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.

Homework Equations

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?

 

[STEMucator]

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 :

p_B(λ) = det(B-λI_n)

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