Chapter 7 - Eigenvalues and Eigenvectors - 7.5 Orthogonal Diagonalization and Quadratic Forms - Problems - Page 474: 9

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