Chapter 7 - Eigenvalues and Eigenvectors - 7.1 The Eigenvalue/Eigenvector Problem - Problems - Page 444: 32

$\lambda_1=i$ $\vec{V}=(i, -1,0, 1)r+(i,-1,1 ,0)s$ $\lambda_2=-i$ $\vec{V}=(-i, -1,0, 1)r+(-i,-1,1 ,0)s$

1. eigenvalues: (A-$\lambda$I)$\vec{V}$=$\vec{0}$ $\begin{bmatrix} -\lambda &1 &0&0 \\ -1& -\lambda&0&0\\ 0&0 &-\lambda&-1\\ 0&0 &1&-\lambda \end{bmatrix}\begin{bmatrix} v_1\\ v_2\\ v_3\\ v_4 \end{bmatrix}=\begin{bmatrix} 0\\ 0\\ 0\\ 0 \end{bmatrix}$ $\begin{vmatrix} -\lambda &1 &0&0 \\ -1& -\lambda&0&0\\ 0&0 &-\lambda&-1\\ 0&0 &1&-\lambda \end{vmatrix}=0$ $\lambda_1=i$ $\lambda_2=-i$ =============================== 2. eigenvectors $\lambda_1=i$ let $B=A-\lambda_1I$ $B= \begin{bmatrix} -\lambda &1 &0&0 \\ -1& -\lambda&0&0\\ 0&0 &-\lambda&-1\\ 0&0 &1&-\lambda \end{bmatrix}$ = $\begin{bmatrix} -i &1 &0&0 \\ -1& -i&0&0\\ 0&0 &-i&-1\\ 0&0 &1&-i \end{bmatrix}$ Then, $B\vec{V}$=$\vec{0}$ Use reduced row echelon form $[B|\vec{0}]$= \[ \left(\begin{array}{@{}cccc|c@{}} 1 & i & 0& 0 &0\\ 0 & 0 & 1 & -i&0 \\ 0 & 0 & 0 & 0&0\\ 0 & 0 & 0 & 0&0\\ \end{array}\right) \] let $r,s$ is a free variable. $\vec{V}=(i, -1,0, 1)r+(i,-1,1 ,0)s$ ============================= $\lambda_1=-i$ let $B=A-\lambda_2I$ $B= \begin{bmatrix} -\lambda_2 &1 &0&0 \\ -1& -\lambda_2&0&0\\ 0&0 &-\lambda_2&-1\\ 0&0 &1&-\lambda_2 \end{bmatrix}$ = $\begin{bmatrix} i &1 &0&0 \\ -1& i&0&0\\ 0&0 &i&-1\\ 0&0 &1&i \end{bmatrix}$ Then, $B\vec{V}$=$\vec{0}$ Use reduced row echelon form $[B|\vec{0}]$= \[ \left(\begin{array}{@{}cccc|c@{}} 1 & - i & 0& 0 &0\\ 0 & 0 & 1 &-i&0 \\ 0 & 0 & 0 & 0&0\\ 0 & 0 & 0 & 0&0\\ \end{array}\right) \] let $r,s$ is a free variable. $\vec{V}=(-i, -1,0, 1)r+(-i,-1,1 ,0)s$