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I have the following system

\begin{align*} x_1'&=-x_1\\ x_2'&=-x_2 \end{align*}

The eigenvalues are $\lambda_{1,2}=-1$.

How to write the eigenvectors?

I've come upto this

\begin{matrix} \begin{pmatrix}0&0\\0&0\end{pmatrix}&\begin{pmatrix}\eta_1\\\eta_2\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix} \end{matrix}

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    $\begingroup$ Looks like any vector $\pmatrix{\eta_1\cr\eta_2\cr}$ is a solution. So what's your problem? But, really, you have separated the variables $x_1$ and $x_2$, so there is no point in treating this as a system. Just solve for $x_1$ and $x_2$ individually. $\endgroup$ – Jyrki Lahtonen Oct 27 '17 at 10:40
  • $\begingroup$ @JyrkiLahtonen Are they linearly independent? $\endgroup$ – Bernhard Listing Oct 27 '17 at 10:42
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    $\begingroup$ You can find linearly independent ones, sure. Any basis will do. But just treat this as two ODEs. Life is much simpler then. $\endgroup$ – Jyrki Lahtonen Oct 27 '17 at 10:43
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    $\begingroup$ Have you noticed that your so-called system is made of two independent ODE, giving you instantly the solutions ? $\endgroup$ – Jean Marie Oct 27 '17 at 10:43
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Don't overthink. The differential matrix of this system is

$$ \begin{pmatrix} -1 &0 \\ 0&-1\end{pmatrix} $$

and, as you have said, the eigenvalues are -1. But notice that this is just $-Id$, where $Id$ is the identity $2\times 2$ matrix, for which every vector is an eigenvector. This why you get that weird system, which is really telling you thst you can take any two linearly independent vectors. For the sake of simplicity, pick $(1,0)$ and $(0,1)$.

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