Chapter 7 - Eigenvalues and Eigenvectors - 7.5 Orthogonal Diagonalization and Quadratic Forms - Problems - Page 474: 6

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1. Find eigenvalues: (A-$\lambda$I)$\vec{V}$=$\vec{0}$ $\begin{bmatrix} 1-\lambda & 2 & 1\\ 2 & 4-\lambda & 2 \\ 1 & 2 & 1-\lambda \end{bmatrix}\begin{bmatrix} v_1\\ v_2 \end{bmatrix}=\begin{bmatrix} 0\\ 0 \end{bmatrix}$ $\begin{bmatrix} 1-\lambda & 2 & 1\\ 2 & 4-\lambda & 2 \\ 1 & 2 & 1-\lambda \end{bmatrix}=0$ $\lambda^3-6\lambda^2=0$ $\lambda_1=0, \lambda_2=0,\lambda=6$ 2. Find eigenvectors: For $\lambda=0$ let $B=A-\lambda_1I$ $B=\begin{bmatrix} 1-\lambda & 2 & 1\\ 2 & 4-\lambda & 2 \\ 1 & 2 & 1-\lambda \end{bmatrix}=\begin{bmatrix} 1 & 2 & 1\\ 2 & 4 & 2\\ 1 & 2 & 1 \end{bmatrix}\begin{bmatrix} v_1\\ v_2 \\ v_3 \end{bmatrix}=\begin{bmatrix} 0\\ 0 \\ 0 \end{bmatrix}$ Let $r,s$ be free variables. $\vec{V}=r(-1,0,1)+s(-2,1,0) \\ E_1=\{(-\frac{1}{\sqrt 2},0,\frac{1}{\sqrt 2});(-\frac{1}{\sqrt 3},\frac{1}{\sqrt 3},-\frac{1}{\sqrt 3})\} \rightarrow dim(E_1)=2$ The eigenvectors span $\{(-\frac{1}{\sqrt 2},0,\frac{1}{\sqrt 2});(-\frac{1}{\sqrt 3},\frac{1}{\sqrt 3},-\frac{1}{\sqrt 3})\} $ in $R$ For $\lambda=6$ let $B=A-\lambda_1I$ $B=\begin{bmatrix} 1-\lambda & 2 & 1\\ 2 & 4-\lambda & 2\\ 1 & 2 & 1-\lambda \end{bmatrix}=\begin{bmatrix} -5 & 2 & 1 \\ 2 & -2 & 2\\ 1 & 2 & -5 \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. $\vec{V}=r(1,2,1) \\ E_2=\{(\frac{1}{\sqrt 6},\frac{2}{\sqrt 6},\frac{1}{\sqrt 6})\} \rightarrow dim(E_2)=1$ The eigenvectors span $\{(\frac{1}{\sqrt 6},\frac{2}{\sqrt 6},\frac{1}{\sqrt 6})\} $ in $R$ Hence, $S=\frac{1}{\sqrt 2}\begin{bmatrix} -\frac{1}{\sqrt 2}& -\frac{1}{\sqrt 3} & \frac{1}{\sqrt 6}\\ 0 & \frac{1}{\sqrt 3} & \frac{2}{\sqrt 6} \\ \frac{1}{\sqrt 2} & -\frac{1}{\sqrt 3} & \frac{1}{\sqrt 6} \end{bmatrix}$ then $S^TAS =\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 6 \end{bmatrix} $