Chapter 7 - Eigenvalues and Eigenvectors - 7.2 General Results for Eigenvalues and Eigenvectors - Problems - Page 452: 16

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1. Find eigenvalues: (A-$\lambda$I)$\vec{V}$=$\vec{0}$ $\begin{bmatrix} -\lambda & -1 & -1 \\ -1 & -\lambda & -1 \\ -1 & -1 & -\lambda \\ \end{bmatrix}\begin{bmatrix} v_1\\ v_2 \\ v_3 \end{bmatrix}=\begin{bmatrix} 0\\ 0 \\ 0 \end{bmatrix}$ $\begin{bmatrix} -\lambda & -1 & -1 \\ -1 & -\lambda & -1 \\ -1 & -1 & -\lambda \\ \end{bmatrix}=0$ $\left (- \lambda \right ) (- \lambda)(-\lambda)=0$ $(\lambda-1)^2(\lambda+2)=0$ $\lambda_1= \lambda_2=1, \lambda_3=-2$ 2. Find eigenvectors: For $\lambda=1$ let $B=A-\lambda_1I$ $B=\begin{bmatrix} -\lambda & -1 & -1 \\ -1 & -\lambda & -1 \\ -1 & -1 & -\lambda \\ \end{bmatrix}=\begin{bmatrix} -1 & -1 & -1 \\ -1 & -1 & -1 \\ -1 & -1 & -1\\ \end{bmatrix}$ Then, $B\vec{V}$=$\vec{0}$ We obtain reduced row echelon form of B: $\begin{bmatrix} -1 & -1 & -1 \\ -1 & -1 & -1 \\ -1 & -1 & -1 \\ \end{bmatrix} \approx \begin{bmatrix} 1 & 1 & 1 | 0 \\ 0 & 0 & 0 | 0\\ 0 & 0 & 0 | 0\\ \end{bmatrix}$ Let $r$ be a free variable. $\rightarrow \vec{V}=r(1,1,1) \\ E_1=\{(1,1,1)\} \rightarrow dim(E_1)=1 $ For $\lambda=-2$ let $B=A-\lambda_1I$ $B=\begin{bmatrix} -\lambda & -1 & -1 \\ -1 & -\lambda & -1 \\ -1 & -1 & -\lambda \\ \end{bmatrix}=\begin{bmatrix} 2 & -1 & -1 \\ -1 & 2 & -1 \\ -1 & -1 & 2\\ \end{bmatrix}$ Then, $B\vec{V}$=$\vec{0}$ We obtain reduced row echelon form of B: $\begin{bmatrix} 2 & -1 & -1 | 0 \\ -1 & 2 & -1 | 0 \\ -1 & -1 & 2 | 0\\ \end{bmatrix} \approx \begin{bmatrix} 1 & 0 & -1 | 0 \\ 0 & 1 & -1 | 0\\ 0 & 0 & 0 | 0\\ \end{bmatrix}$ Let $r$ and $s$ be free variables. $\rightarrow \vec{V}=r(-1,1,0)+s(-1,0,1) \\ E_1=\{(-1,1,0),(-1,0,1)\} \rightarrow dim(E_2)=2 $ Hence, matrix $A$ is nondefective.