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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.

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    $\begingroup$ By the way: the eigenvalues of a matrix form its spectrum. The plural of spectrum is spectrums or spectra, not spectres. $\endgroup$ – Ben Grossmann May 30 '17 at 10:54
  • $\begingroup$ See this post $\endgroup$ – Ben Grossmann May 30 '17 at 12:26
  • $\begingroup$ okay, thanks :) $\endgroup$ – ivana14 May 30 '17 at 12:52
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It's not clear what you mean by "finding the eigenvalues of unknown matrices", but clearly you're misunderstanding how these proofs are meant to go. In any case: to show that two matrices have the same eigenvalues, the most convenient approach is often to show that they have the same characteristic polynomial.

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.

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  • $\begingroup$ Could you please elaborate on the 1st approach a bit more? I haven't really learned that method, so I am not sure how to use it? Thanks $\endgroup$ – ivana14 May 30 '17 at 12:18

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