# Determine the eigenvectors

Let $f'(u_1)<0$ and $c>0$. Consider the matrix $$\begin{pmatrix}0 & 1 & 0\\-f'(u_1) & -c & 1\\0 & 0 & 0\end{pmatrix}.$$ Its eigenvalues are $$\lambda_1=0,~~~~~\lambda_2=-\frac{c}{2}+\sqrt{\frac{c^2}{4}-f'(u_1)},~~~~~\lambda_3=-\frac{c}{2}-\sqrt{\frac{c^2}{4}-f'(u_1)}.$$

The eigenvector $\Lambda_1$ that belongs to the eigenvalue $\lambda_1$ is given by $$\Lambda_1=(-1,0,-f'(u_1)).$$ Now I am searching the eigenvectors $\Lambda_2$ (belonging to $\lambda_2$) and $\Lambda_3$ (belonging to $\lambda_3$). Unfortunately, I was not successfull in finding them.

In the general case, $\lambda_2$ and $\lambda_3$ are both nonzero. When $\lambda$ is one of them you know that $$\left|\begin{matrix}-\lambda & 1 & 0 \\ -f'(u_1) & -c-\lambda & 1 \\ 0 & 0 & -\lambda \end{matrix}\right| = -\lambda \left|\begin{matrix}-\lambda & 1 \\ -f'(u_1) & -c-\lambda \end{matrix} \right| = 0$$
so the rows of the $2\times 2$ matrix are linearly dependent and if we find a vector that maps to something with $0$ in the first component it will also map to something with $0$ in the second component.
Thus, an eigenvector corresponding to $\lambda_i$ will be $(1,\lambda_i,0)$.
In the cases where one of $\lambda_2$ and $\lambda_3$ is zero, the above reasoning doesn't work, so in principle you'll have to handle those specially. Fortunately, this only happens if $-f'(u_1)=0$, in which case a special-case analysis should be simple to conduct.
Another special case is when $\lambda_2=\lambda_3$, in which case a further analysis is necessary find out whether its eigenspace has dimension one or two.
• So the eigenvector $\Lambda_2$ belonging to $\lambda_2$ is $\Lambda_2=(1,-c-\lambda_3,0)$ and, similarly, the eigenvector $\Lambda_3$ belonging to $\lambda_3$ is given by $\Lambda_3=(1,-c-\lambda_2,0)$? – M. Meyer Nov 19 '15 at 10:12
• @M.Meyer: I had a bit of a braino there, confusing the column space of $A-\lambda I$ with its null space. Answer updated and hopefully it works better now. – Henning Makholm Nov 19 '15 at 10:56