Eigenvalues and eigenvectors of $4\times 4$ matrix Let M the $4\times 4$ matrix: 
$M = \begin{pmatrix}
0 &-2  & 4 &-2 \\ 
1 &1  &-2  &-1 \\ 
0 &0  &0  &0 \\ 
1 &-1  &2  &-3 
\end{pmatrix}  $  and $M^n = (-1)^n2^{n-2}M^2$
 I determined 
$$e^M = I + M + M^2 \times \sum_{n=2}^{\infty}\frac{(-1)^n \times 2^n}{n!} $$
but can i use $e^M $ to determine Eigenvalues and eigenvectors of this matrix ? 
 A: The relationship you have given $M^n = (-1)^n2^{n-2}M^2$ can give you very easily the characteristic polynomial, that needs to be of order 4. 
$M^4=4M²$, and $M^3=-2M²$, thus the $M^4+2M^3=0$...
Of course I don't know where you got this relationship: either you found it by hard work, and maybe this was easier to compute directly the characteristic polynomial like in the other answer, or it was given to you, and my answer is a tad quicker. 
A: Since the characteristic polynomial of $M$ is $x^4+2x^3$, the eigenvalues are $-2$ and $0$. Its now easy to compute the eigenvectors.
A: As @5xum already said: 4x4 is not very large. Furthermore you can notice the row of zeros (which is really nice).

$$\det{(M-I\lambda)} = \det\begin{pmatrix}
0-\lambda &-2  & \color{red}4 &-2 \\ 
1 &1 - \lambda  &\color{red}{-2}  &-1 \\ 
\color{red}0 &\color{red}0  &\color{red}{-\lambda}  &\color{red}0 \\ 
1 &-1  &\color{red}2  &-3-\lambda 
\end{pmatrix}  $$

$$
=-\lambda \cdot \det\begin{pmatrix}
0-\lambda &-2 &-2 \\ 
1 &1 - \lambda &-1 \\ 
1 &-1  &-3-\lambda 
\end{pmatrix}  $$

This can be easily calculated:
$$ = -\lambda \cdot \left( -\lambda \cdot (1 - \lambda) \cdot (-3-\lambda) + (-2) \cdot (- 1) \cdot  1 +  (-2) \cdot 1 \cdot (-1)\right ....)$$
