2
$\begingroup$

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 $BA$.

  • 2
    Is this homework? what have you tried?2012-12-19
  • 0
    A [related question](http://math.stackexchange.com/q/94926/11619). 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).2012-12-19

2 Answers 2

6

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

  • 0
    I think you mean $Bv$2012-12-19
  • 0
    @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.2012-12-19
  • 0
    right. I initially thought you wanna ask what's the eigenvector of BA.2012-12-19
3

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.