Chapter 7 - Eigenvalues and Eigenvectors - 7.2 General Results for Eigenvalues and Eigenvectors - Problems - Page 452: 14

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1. Find eigenvalues: (A-$\lambda$I)$\vec{V}$=$\vec{0}$ $\begin{bmatrix} 1-\lambda & -1 & 2 \\ 1 & -1-\lambda & 2 \\ 1 & -1 & 2-\lambda \\ \end{bmatrix}\begin{bmatrix} v_1\\ v_2 \\ v_3 \end{bmatrix}=\begin{bmatrix} 0\\ 0 \\ 0 \end{bmatrix}$ $\begin{bmatrix} 1-\lambda & -1 & 2 \\ 1 & -1-\lambda & 2 \\ 1 & -1 & 2-\lambda \\ \end{bmatrix}=0$ $\left (1- \lambda \right ) (-1- \lambda)(2-\lambda)=0$ $(\lambda-2).\lambda^2=0$ $\lambda_1= \lambda_2=2,\lambda_3=0$ 2. Find eigenvectors: For $\lambda=2$ let $B=A-\lambda_1I$ $B=\begin{bmatrix} 1-\lambda & -1 & 2 \\ 1 & -1-\lambda & 2 \\ 1 & -1 & 2-\lambda \\ \end{bmatrix}=\begin{bmatrix} -1 & -1 & 2 \\ 1 & -3 & 2 \\ 1 & -1 & 0 \\ \end{bmatrix}$ Then, $B\vec{V}$=$\vec{0}$ We obtain reduced row echelon form of B: $\begin{bmatrix} -1 & -1 & 2 |0 \\ 1 & -3 & 2 |0 \\ 1 & -1 & 0 | 0\\ \end{bmatrix}\approx\approx \begin{bmatrix} 1 & 0 & -1 | 0 \\ 0 & 1 & -1 | 0\\ 0 & 0 & 0 | 0\\ \end{bmatrix}$ Let $r$ be a free variable. $\rightarrow \vec{V}=r(1,1,1) \\ E_1=\{(1,1,1)\} \rightarrow dim(E_1)=1 $ For $\lambda=0$ let $B=A-\lambda_1I$ $B=\begin{bmatrix} 1-\lambda & -1 & 2 \\ 1 & -1-\lambda & 2 \\ 1 & -1 & 2-\lambda \\ \end{bmatrix}=\begin{bmatrix} 1 & -1 & 2 \\ 1 & -1 & 2 \\ 1 & -1 & 2 \\ \end{bmatrix}$ Then, $B\vec{V}$=$\vec{0}$ We obtain reduced row echelon form of B: $\begin{bmatrix} -1 & -1 & 2 |0 \\ 1 & -1 & 2 |0 \\ 1 & -1 & 2 | 0\\ \end{bmatrix}\approx\approx \begin{bmatrix} 1 & -1 & 2 | 0 \\ 0 & 0 & 0 | 0\\ 0 & 0 & 0 | 0\\ \end{bmatrix}$ Let $r,s$ be free variables. $\rightarrow \vec{V}=r(1,1,0) +s(-2,0,1)\\ E_2=\{(1,1,0),(-2,0,1)\} \rightarrow dim(E_2)=2$ Hence, matrix $A$ is non-defective.