What is the relation between eigen vectors of AB and BA? For two matrices $A$ and $B$ if the eigen vector of $AB$ is $X$ and that of $BA$ is $Y$, What is the relation between $X$ and $Y$? Can I prove $BX =Y$?
 A: If $v$ is an eigenvector of $AB$ to the eigenvalue $\lambda\neq0$, then $Bv\ne0$
and $$\lambda Bv=B(ABv)=(BA)Bv,$$ which means $Bv$ is an eigenvector for $BA$ with the same eigenvalue.
But if $\lambda=0$ then
$0=\det(AB)=\det(BA)$, so $0$ is an eigenvalue of $BA$ as well.
A: I take $X$ to be the set of eigenvectors for $AB$, and similarly for $Y$.
If $B$ is invertible then $AB v = \lambda v$ implies $B A (B v) = \lambda (B v)$. Conversely, $B A u = \lambda u$ implies $A B (B^{-1} u) = \lambda (B^{-1} u)$, so $B X = Y$. 
If $A$ is invertible, a similar argument shows that $A Y = X$.
Now note the example
$$
A = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix},
\qquad
B = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}.
$$
Here $A$ is invertible, while $B$ is not. We have
$$
A B =
\begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix},
\qquad
B A =
\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}.
$$
So $X$ consists of the multiples of $e_2$, and $Y$ of the multiples of $e_{1}$, but $B X = \{ 0 \} \ne Y$. (However, $A Y = X$.)
Consider also a case when neither matrix is invertible,
$$
A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix},
\qquad
B = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}.
$$
Then $AB = 0$, so $X = V$, the whole space, while $BA = A$, so $Y$ consists of the multiples of $e_1$ only. 
