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.


1 Answer 1


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.

  • $\begingroup$ 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 $\Lambda_3=(1,-c-\lambda_2,0)$? $\endgroup$
    – M. Meyer
    Commented Nov 19, 2015 at 10:12
  • $\begingroup$ @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. $\endgroup$ Commented Nov 19, 2015 at 10:56

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .