Eigenvalues of negative companion matrix

Question

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:

1. 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.

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

Answer 1

The problem, as stated, is not solvable (confirmed by the professor who assigned it).

The characteristic polynomial is correct as written, and there is no general way to find the eigenvalues of this matrix.

Answer 2

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.