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\begin{align} x'(t)&=x(1-10y)\\ y'(t)&=y(1-10x) \end{align}

One of the equilibrium points of this system is $(0,0)$. I was trying to draw a phase portrait with this point. Now I figured this is an $\textit{unstable node}$ as the linearized system has two repeated positive real eigenvalues. Now I need an eigenvector corresponding to it to get a parallel trajectory through $(0,0)$ to initiate the sketch of the phase portrait. But I've got stuck here while trying to get the eigenvector.

\begin{align} \begin{pmatrix} 0 & 0\\ 0 & 0 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}&=\begin{pmatrix} 0\\0 \end{pmatrix} \end{align}

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Any vector satisfies $I\cdot v = 1\cdot v$, that means the whole space is the eigenspace, any basis is an eigenvector basis.

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