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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 ?

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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.

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  • $\begingroup$ +1 for answering the actual question that was asked. $\endgroup$ – amd Sep 24 '18 at 19:37
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Since the characteristic polynomial of $M$ is $x^4+2x^3$, the eigenvalues are $-2$ and $0$. Its now easy to compute the eigenvectors.

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  • $\begingroup$ how did you find $x^4 + 2x^3$ ? $\endgroup$ – KEVIN DLL Sep 24 '18 at 8:23
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    $\begingroup$ @KEVINDLL I computed the characteristic polynomial of $M$, which is$$\det\begin{pmatrix}-x & -2 & 4 & -2 \\ 1 & 1-x & -2 & -1 \\ 0 & 0 & -x & 0 \\ 1 & -1 & 2 & -x-3\end{pmatrix}.$$ $\endgroup$ – José Carlos Santos Sep 24 '18 at 8:25
  • $\begingroup$ but the matrix is very large and you found the result in less than 10 minutes ? or maybe i mean what way did you use ? $\endgroup$ – KEVIN DLL Sep 24 '18 at 8:27
  • $\begingroup$ @KEVINDLL $4\times 4$ is not "very large". If you expand the fourth row of the matrix, you are left with one $3\times 3$ determinant to simplify, which can be done in under $10$ minutes easily (with some practice, of course) $\endgroup$ – 5xum Sep 24 '18 at 8:29
  • $\begingroup$ @KEVINDLL I wouldn't say that a $4\times4$ matrix is very large. Besides, those $0$'s on the third row are quit helpful. They reduce the problem of computing that determinant to the problem of computing the determinant of a $3\times3$ matrix. $\endgroup$ – José Carlos Santos Sep 24 '18 at 8:30
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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:

$$ = -\lambda \cdot \left( -\lambda \cdot (1 - \lambda) \cdot (-3-\lambda) + (-2) \cdot (- 1) \cdot 1 + (-2) \cdot 1 \cdot (-1)\right ....)$$

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