91Ó°ÊÓ

For a matrix A, all possible values of$\mathit{λ}$which satisfies the equation$\left|A-\mathrm{λ}I\right|\mathbf{=}\mathbf{0}$, where$\mathit{I}$is the identity matrix, are considered as the eigenvalues of a matrix.

Given matrix is $A=\left(\begin{array}{cc}5& -4\\ -4& 5\end{array}\right)$. The characteristic equation is $\left|A-\mathrm{λ}I\right|=0$.

Solve for $\mathrm{λ}$as follows:

$\begin{array}{rcl}\left|\left(\begin{array}{cc}5& -4\\ -4& 5\end{array}\right)-\mathrm{λ}\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)\right|& =& 0\\ \left|\begin{array}{cc}5-\mathrm{λ}& -4\\ -4& 5-\mathrm{λ}\end{array}\right|& =& 0\\ \left(5-\mathrm{λ}\right)\left(5-\mathrm{λ}\right)-\left(-4\right)×\left(-4\right)& =& 0\\ 25-10\mathrm{λ}+{\mathrm{λ}}^2-16& =& 0\end{array}$

$\begin{array}{rcl}{\mathrm{λ}}^2-10\mathrm{λ}+9& =& 0\\ \left(\mathrm{λ}-9\right)\left(\mathrm{λ}-1\right)& =& 0\end{array}$

The above equation implies that $\mathrm{λ}=9$or $\mathrm{λ}=1$.

Therefore, the eigenvalues are 1 and 9.

Now, find the eigenvectors for each eigenvalue by solving role="math" localid="1664286559670" $\left(A-\mathrm{λ}I\right)X=0$for X where $X=\left(\begin{array}{c}{x}_1\\ x_2\end{array}\right)$.

For $\mathrm{λ}=1$,

$\begin{array}{rcl}\left(A-I\right)X& =& 0\\ \left(\begin{array}{cc}5-1& -4\\ -4& 5-1\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)& =& \left(\begin{array}{c}0\\ 0\end{array}\right)\\ \left(\begin{array}{cc}4& -4\\ -4& 4\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)& =& \left(\begin{array}{c}0\\ 0\end{array}\right)\\ 4x-4y& =& 0\\ x& =& y\end{array}$

The eigenvectors for each eigenvalue by solving for X where $X=\left(\begin{array}{c}{x}_1\\ x_2\end{array}\right)$.

$X=x\left(\begin{array}{c}1\\ 1\end{array}\right)$

Therefore, the eigenvector corresponding to the eigenvalue 1 is $\left(\begin{array}{c}1\\ 1\end{array}\right)$.

$\begin{array}{rcl}For\mathrm{λ}& =& 9\\ \left(A-I\right)X& =& 0\\ \left(\begin{array}{cc}5-9& -4\\ -4& 5-9\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)& =& \left(\begin{array}{c}0\\ 0\end{array}\right)\\ \left(\begin{array}{cc}-4& -4\\ 4& 4\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)& =& \left(\begin{array}{c}0\\ 0\end{array}\right)\\ -x-y& =& 0\\ y& =& -x\\ X& =& \left(\begin{array}{c}x\\ -x\end{array}\right)\\ X& =& x\left(\begin{array}{c}1\\ -1\end{array}\right)\end{array}$

Therefore, the eigenvector corresponding to the eigenvalue 9 is $\left(\begin{array}{c}1\\ -1\end{array}\right)$.