Question

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

$\left[\begin{array}{cc}1& 0\\ 3& -2\end{array}\right]$

Short answer

The eigenvalues of the matrix are, $\lambda =1, -2$

The eigenvector corresponding to $\lambda =1$is$\left[\begin{array}{c}1\\ 1\end{array}\right]$

The eigenvector corresponding to$\lambda =-2$ is .role="math" localid="1664274394944" $\left[\begin{array}{c}0\\ 1\end{array}\right]$

Step-by-step solution

To find the eigenvalues of matrix

Let $\left[\begin{array}{cc}1& 0\\ 3& -2\end{array}\right]$be the matrix.

The characteristic equation is given by,

$\begin{array}{rcl}\left|M-\lambda I\right|& =& 0\\ \left|\begin{array}{cc}1-\lambda & 0\\ 3& -2-\lambda \end{array}\right|& =& 0\\ \left(1-\lambda \right)\left(-2-\lambda \right)& =& 0\\ \lambda & =& 1,-2\end{array}$

Thus, the eigenvalues of the matrix are,$\lambda =1,-2$

To find the eigenvectors of matrix

Now, to find eigenvectors.

Consider $\left(M-\lambda I\right)X=0$

For $\lambda =1$

$\begin{array}{rcl}\left(M-I\right)X& =& 0\\ \left[\begin{array}{cc}0& 0\\ 3& -3\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]& =& 0\left[\frac{R_2}{3}\right]\\ \left[\begin{array}{cc}0& 0\\ & -1\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]& =& 0\\ x-y& =& 0\\ x& =& y\\ & & \end{array}$

Therefore,

$$\begin{gathered} X=\left[\begin{array}{c}x\\ y\end{array}\right] \\ X=x\left[\begin{array}{c}1\\ 1\end{array}\right] \end{gathered}$$

Hence, the eigenvector corresponding to $\lambda =1$is$\left[\begin{array}{c}1\\ 1\end{array}\right]$.

For $\lambda =-2$

$\begin{array}{rcl}\left|M+2I\right|X& =& 0\\ \left[\begin{array}{cc}3& 0\\ 3& 0\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]& =& 0\left(R_2-R_1\right)\\ \left[\begin{array}{cc}3& 0\\ 0& 0\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]& =& 0\left(\frac{R_1}{3}\right)\\ \left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]& =& 0\\ x& =& 0\\ X& =& \left[\begin{array}{c}0\\ y\end{array}\right]\\ X& =& y\left[\begin{array}{c}0\\ 1\end{array}\right]\end{array}$

Hence, the eigenvector corresponding to$\lambda =-2$ is$\left[\begin{array}{c}0\\ 1\end{array}\right]$ .