0
$\begingroup$

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?

$\endgroup$
  • 2
    $\begingroup$ 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. $\endgroup$ – Stefano Mar 30 '14 at 18:44
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Roger Burt Mar 30 '14 at 22:03
  • $\begingroup$ Well, the point is I'm not sure why you're looking for any polynomial roots here. $\endgroup$ – Omnomnomnom Mar 30 '14 at 22:06
  • $\begingroup$ The roots of the polynomial are the eigenvalues of the matrix. The question asks for these eigenvalues. $\endgroup$ – Roger Burt Mar 30 '14 at 22:13
  • $\begingroup$ 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. $\endgroup$ – Omnomnomnom Mar 30 '14 at 22:17
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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