I am interested in checking the relations between the eigenvalues and eigenvectors of the two products $A B $ and $ B A$ of two $n \times n$ matrices $A$ and $B$.
First, we note that $\mbox{det}(A B) = \mbox{det}(B A)$, which implies that the product of the eigenvalues of $A B$ is equal to the product of the eigenvalues of $B A$. But we can say more about it.
Let us first consider the case when $\lambda \neq 0$ is an eigenvalue of $A B$ with the corresponding eigenvector $v$. Then by definition, $$ (A B) (v) = \lambda v, \ \ \left[ v \in \mathbf{R}^n, v \neq 0 \right] \tag{1} $$
We claim that $(\lambda, B v)$ is an eigenpair for the matrix $B A$. This can be established as follows.
$$ \lambda (B v) = B (\lambda v) = B [A B v] = (B A) (B v) $$
Thus, $$ (B A) (B v) = \lambda (B v) \tag{2} $$ and furthermore, $B v \neq 0$ because if $B v = 0$, then (1) gives $\lambda v = 0$, which cannot happen since $\lambda \neq 0$ and $v \neq 0$.
Hence, (2) establishes that $(\lambda, B v)$ is an eigenpair for $B A$.
In a very similar manner, by interchanging the roles of $A$ and $B$, we can establish the following:
If $\mu \neq 0$ is an eigenvalue for $B A$ with the eigenvector $w$, then $(\mu, A w)$ is an eigenpair for $A B$.
We consider the scenario when $\lambda = 0$ is an eigenvalue for $A B$ with the corresponding eigenvector $v$.
In this case, $\mbox{det}(A B) = 0$ which shows that $\mbox{det}(B A) = 0$.
Hence, we can conclude that $\lambda$ is also an eigenvalue of $B A$.
Does it follow that $B v$ is an eigenvector for $B A$ corresponding to $\lambda = 0$ as the eigenvalue?
We can establish that $(B A) (B v) = B [A B v] = B (0 v) = 0$ but I have a problem in showing that $B v \neq 0$. Can you throw some light on this?
If we consider a general scenario of matrices $A$ and $B$ of dimensions $(m \times n)$ and $(n \times m)$, then $A B$ will have size $(m \times m)$ and $(n \times n)$ respectively. To what extent we can extend the above calculations for the rectangular matrices $A$ and $B$? (I am yet to try calculations on these, sorry! Some insight on this is highly appreciated!)
