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Let $A,B \in M(n,\mathbb{C})$ be two $n\times n$ matrices. I would like know how to prove that eigen-value of $AB$ is the same as the eigen-values of $BA$.

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Is this homework? what have you tried? –  yohBS Dec 19 '12 at 12:08
    
A related question. Not an exact duplicate, because there it was assumed that $B$ is invertible. Studying the answers given there will get you started anyway (more or less along the lines of Matt Pressland's +1 answer). –  Jyrki Lahtonen Dec 19 '12 at 14:10
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2 Answers 2

Hint: let $v$ be an eigenvector of $AB$ with eigenvalue $\lambda$. What is $BABv$?

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I think you mean $Bv$ –  sun Dec 19 '12 at 13:52
    
@sun Nope, although that would also be a good question to ask. Answering the question I asked should lead to an answer to your suggested modification, if I've understood correctly. –  Matt Pressland Dec 19 '12 at 13:59
    
right. I initially thought you wanna ask what's the eigenvector of BA. –  sun Dec 19 '12 at 14:15
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you can prove $|\lambda I-AB|=|\lambda I-BA|$ by computing the determinant of following $$ \left( \begin{array}{cc} I & A \\ B & I \\ \end{array} \right) $$ in two diffeerent ways.

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