Chapter 7 - Eigenvalues and Eigenvectors - 7.1 The Eigenvalue/Eigenvector Problem - Problems - Page 444: 31

$\lambda_1=16$ $\vec{V}=(0.5484, 0.4194, 1.1290, 1)r$ $\lambda_2=-2$ $\vec{V}=(-1,1,-1,1)s$ $\lambda_3=0$ $\vec{V}=(1, -2, 1, 0)q + (2, -3, 0,1)p$

1. eigenvalues: (A-$\lambda$I)$\vec{V}$=$\vec{0}$ $\begin{bmatrix} 1-\lambda &2 &3&4 \\ 4& 3-\lambda&2&1\\ 4&5 &6-\lambda&7\\ 7&6&5&4-\lambda \end{bmatrix}\begin{bmatrix} v_1\\ v_2\\ v_3\\ v_4 \end{bmatrix}=\begin{bmatrix} 0\\ 0\\ 0\\ 0 \end{bmatrix}$ $\begin{vmatrix} 1-\lambda &2 &3&4 \\ 4& 3-\lambda&2&1\\ 4&5 &6-\lambda&7\\ 7&6&5&4-\lambda \end{vmatrix}=0$ $\lambda_1=16$ $\lambda_2=-2$ $\lambda_3=0$ =============================== 2. eigenvectors $\lambda_1=16$ let $B=A-\lambda_1I$ $B= \begin{bmatrix} 1-\lambda_1 &2 &3&4 \\ 4& 3-\lambda_1 &2&1\\ 4&5 &6-\lambda_1 &7\\ 7&6&5&4-\lambda_1 \end{bmatrix}$ = $\begin{bmatrix} -15 &2 &3&4 \\ 4& -13&2&1\\ 4&5 &-10&7\\ 7&6&5&-12 \end{bmatrix}$ Then, $B\vec{V}$=$\vec{0}$ Use reduced row echelon form $[B|\vec{0}]$= \[ \left(\begin{array}{@{}cccc|c@{}} 1 & 0 & 0& -0.54840&0\\ 0 & 1 & 0 & -0.4194&0\\ 0 & 0 & 1 & -1.1290&0\\ 0 & 0 & 0 & 0&0\\ \end{array}\right) \] let $r$ is a free variable. $\vec{V}=(0.5484, 0.4194, 1.1290, 1)r$ ============================= $\lambda_2=-2$ let $B=A-\lambda_2I$ $B= \begin{bmatrix} 1-\lambda_2 &2 &3&4 \\ 4& 3-\lambda_2&2&1\\ 4&5 &6-\lambda_2&7\\ 7&6&5&4-\lambda_2 \end{bmatrix}$ = $\begin{bmatrix} 3 &2 &3&4 \\ 4&5&2&1\\ 4&5 &8&7\\ 7&6&5&6 \end{bmatrix}$ Then, $B\vec{V}$=$\vec{0}$ Use reduced row echelon form $[B|\vec{0}]$= \[ \left(\begin{array}{@{}cccc|c@{}} 1 & 0 & 0& 1&0\\ 0 & 1 & 0 & -1&0\\ 0 & 0 & 1 & 1&0\\ 0 & 0 & 0 & 0&0\\ \end{array}\right) \] let $s$ is a free variable. $\vec{V}=(-1,1,-1,1)s$ ============================= $\lambda_3=0$ let $B=A-\lambda_3I$ $B= \begin{bmatrix} 1-\lambda_3 &2 &3&4 \\ 4& 3-\lambda_3&2&1\\ 4&5 &6-\lambda_3&7\\ 7&6&5&4-\lambda_3 \end{bmatrix}$ = $\begin{bmatrix} 1&2 &3&4 \\ 4& 3&2&1\\ 4&5 &6&7\\ 7&6&5&4 \end{bmatrix}$ Then, $B\vec{V}$=$\vec{0}$ Use reduced row echelon form $[B|\vec{0}]$= \[ \left(\begin{array}{@{}cccc|c@{}} 1 & 0 & -1& -2&0\\ 0 & 1 &2 & 3&0\\ 0 & 0 & 0 & 0&0\\ 0 & 0 & 0 & 0&0\\ \end{array}\right) \] let $p,q$ is a free variable. $\vec{V}=(1, -2, 1, 0)q + (2, -3, 0,1)p$