eEigenVector

Example Find the eigen value and eigen vector of $A$

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$\displaystyle A = \left(\begin{array}{rrr} 0&1&1\\ 1&0&1\\ 1&1&0 \end{array}\right)$

Ansewr $\Phi_{A}(t) = \left \vert\begin{array}{rrr} -t & 1 & 1\\ 1 & -t & 1\\ 1 & 1 & -t \end{array}\right \vert = -(t+1)^{2}(t-2).$ Then the eigen vector of $A$ is $\lambda = 2,-1$ ．Now we find the eigen vector of $A$ ．

The eigen vector corresponding to $\lambda = 2$ satisfies $(A - 2I){\mathbf x} = {\bf0}$ . So, we solve this using the reduction of the matrix $A$ .

$\displaystyle A - 2I$	$\displaystyle =$	$\displaystyle \left(\begin{array}{rrr} -2&1 & 1\\ 1& -2 & 1\\ 1&1&-2 \end{arr... ...htarrow} \left(\begin{array}{rrr} 1&1&-2\\ 1&-2&1\\ -2&1&1 \end{array}\right)$
	$\displaystyle \stackrel{\begin{array}{cc} {}^{-R_{1} + R_{2}}\\ {}^{2R_{1} + R_{3}} \end{array}}{\longrightarrow}$	$\displaystyle \left(\begin{array}{rrr} 1&1&-2\\ 0&-3&3\\ 0 & 3&-3 \end{array}... ...rrow} \left(\begin{array}{rrr} 1&1&-2\\ 0&1&-1\\ 0 & 0&0 \end{array}\right) .$

Note that the degree of freedom is 1. Thus, we set $x_{3} = \alpha$ . Then

$\displaystyle {\mathbf x} = \left(\begin{array}{c} x_{1}\\ x_{2}\\ x_{3} \end... ...alpha \left(\begin{array}{c} 1\\ 1\\ 1 \end{array}\right) \ (\alpha \neq 0).$

Next we find the eigen vector corresponding to $\lambda = -1$ . To do so, we solve $(A - 3I){\mathbf x} = {\bf0}$ . Then

$\displaystyle A+ I = \left(\begin{array}{rrr} 1&1&1\\ 1&1&1\\ 1&1&1 \end{arra... ...htarrow} \left(\begin{array}{rrr} 1&1&1\\ 0&0&0\\ 0 & 0&0 \end{array}\right)$

Then the degree of freedom is 2. Thus , we can set $x_{2} = \beta \neq 0, x_{3} = \gamma \neq 0$ . Then

$\displaystyle {\mathbf x} = \left(\begin{array}{c} -x_{2} - x_{3}\\ x_{2}\\ x... ...array}\right) + \gamma \left(\begin{array}{r} -1\\ 0\\ 1 \end{array}\right).$

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