# Prove AB and BA have the same characteristic polynomial

IniquiTrance

## Homework Statement

Show that for any square matrices of the same size, A, B, that AB and BA have the same characteristic polynomial.

## The Attempt at a Solution

I understand how to do this if either A or B is invertible, since they would be similar then. I saw a proof that states to take $A = A + \epsilon I$, so that $det(A+\epsilon I) \neq 0$, and then we have,

$$det \left((A+\epsilon I) B - \lambda I \right) = det \left((B(A+\epsilon I) - \lambda I\right)$$

Since then we can use the similarity argument. We then take $\epsilon\rightarrow 0$.

I'm wondering if someone could please provide an accessible, but sound explanation of why we are allowed to use this $\epsilon \rightarrow 0$ argument.

Thanks!

voko
Can you see that the entities on either side of the equation are multivariate polynomials with respect to epsilon and lambda?

IniquiTrance
Hmm, somewhat. Could you please provide greater insight?

voko
Hmm, somewhat. Could you please provide greater insight?

"Somewhat" is not good. If you really don't see it, take a simple 2x2 case and work it out.

If you do see that, then what do you know of continuity of polynomials? What is $\lim_{\epsilon \rightarrow 0} P(\epsilon, \lambda)$, if $P(\epsilon, \lambda)$ is a polynomial?

Homework Helper
Hmm are you sure that's the right question? A would need some inverse C such that :

C-1AC = B

Then you could start by saying :

pB(λ) = det(B-λIn)

Then you should use what you know about B and the identity matrix In to massage the expression into pA(λ) [ Hint : Think about inverses and how they work! ]