Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I would like to find the roots of a polynomial using its companion matrix.

The polynomial is ${p(x) = x^4-10x^2+9}$

The companion matrix $M$ is

$M={\left[ \begin{array}{cccc} 0 & 0 & 0 & -9 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 10 \\ 0 & 0 & 1 & 0 \end{array} \right]}$

A theorem says that the eigenvalues of $M$ are the roots of $p(x)$. I tried to find the characteristic polynomial of $M$ but it turned out to be $p(x)$. What should I do to obtain the eigenvalues of $M$?

share|cite|improve this question
Note $p(x) = x^4-10x^2+9 = x^4-10x^2+25 - 16 = (x^2 -5)^2 - 4^2 = ((x^2 - 5)-4)((x^2-5)+4) = (x^2 -9)(x^2-1) = (x-3)(x+3)(x-1)(x+1)$, so the zeros of $p$ are $\pm1, \pm 3$. Maybe you already knew that and you just want to figure out the companion matrix machinery, but they are the zeros for anyone that wants to check their answer. – Michael Albanese Feb 3 '13 at 13:48
I don't know where it came from, but I'd say that this is a pretty goofy problem. The point, I suppose, is that computing eigenvalues is a good way to find roots of polynomials (because software for finding eigenvalues is widely available and very well developed -- see packages like Linpack, etc.) That's all very valid. But, then, in this case, finding the eigenvalues is more difficult than finding the roots. It would have been much better to choose a polynomial whose roots are not so obvious. – bubba Feb 4 '13 at 6:04

Since $p(x)$ is biquadratic, then if $\alpha$ is root, it follows that $-\alpha$ also is a root.

Looking at $M$ you can notice that if you sum the entries of each column, you'll always get $1$. This implies $1$ is an eigenvalue. (Do you know why?).

You have two roots now.

Continue with long division to find the remaining roots.

If you want to use the matrix to find all eigenvalues, recall that $\det (M)$ is the product of all eigenvalues. You can easily compute $\det (M)$ through expansion along the fourth column to find $\det (M)=9$.

Use the first sentence in my answer again to find the other eigenvalues.

share|cite|improve this answer
Very nice -- this gives a heuristic for how one might approach the problem without numerical software. – Assad Ebrahim Feb 1 '14 at 21:15

It's no surprise you got the characteristic equation from the companion matrix, because the equation came from that precise matrix.

If you want to find roots of a polynomial using a companion matrix, you would use one of the specialized numerical methods for computing the eigenvalue of a matrix. Such methods do not use the characteristic polynomial and are tailored to the symmetry of the matrix. See, for example, Numerical Recipes (the link goes to the Householder Method, which is not applicable to your matrix, but gives you the general idea of what's involved).

Typically, you would use a package to find eigenvalues numerically because in general, such routines are very complex and not worth trying to figure out on your own. LINPACK, Matlab, Mathematica, etc., all lhave such routines for general matrices.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.