 1\\ 5&-1 & -3\\ -1&0&1 \end{pmatrix}\begin{pmatrix} 0.58 \\ -1 \\ 0.13 \end{pmatrix}=\begin{pmatrix} 2.03 \\ -3.51\\ 0.45 \end{pmatrix}$

$x_{10}=\frac{1}{3.51}\begin{pmatrix} 2.03 \\ -3.51 \\ 0.45 \end{pmatrix}=\begin{pmatrix} 0.58 \\ -1\\ 0.13 \end{pmatrix}$

The dominant eigenvalue of D^- $3.51$

The magnitude of the smallest eigenvalues of $D=\frac{1}{3.51}=0.2849$

and the corresponding eigenvector of

$D=\begin{pmatrix} 0.58\\ -1 \\ 0.13 \end{pmatrix}$