Help with a differential equation system Given $x' = -x$ and $y' = -4x^3+y$, we want to linearize and show phase portrait at origin.
So I make system $\vec{Y}' = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}\vec{Y}$ by just scrapping the $-4x^3$ term. But now we have repeated $0$ eigenvalue, so I try to find an eigenvector.
$\left[ \begin{pmatrix} -1 & 0 \\ 0 &1 \end{pmatrix} - \begin{pmatrix} 0 & 0 \\ 0 & 0\end{pmatrix} \right]\begin{pmatrix}v_1 \\ v_2\end{pmatrix} = \begin{pmatrix}0 \\ 0\end{pmatrix} \implies v_1 = v_2 = 0$. So $\vec{v} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$?
Unless I am mistaken. What kind of eigenvector is this? I can't think of how to draw a phase portrait, thanks!

The assertion that the eigenvalues of the matrix $\vec Y'$ are both zero is erroneous.  However, we have:
The eigenvectors of the matrix
$\vec Y' = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} \tag 1$
are $(1, 0)^T$, with eigenvalue $-1$, and $(0, 1)$, with eigenvalue $1$, as is easily checked, e.g.
$\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = -1\begin{pmatrix} 1 \\ 0 \end{pmatrix}, \tag 2$
with a similar calculation for eigenvcector $(0, 1)$.  Thus the point $(0, 0)$ is a saddle, as corroborated by the phase portrait which is easily drawn.
