# Prove AB and BA have the same characteristic polynomial

• IniquiTrance
In summary, for any square matrices A and B of the same size, the characteristic polynomial of AB and BA are the same. This can be proven by taking A = A + \epsilon I and using the similarity argument. By taking \epsilon \rightarrow 0, we can use the continuity of polynomials to show that the limit of P(\epsilon, \lambda) is equal to the characteristic polynomial of both AB and BA. This is possible because the entities on either side of the equation are multivariate polynomials with respect to epsilon and lambda. Thus, the characteristic polynomial of AB and BA are equivalent.

## 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!

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

Hmm, somewhat. Could you please provide greater insight?

IniquiTrance said:
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?

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! ]