Suppose $ A $ is an $ m\times n $ complex matrix and $ B $ an $ n\times m $ complex matrix. Please give an example of matrices $ A, B $ such that $ 0 $ is an eigenvalue of $ AB $ but not of $ BA $.
For the case $ m=n $ please see: Are the eigenvalues of $AB$ equal to the eigenvalues of $BA$? (Citation needed!)
So we need to find in matrices where $ n\neq m $. However, I find out this might not be easy because I don't have an efficient way to carry out. Any help here? Thanks~
Take $A=\begin{pmatrix} 1& 0& 0\end{pmatrix}$ and $B=\begin{pmatrix}1\\0\\0\end{pmatrix}$. Then $AB=1$ while $BA=\begin{pmatrix} 1&0&0\\0&0&0\\0&0&0\end{pmatrix}$.
In fact, you can generalize this example. Take $A=u^*$ and $B=u$ where $u\in\mathbb C^n$ is nonzero. Then $A=u^*u=||u||^2\neq 0$ but $BA=uu^*\in \mathbb C^{n\times n}$ is a diagonalizable matrix that has rank equal to $1$, and so if $n>1$, certainly has $n-1>0$ zero eigenvalues.
Take$$A=\begin{pmatrix}1\\1\end{pmatrix}\text{ and }B=\begin{pmatrix}1&1\end{pmatrix}.$$Then$$AB=\begin{pmatrix}1&1\\1&1\end{pmatrix}\text{ and }BA=(2).$$
If $m<n$ then $BA$ will have determinant zero, and so have zero as an eigenvalue. So all you have to do is to fix it so that $AB$ is nonsingular.
