Chapter 7 - Eigenvalues and Eigenvectors - 7.5 Orthogonal Diagonalization and Quadratic Forms - Problems - Page 474: 13

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