Chapter 7 - Eigenvalues and Eigenvectors - 7.1 The Eigenvalue/Eigenvector Problem - Problems - Page 444: 23

$\lambda_1=3$ $\vec{V}=(-1,1,0)r$ $\lambda_2=1$ $\vec{V}=(-0.6,1,1)r$ $\lambda_3=-1$ $\vec{V}=(-0.2,0.75,1)w$

1. eigenvalues: (A-$\lambda$I)$\vec{V}$=$\vec{0}$ $\begin{bmatrix} 6-\lambda &3 &-4 \\ -5& -2-\lambda&2\\ 0&0 &-1-\lambda \end{bmatrix}\begin{bmatrix} v_1\\ v_2\\ v_3 \end{bmatrix}=\begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix}$ $\begin{vmatrix} 6-\lambda &3 &-4 \\ -5& -2-\lambda&2\\ 0&0 &-1-\lambda \end{vmatrix}=0$ $\lambda_1=3$ $\lambda_2=1$ $\lambda_3=-1$ =============================== 2. eigenvectors $\lambda_1=3$ let $B=A-\lambda_1I$ $B= \begin{bmatrix} 6-\lambda_1 &3 &-4 \\ -5& -2-\lambda_1&2\\ 0&0 &-1-\lambda_1 \end{bmatrix}$ = $\begin{bmatrix} 3&3&-4 \\ -5 &-5 &2\\ 0&0 &-4 \end{bmatrix}$ Then, $B\vec{V}$=$\vec{0}$ Use reduced row echelon form $[B|\vec{0}]$= \[ \left(\begin{array}{@{}ccc|c@{}} 1 & 1 & 0 & 0 \\ 0 & 0& 1 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right) \] Let $r$ is free variable. Then $\vec{V}=(-1,1,0)r$ ============================= $\lambda_2=1$ let $B=A-\lambda_2I$ $B= \begin{bmatrix} 6-\lambda_2 &3 &-4 \\ -5& -2-\lambda_2&2\\ 0&0 &-1-\lambda_2 \end{bmatrix}$ = $\begin{bmatrix} 5&3&-4 \\ -5 &-3 &2\\ 0&0 &-2 \end{bmatrix}$ Then, $B\vec{V}$=$\vec{0}$ Use reduced row echelon form $[B|\vec{0}]$= \[ \left(\begin{array}{@{}ccc|c@{}} 1 & 0.6 & 0 & 0 \\ 0 & 0 &1& 0 \\ 0 & 0 & 0 & 0 \end{array}\right) \] let $s$ is a free variable. $\vec{V}=(-0.6,1,1)r$ ============================= $\lambda_3=-1$ let $B=A-\lambda_3I$ $B= \begin{bmatrix} 6-\lambda_3 &3 &-4 \\ -5& -2-\lambda_3&2\\ 0&0 &-1-\lambda_3 \end{bmatrix}$ = $\begin{bmatrix} 7&3&-4 \\ -5 &-1 &2\\ 0&0 &0 \end{bmatrix}$ Then, $B\vec{V}$=$\vec{0}$ Use reduced row echelon form $[B|\vec{0}]$= \[ \left(\begin{array}{@{}ccc|c@{}} 1 & 0.6 & -0.25 & 0 \\ 0 & 1 &-0.75& 0 \\ 0 & 0 & 0 & 0 \end{array}\right) \] let $w$ is a free variable. $\vec{V}=(-0.2,0.75,1)w$