Question

Find eigenvalues and eigenvectors of the matrices in the following problems.

$\left(\begin{array}{cc}5& -4\\ -4& 5\end{array}\right)$

Short answer

The eigenvector for the eigenvalue 1 is$\left(\begin{array}{c}1\\ 1\end{array}\right)$ and the eigenvector for the eigenvalue 9 is$\left(\begin{array}{c}1\\ -1\end{array}\right)$ .

Step-by-step solution

Definition of eigenvalue

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

Find the eigenvalues and the eigenvectors

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

Solve for $\lambda$as follows:

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

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

The above equation implies that $\lambda =9$or $\lambda =1$.

Therefore, the eigenvalues are 1 and 9.

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

For $\lambda =1$,

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

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 $\left(\begin{array}{c}1\\ 1\end{array}\right)$.

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

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