Question

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

$\left(\begin{array}{cc}5& 1\\ 4& 2\end{array}\right)$

Short answer

The eigenvector for the eigenvalue 1 is$\left(\begin{array}{c}1\\ -4\end{array}\right)$ and the eigenvector for the eigenvalue 6 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 I is the identity matrix, are considered as the eigenvalue of a matrix.

Find the eigenvalues and the eigenvectors

Given matrix is $A=\left(\begin{array}{cc}5& 1\\ 4& 2\end{array}\right)$ . The characteristic equation is $\left|A-\lambda I\right|=0$.

Solve for $\lambda$ as follows:

$\begin{array}{rcl}\left|\left(\begin{array}{cc}5& 1\\ 4& 2\end{array}\right)-\lambda \left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)\right|& =& 0\\ \left|\begin{array}{cc}5-\lambda & 1\\ 4& \left|2-\lambda \right|\end{array}\right|& =& 0\\ \left(5-\lambda \right)\left(2-\lambda \right)-4& =& 0\\ 10-7\lambda +{\lambda }^2-4& =& 0\end{array}$

Solve further as follows:

$$\begin{gathered} {\lambda }^2-7\lambda +6=0 \\ \left(\lambda -6\right)\left(\lambda -1\right)=0 \\ \lambda =1,6 \end{gathered}$$

Therefore, the eigenvalues are 1 and 6 .

Now, find the eigenvectors for each eigenvalue by solving $\left(A-\lambda I\right)X=0$ for X where localid="1664282164292" $X=\left(\begin{array}{c}{x}_1\\ x_2\end{array}\right)$ For $\lambda =1$

$\begin{array}{rcl}\left(A-I\right)X& =& 0\\ \left(\begin{array}{cc}5-1& 1\\ 4& 2-1\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)& =& \left(\begin{array}{c}0\\ 0\end{array}\right)\\ \left(\begin{array}{cc}4& 1\\ 4& 1\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)& =& \left(\begin{array}{c}0\\ 0\end{array}\right)\end{array}$

The above equation implies that,

$\begin{array}{rcl}4x+y& =& 0\\ y& =& -4x\end{array}$

$$\begin{gathered} X=\left(\begin{array}{c}x\\ -4x\end{array}\right) \\ X=x\left(\begin{array}{c}1\\ -4\end{array}\right) \end{gathered}$$

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

For $\lambda =6$

$\begin{array}{rcl}\left(A-I\right)X& =& 0\\ \left(\begin{array}{cc}5-6& 1\\ 4& 2-6\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)& =& \left(\begin{array}{c}0\\ 0\end{array}\right)\\ \left(\begin{array}{cc}-1& 1\\ 4& -4\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)& =& \left(\begin{array}{c}0\\ 0\end{array}\right)\\ -x+y& =& 0\\ y& =& x\end{array}$

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

Therefore, the eigenvector corresponding to the eigenvalue 6 is$\left(\begin{array}{c}1\\ 1\end{array}\right)$.