Chapter 7 - Eigenvalues and Eigenvectors - 7.5 Orthogonal Diagonalization and Quadratic Forms - Problems - Page 474: 5

See below

1. Find eigenvalues: (A-$\lambda$I)$\vec{V}$=$\vec{0}$ $\begin{bmatrix} -\lambda & 0 & 3\\ 0 & -2-\lambda & 0 \\ 3 & 0 & -\lambda \end{bmatrix}\begin{bmatrix} v_1\\ v_2 \end{bmatrix}=\begin{bmatrix} 0\\ 0 \end{bmatrix}$ $\begin{bmatrix} -\lambda & 0 & 3\\ 0 & -2-\lambda & 0 \\ 3 & 0 & -\lambda \end{bmatrix}=0$ $(\lambda+2)(\lambda-3)(\lambda +3)=0$ $\lambda_1=-3, \lambda_2=-2,\lambda=3$ 2. Find eigenvectors: For $\lambda=-3$ let $B=A-\lambda_1I$ $B=\begin{bmatrix} -\lambda & 0 & 3\\ 0 & -2-\lambda & 0 \\ 3 & 0 & -\lambda \end{bmatrix}=\begin{bmatrix} 3 & 0 & 3\\ 0 & 1 & 0\\ 3 & 0 & 3 \end{bmatrix}\begin{bmatrix} v_1\\ v_2 \\ v_3 \end{bmatrix}=\begin{bmatrix} 0\\ 0 \\ 0 \end{bmatrix}$ Let $r$ be a free variable. $\vec{V}=r(-1,0,1) \\ E_1=\{(-1,0,1)\} \rightarrow dim(E_1)=1$ The eigenvectors span $\{-1,0,1)\}$ in $R$ For $\lambda=-2$ let $B=A-\lambda_1I$ $B=\begin{bmatrix} -\lambda & 0 & 3\\ 0 & -2-\lambda & 0 \\ 3 & 0 & -\lambda \end{bmatrix}=\begin{bmatrix} 2 & 0 & 3 \\ 0 & 0 & 0\\ 3 & 0 & 2 \end{bmatrix}\begin{bmatrix} v_1\\ v_2 \\ v_3 \end{bmatrix}=\begin{bmatrix} 0\\ 0 \\ \end{bmatrix} $ Let $s$ be a free variable. $\vec{V}=r(0,1,0) \\ E_2=\{(0,1,0)\} \rightarrow dim(E_2)=1$ The eigenvectors span $\{(0,1,0)\}$ in $R$ For $\lambda=3$ let $B=A-\lambda_1I$ $B=\begin{bmatrix} -\lambda & 0 & 3\\ 0 & -2-\lambda & 0 \\ 3 & 0 & -\lambda \end{bmatrix}=\begin{bmatrix} -3 & 0 & 3 \\ 0 & -5 & 0\\ 3 & 0 & -3 \end{bmatrix}\begin{bmatrix} v_1\\ v_2 \\ v_3 \end{bmatrix}=\begin{bmatrix} 0\\ 0 \\ \end{bmatrix} $ Let $t$ be a free variable. $\vec{V}=t(1,0,1) \\ E_3=\{(1,0,1)\} \rightarrow dim(E_3)=1$ The eigenvectors span $\{(1,0,1)\}$ in $R$ Hence, $S=\frac{1}{\sqrt 2}\begin{bmatrix} -1 & 0 & 1\\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix} =\begin{bmatrix} -\frac{1}{\sqrt 2} & 0 & \frac{1}{\sqrt 2} \\ 0 & 1 & 0 \\ \frac{1}{\sqrt 2} & 0 & \frac{1}{\sqrt 2} \end{bmatrix} $