# Eigenvalues and eigenvectors of a $2 \times 2$ matrix.

Find the eigenvalues and eigenvectors of $\begin{bmatrix} 1 & 4 \\ 3 & 2\end{bmatrix}$

$\begin{bmatrix} 1 & 4 \\ 3 & 2\end{bmatrix} \begin{bmatrix} a \\ b\end{bmatrix} = \lambda \begin{bmatrix} a \\ b\end{bmatrix} \implies a+4b = \lambda a, ~ 3a+2b = \lambda b \implies \lambda = -2, 5.$

How do I find the eigenvectors?

• Find the nullspace of $\lambda I-\left(\begin{matrix}1&4\\3&2\end{matrix}\right)$. – Element118 Dec 3 '15 at 5:25

Substitute the values of $\lambda$ into your equations. First, $\lambda = -2$ gives the system $\{a+4b=-2a,3a+2b=-2b\}$, which reduces to $\{3a+4b=0\}$.
Now, if $3a+4b=0$, then we must have $b=-\frac34a$. We can put $a$ and $b$ in terms of a parameter $t$ by letting $a=t$, thus: $$\begin{bmatrix}a\\b\\ \end{bmatrix}=\begin{bmatrix}1\\-\frac34\\ \end{bmatrix}t$$
So the eigenvector associated with $\lambda=-2$ is $\begin{bmatrix}1\\-\frac34\\ \end{bmatrix}$. We can scale the eigenvector by any nonzero factor, so just to make it a bit cleaner, we can rewrite it as $\begin{bmatrix}4\\-3\\ \end{bmatrix}$. You can verify that this is indeed an eigenvector by multiplying it by the original matrix.
You have to solve the following $\text{Kern}(A-\lambda Id)$,where $A$ is your original matrix, once with $\lambda=-2$ and once with $\lambda=5$. You have then to extract a basis of these subspaces and these will be your eigenvectors.