Eigen values of AB and BA 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.
 A: 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.
A: Here is a way that uses matrix computations (that works for all $n,m$):
Let $R_1 = \begin{bmatrix} I & 0 \\ -A & I \end{bmatrix} $, $R_2 = \begin{bmatrix} I & -B \\ 0 & I \end{bmatrix}$,
$M=\begin{bmatrix} I & B \\ A & I \end{bmatrix}$.
Note that 
$R_1 M = \begin{bmatrix} I & B \\ 0 & I-AB \end{bmatrix}$,
$R_2 M = \begin{bmatrix} I-BA & 0 \\ A & I \end{bmatrix}$,
and $\det R_k = 1$.
Consequently, we have $\det (I-AB) = \det(I-BA)$ for all appropriately sized $A,B$.
Now suppose $\lambda \neq 0$, and replace $A$ by ${1 \over \lambda } A$, which gives
$\det (I-{1 \over \lambda }AB) = \det(I-{1 \over \lambda }BA)$, or
${1 \over \lambda^m } \det (\lambda I-AB) = {1 \over \lambda^n } \det (\lambda I-BA) $.
Note that $p_{AB}(\lambda)= \det (\lambda I-AB) $ and
$p_{BA}(\lambda)= \det (\lambda I-BA) $ are both polynomials, hence it follows that $(\lambda-\mu)^k$ divides $p_{AB}$ iff it divides $P_{BA}$,
and so the non-zero eigenvalues of $AB$ and $BA$ have the same algebraic  multiplicity.
(Note: for a truly 'matrix computation-free' proof, one could show that
$\dim \ker (\lambda I-AB)^k = \dim \ker (\lambda I-BA)^k$ for all $\lambda \neq 0$, and $k$, but this is far more tedious.)
A: If $n=m$: We want to show that characteristic polynomial of $AB=$ the characteristic polynomial of $BA$, i.e $det(AB-xI_n)=\det(BA-xI_n)$  


*

*First case : if $A$ is invertible: 
$$\det(AB-xI_n)=\det(A(B-xA^{-1}))=\det(A)\det(B-xA^{-1})=\det(B-xA^{-1})\det(A)=\det((B-xA^{-1})A)=\det(BA-xI_n)$$

*Second case: if $A$ is  invertible:
Let $\{0,\lambda_2,\lambda_3,\ldots,\lambda_r\}$ the eigenvalues of $A$, with 
$0<|\lambda_1|<\ldots <|\lambda_r|$, now remarking that if$ \alpha\in \Bbb R$  such that $0<\alpha<|\lambda_1|$ the $x$ is not an eigenvalue of $A$ and $A-\alpha I_n$ is invertible, it is easy to show that there is $N\in \Bbb N$ such that for all $p\leq N$, $0<\dfrac{1}{p}<|\lambda_1|$, then for all $p\geq N$, the matrix  $A_p=A-\dfrac{1}{p}I_n$ is invertible. By the first case for all $p\geq N$, $\det(A_pB-xI_n)=\det(BA_p-I_n)$ passing to the limit as $p\to \infty$ (and using the fact that $\det$ is continuous and $A_p\to A$ ) we obtain $\det(AB-xI_n)=\det(BA-xI_n)$. 

