# equality of spectrums [duplicate]

Let $A$ and $B$ be both square matrices with the same dimension. Prove that the following is correct:

$S_p(AB)=S_p(BA)$ , where $S_p$ is the set of eigenvalues.

I am really confused by how I am supposed to find the eigenvalues of these unknown matrices. So if anyone could explain it a bit, I would really appreciate it.

• By the way: the eigenvalues of a matrix form its spectrum. The plural of spectrum is spectrums or spectra, not spectres. – Ben Grossmann May 30 '17 at 10:54
• See this post – Ben Grossmann May 30 '17 at 12:26
• okay, thanks :) – ivana14 May 30 '17 at 12:52

Approach 1: Using either Sylvester's determinant identity or something equivalent, one can show that for all $t \neq 0$, we have $$\det(tI - AB) = \det(tI - BA)$$ if the polynomials are equal for all $t \neq 0$, then they must be the same polynomial.
Approach 2: Note that $AB$ is similar to $BA$ whenever $B$ is invertible (why?). It follows that $\det(t I - BA) = \det(t I - AB)$. When $B$ is not invertible, we note that $$\det(t I - BA) = \lim_{\epsilon \to 0} \det(t I - (B + \epsilon I)A) = \lim_{\epsilon \to 0} \det(t I - A(B + \epsilon I)) = \det(t I - AB)$$ since $B + \epsilon I$ is invertible when $|\epsilon|$ is sufficiently small and non-zero.