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 [itex]A = A + \epsilon I[/itex], so that [itex]det(A+\epsilon I) \neq 0[/itex], and then we have,
[tex] det \left((A+\epsilon I) B - \lambda I \right) = det \left((B(A+\epsilon I) - \lambda I\right)[/tex]
Since then we can use the similarity argument. We then take [itex]\epsilon\rightarrow 0[/itex].
I'm wondering if someone could please provide an accessible, but sound explanation of why we are allowed to use this [itex]\epsilon \rightarrow 0 [/itex] argument.