Question

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

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

Short answer

The eigenvector for the eigenvalue 0 is$\left(\begin{array}{c}1\\ -2\end{array}\right)$ and the eigenvector for the eigenvalue 5 is $\left(\begin{array}{c}2\\ 1\end{array}\right)$.

Step-by-step solution

Definition of eigenvalue

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

Find the eigenvalues and the eigenvectors

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

Solve for $\lambda$as follows:

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

Solve further as follows,

$\begin{array}{rcl}{\lambda }^2-5\lambda & =& 0\\ \lambda \left(\lambda -5\right)& =& 0\\ \lambda & =& 0,5\end{array}$

Therefore, the eigenvalues are 0 and 5.

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

For $\lambda =0$,

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

The above equation implies that $4x+2y=0$which is equivalent to role="math" localid="1664344521227" $4x=-2y$. Which becomes:

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

Therefore, the eigenvector corresponding to the eigenvalue 0 is $\left(\begin{array}{c}1\\ -2\end{array}\right)$.

For $\lambda =5$

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

The above equation implies that $-x+2y=0$which is equivalent to $x=2y$. Which becomes:

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

Therefore, the eigenvector corresponding to the eigenvalue 5 is $\left(\begin{array}{c}2\\ 1\end{array}\right)$.