AB and BA have identical nonsingular Jordan blocks If A and B are square matrices of the same size I know how to show that AB and BA have the same eigenvalues and characteristic polynomials. But I want to show that they have identical nonsingular Jordan Blocks. 
I am not really sure how to proceed. Maybe some argument about the dimension of the kernel of AB and BA?
 A: First, let's prove it for invertible $AB$: If $AB$ is invertible, then so is $A$ and $B$. Let \begin{equation}AB=P^{-1}JP.\end{equation} Multiply with $B$ on the left, and $B^{-1}$ on the right: \begin{equation}BA=BP^{-1}JPB^{-1}.\end{equation} So $AB$ and $BA$ have the same Jordan normal form. In fact this works also if only one of $A$ or $B$ is invertible.
So that only leaves the case where both $A$ and $B$ are singular...

OK, let's suppose that both $A$ and $B$ are singular - after consulting Matrix Theory by Zhang, I can provide the following:
Consider this identity:
$$\begin{bmatrix} I & -A \\ 0 & I\end{bmatrix}\begin{bmatrix} AB & 0 \\ B & 0\end{bmatrix}\begin{bmatrix} I & A \\ 0 & I\end{bmatrix} = \begin{bmatrix} 0 & 0 \\ B & BA\end{bmatrix},$$ which shows that $$\begin{bmatrix} AB & 0 \\ B & 0\end{bmatrix} \text{ and } \begin{bmatrix} 0 & 0 \\ B & BA\end{bmatrix}$$ are similar. 
Now I make use of the following theorem: Let $M$ be a $n \times n$ matrix with eigenvalues $\lambda_1,\lambda_2,\ldots,\lambda_r$. Let $J$ be a Jordan canonical matrix similar to $M$. The number of simple Jordan blocks $J_m(\lambda_i)$ in $J$, with $m \geq k$, is
$$\text{rank}(M - \lambda_iI)^{k-1} - \text{rank}(M - \lambda_iI)^{k},\quad k=1,2,\ldots$$ (Theorem 5.14 in Cullen's Matrices and Linear Transformations).
First the characteristic polynomials of $AB$ and $\begin{bmatrix} AB & 0 \\ B & 0\end{bmatrix}$ are the same (follows from determinant of a block triangular matrix is the same as the product of the determinants of the diagonal blocks). Likewise $BA$ and $\begin{bmatrix} 0 & 0 \\ B & BA\end{bmatrix}$ have the same characteristic polynomials. So all the matrices involved have the same eigenvalues.
Now assume $A$ and $B$ are $n \times n$ matrices. Notice that for any non-zero eigenvalue $\lambda$ we have $$\text{rank}\left (\begin{bmatrix} AB & 0 \\ B & 0\end{bmatrix} - \lambda I\right ) = \text{rank}(AB - \lambda I) + n.$$ Also $$\text{rank}\left (\begin{bmatrix} AB & 0 \\ B & 0\end{bmatrix} - \lambda  I\right )^k = \text{rank}\left (\begin{bmatrix} (AB-\lambda I)^k & 0 \\ D & (-1)^k \lambda^k I\end{bmatrix} \right ) = \text{rank}(AB - \lambda I)^k + n,$$ where the matrix $D$ is some matrix resulting from the multiplication.
Combining this with the theorem mentioned, it follows that $AB$ and $\begin{bmatrix} AB & 0 \\ B & 0\end{bmatrix}$ have the same nonsingular Jordan blocks. 
A similar argument will show that $BA$ and $\begin{bmatrix} 0 & 0 \\ B & BA\end{bmatrix}$ have the same nonsingular Jordan blocks. Combining this with the similarity between the block matrices yields the desired result.  
A: Not totally sure about this approach yet... work in progress

Suppose that $\lambda \neq 0$ is an eigenvalue of $AB$.  Let $v$ be an associated eigenvector (noting that $Bv \neq 0$). Then, we have
$$
(BA - \lambda I)(Bv) = 
BAB v - \lambda(Bv) = \\
B[AB - \lambda I]v = 0
$$
In other words, we have
$$
(AB - \lambda I)v = 0 \implies (BA - \lambda I)(Bv) = 0
$$
Similarly, we can note that if $(AB - \lambda I)^n v = 0$, then $(BA - \lambda I)^n (Bv) = 0$.  Thus, $AB$ and $BA$ have the same generalized eigenvectors.  The conclusion would follow.
