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?
 A: 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.
A: 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.
The other eigenvector is handled similarly. In general, all these calculations would typically be done with matrices rather than writing out the equations explicitly.
