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

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

Given matrix is $A=\left(\begin{array}{ccc}3& 0& -2\\ 0& 4& 0\\ -2& 0& 3\end{array}\right)$. The characteristic equation is $\left|A-\lambda I\right|=0$ .

Solve for $\lambda$as follows:

$\begin{array}{rcl}\left|\left(\begin{array}{ccc}3& 0& -2\\ 0& 4& 0\\ -2& 0& 3\end{array}\right)-\lambda \left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)\right|& =& 0\\ \left|\begin{array}{ccc}3-\lambda & 0& -2\\ 0& 4-\lambda & 0\\ -2& 0& 3-\lambda \end{array}\right|& =& 0\\ \left(3-\lambda \right)\left(4-\lambda \right)\left(3-\lambda \right)-\left(-2\right)\left[\left(-2\right)\times \left(4-\lambda \right)\right]& =& 0\\ \left(4-\lambda \right)\left(9-6\lambda +{\lambda }^2-4\right)& =& 0\end{array}$

Solve further,

role="math" localid="1664346025425" $$\begin{gathered} \left(4-\lambda \right)\left({\lambda }^2-6\lambda +5\right)=0 \\ \left(4-\lambda \right)\left(\lambda -5\right)\left(\lambda -1\right)=0 \\ \lambda =1,5,4 \end{gathered}$$

Therefore, the eigenvalues are 1, 4, and 5.

Now, find the eigenvectors for each eigenvalue by solving $\left(A-\lambda I\right)X=0$ for Xwhere $X=\left(\begin{array}{c}x\\ y\\ z\end{array}\right)$.

For $\lambda =1$:

$\begin{array}{rcl}\left(A-I\right)X& =& 0\\ \left|\begin{array}{ccc}3-1& 0& -2\\ 0& 4-1& 0\\ -2& 0& 3-1\end{array}\right|\left(\begin{array}{c}x\\ y\\ z\end{array}\right)& =& 0\\ \left(\begin{array}{ccc}2& 0& -2\\ 0& 3& 0\\ -2& 0& 2\end{array}\right)\left(\begin{array}{c}x\\ y\\ z\end{array}\right)& =& 0\\ 3y& =& 0,y-0,2x-2z=0\\ x& =& z\end{array}$

Now, we have $X=\left(\begin{array}{c}x\\ y\\ z\end{array}\right)$which becomes:

$$\begin{gathered} X=\left(\begin{array}{c}x\\ 0\\ x\end{array}\right) \\ X=x\left(\begin{array}{c}1\\ 0\\ 1\end{array}\right) \end{gathered}$$

Therefore, the eigenvector corresponding to the eigenvalue 1 is $\left(\begin{array}{c}1\\ 0\\ 1\end{array}\right)$.

For $\lambda =5$,

$\begin{array}{rcl}\left|A-I\right|X& =& 0\\ \left|\begin{array}{ccc}3-5& 0& -2\\ 0& 4-5& 0\\ -2& 0& -2\end{array}\right|\left(\begin{array}{c}x\\ y\\ z\end{array}\right)& =& 0\\ \left(\begin{array}{ccc}-2& 0& -2\\ 0& -1& 0\\ -2& 0& -2\end{array}\right)\left(\begin{array}{c}x\\ y\\ z\end{array}\right)& =& 0\\ -y& =& 0,y=0,-2x-2z=0\\ x& =& -z\end{array}$
Now, we have $X=\left(\begin{array}{c}x\\ y\\ z\end{array}\right)$which becomes:

$$\begin{gathered} X=\left(\begin{array}{c}x\\ 0\\ -x\end{array}\right) \\ X=x\left(\begin{array}{c}1\\ 0\\ -1\end{array}\right) \end{gathered}$$

For $\lambda =4$

$\begin{array}{rcl}\left|A-I\right|X& =& 0\\ \left|\begin{array}{ccc}3-4& 0& -2\\ 0& 4-4& 0\\ -2& 0& 3-4\end{array}\right|\left(\begin{array}{c}x\\ y\\ z\end{array}\right)& =& 0\\ \left|\begin{array}{ccc}-1& 0& -2\\ 0& 0& 0\\ -2& 0& -1\end{array}\right|\left(\begin{array}{c}x\\ y\\ z\end{array}\right)& =& 0\\ -x-2z& =& 0\\ x& =& -2z\end{array}$

$$\begin{gathered} X=\left(\begin{array}{c}-2z\\ y\\ z\end{array}\right) \\ X=z\left(\begin{array}{c}-2\\ 0\\ 1\end{array}\right)+y\left(\begin{array}{c}0\\ 1\\ 0\end{array}\right) \end{gathered}$$