TITLE: Eigen values of AB and BA
QUESTION [6 upvotes]: let A be a linear transformation from $R^n$ to $R^m$, and B be a linear transformation from $R^m$ to $R^n$, it's easy to show that AB and BA has same eigen-value(except $0$).
But my question is  how to show that the multiplicity of eigen-values are the same? can anyone give a proof just from the theory of linear transformation? I mean, without matrix computation.

REPLY [5 votes]: Just in case you want the result for geometric multiplicity (dimension of eigenspaces) instead of (or in addition to) algebraic multiplicity, here's a matrix-free proof of that result.
Consider any non-zero eigenvalue $\lambda$ of $AB$ and the corresponding eigenspace $E$.  Also, let $F$ be the eigenspace of $BA$ for the same eigenvalue $\lambda$.  (If you didn't already know that $\lambda$ is an eigenvalue of $BA$, then I'd allow the possibility that $F=\{0\}$; that wouldn't damage the following argument.)  For any $x\in E$, we have $ABx=\lambda x$, and therefore, applying $B$ to both sides, we also have $BABx=\lambda Bx$.  This says that $Bx\in F$.  So we've shown that $B$ maps $E$ into $F$.  Furthermore, this map is injective; non-zero vectors $x\in E$ have non-zero images $Bx\in F$, because $ABx=\lambda x\neq0$.  (This is where we use that $\lambda\neq0$.)  Since we have an injective linear map from $E$ into $F$, it follows that $\dim(E)\leq\dim(F)$.  A symmetrical argument, interchanging the roles of $A$ and $B$) shows that $\dim(F)\leq\dim(E)$.  Therefore the two dimensions are equal.