Eigenvalues of negative companion matrix Here's a homework question I've been stuck on for a while.


Given
$$A = \begin{bmatrix}
0 & 0 & 0 & \cdots & 0 & a_0 \\
-1 & 0 & 0 & \cdots & 0 & a_1 \\
0 & -1 & 0 & \cdots & 0 & a_2 \\
\vdots & \vdots & \vdots &\ddots & \vdots & \vdots \\
0 & 0 & 0 & \cdots & 0 & a_{n-2} \\
0 & 0 & 0 & \cdots & -1 & a_{n-1} \end{bmatrix}$$
Find the characteristic polynomial of $A$ and its eigenvalues.

The characteristic polynomial didn't seem too bad. I calculated a few of them for low values of $n$ and then proved the formula using induction. I think it's:
$$f(\lambda) = a_0 - a_1 \lambda + a_2 \lambda^2 -a_3 \lambda^3 + \cdots + (-1)^{n-1}a_{n-1}\lambda^{n-1} + (-1)^n\lambda^n$$
But then there is the problem of the eigenvalues. This polynomial seems hopelessly general in terms of actually finding its roots in terms of the $a_i$s.
Questions:


*

*Is the characteristic polynomial correct? I've checked it a few times in hopes of finding something wrong, but I would love for this to be the problem. That would make my life a lot easier.

*If it's correct, how do I find its roots? Is such a thing even possible?

Hint: I'm not sure why you need the roots of either polynomial.  Given $f(\lambda) = \lambda^n + a_{n-1}\lambda^{n-1} + \cdots + a_0$, consider $f(-\lambda)$.
In general, something along these lines happens to the characteristic polynomial of a matrix whenever you multiply that matrix by a scalar.
