# Eigenvalues of negative companion matrix

Here's a homework question I've been stuck on for a while.

Given

$A = \left[ \begin{array}{cccccc} 0 & 0 & 0 & \cdots & 0 & a_0 \\ -1 & 0 & 0 & \cdots & 0 & a_1 \\ 0 & -1 & 0 & \cdots & 0 & a_2 \\ \vdots & \vdots & \vdots &\ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & 0 & a_{n-2} \\ 0 & 0 & 0 & \cdots & -1 & a_{n-1} \end{array} \right]$

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?

• This is the example provided to show that, given a polynomial, you can always find a matrix with characteristic polynomial equal to the one given. For the eigenvalues the problem is the same as finding all the roots of a given polynomial There is no general way. – Stefano Mar 30 '14 at 18:44

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 $f(-\lambda)$.
• I am not sure what you mean by 'either' polynomial. Only the matrix $A$ was given, and only one characteristic polynomial is associated with it. – Roger Burt Mar 30 '14 at 22:03
• Ah, okay. What I meant to say is that you shouldn't find the roots in terms of the $a_i$, find the roots in terms of the roots of the polynomial $\lambda^n + a_{n-1}\lambda^{n-1} + \cdots + a_0$, which are the eigenvalues of the characteristic matrix. – Omnomnomnom Mar 30 '14 at 22:17