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 Iis the identity matrix, are considered as the eigenvalues of a matrix.

Given matrix is $A=\left(\begin{array}{ccc}3& 0& 1\\ 0& 3& 1\\ 1& 1& 2\end{array}\right)$. The characteristic equation isrole="math" localid="1664351117294" $\left|A-\mathrm{λ}I\right|=0$.

Solve for $\mathrm{λ}$as follows:

$\begin{array}{rcl}\left|\begin{array}{c}\left(\begin{array}{ccc}3& 0& 1\\ 0& 3& 1\\ 1& 1& 2\end{array}\right)-\mathrm{λ}\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)\end{array}\right|& =& 0\\ \left|\begin{array}{ccc}3-\mathrm{λ}& 0& 1\\ 0& 3-\mathrm{λ}& 1\\ 1& 1& 2-\mathrm{λ}\end{array}\right|& =& 0\\ \left(3-\mathrm{λ}\right)\left[\left(3-\mathrm{λ}\right)\left(2-\mathrm{λ}\right)-1\right]+1\left[-\left(3-\mathrm{λ}\right)\right]& =& 0\\ \left(3-\mathrm{λ}\right)\left(6-5\mathrm{λ}+{\mathrm{λ}}^2-1-1\right)& =& 0\end{array}$

Solve further,

$$\begin{gathered} \left(3-\mathrm{λ}\right)\left({\mathrm{λ}}^2-5\mathrm{λ}+4\right)=0 \\ \left(3-\mathrm{λ}\right)\left(\mathrm{λ}-4\right)\left(\mathrm{λ}-1\right)=0 \end{gathered}$$

The above equation implies that , role="math" localid="1664351145637" $\mathrm{λ}=1,\mathrm{λ}=3,\mathrm{λ}=4$.

Therefore, the eigenvalues are 1, 3, and 4.

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

For $\mathrm{λ}=1$,

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

The above equation impliesrole="math" localid="1664351166717" $2x+z=0$that which is equivalent to $z=-2x$and $2y+z=0$which is equivalent to $z=-2y$, and above two equation implies that role="math" localid="1664350378471" $x=y$. Now, we have $X=\left(\begin{array}{c}x\\ y\\ z\end{array}\right)$which becomes:

$$\begin{gathered} X=\left(\begin{array}{c}x\\ x\\ -2x\end{array}\right) \\ X=x\left(\begin{array}{c}1\\ 1\\ -2\end{array}\right) \end{gathered}$$

Therefore, the eigenvector corresponding to the eigenvalue 1 is $\left(\begin{array}{c}1\\ 1\\ -2\end{array}\right)$.

For $\mathrm{λ}=3$,

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

The above equation implies that role="math" localid="1664351154758" $z=0$, and role="math" localid="1664351176940" $x+y-z=0$which is equivalent to role="math" localid="1664351190094" $x+y=0⇒y=-x$. Now, we have $X=\left(\begin{array}{c}x\\ y\\ z\end{array}\right)$which becomes:

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

Therefore, the eigenvector corresponding to the eigenvalue 3 is $\left(\begin{array}{c}1\\ -1\\ 0\end{array}\right)$.

For $\mathrm{λ}=4$,

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

The above equation implies that $-x+z=0$which is equivalent to $z=x$ and $-y+z=0$which is equivalent to $y=z$and the above two equations implies $x=y$ . Now, we have $X=\left(\begin{array}{c}x\\ y\\ z\end{array}\right)$which becomes:

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

Therefore, the eigenvector corresponding to the eigenvalue 4 is $\left(\begin{array}{c}1\\ 1\\ 1\end{array}\right)$.