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?
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.
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.
I think the right answer here avoids getting in the weeds of Jordan Forms and instead involves a small refinement of the notion of $T-$invariant subspaces.
The basic insight is $B(AB)^n = (BA)^n B$ and $A(BA)^n = (AB)^n A$ and write $\mathbb F^n = V = \text{image }(AB)^n \oplus \text{ker }(AB)^n= \text{image }(BA)^n \oplus \text{ker } (BA)^n$
The standard commutativity argument tells us e.g. $(AB)$ commutes with $(AB)^n$ so for $\mathbf z \in \text{image }(AB)^n$ we have $\big((AB)\mathbf z\big) = (AB)(AB)^n \mathbf z' = (AB)^n (AB)\mathbf z'\implies \big((AB)\mathbf z\big) \in \text{image }(AB)^n$, i.e. $\text{image }(AB)^n$ is $(AB)$ invariant; $\text{ker }(AB)^n$ is also $(AB)$ invariant [check $(AB)^n (AB)\mathbf z =(AB)(AB)^n\mathbf z=\mathbf 0$ for $\mathbf z\in\text{ker }(AB)^n$]. Notice this implies $(AB)_{\vert \text{ker }(AB)^n}$ is nilpotent and $(AB)_{\vert \text{image }(AB)^n}$ is invertible-- and the latter is what we want to focus on.
$W_1:= \text{image }(AB)^n$ and $W_2:= \text{image }(BA)^n$ with respective bases $\mathbf B_1$ and $\mathbf B_2$
note that $r =\dim W_1 = \dim W_2$, which can be justified e.g. by the following rank inequalities:
$\text{rank}\big((AB)^{n}\big)=\text{rank}\big((AB)^{n+1}\big)=\text{rank}\big(A(BA)^{n}B\big) \leq \text{rank}\big((BA)^{n}\big)$ and by nearly identical argument
$\text{rank}\big((BA)^{n}\big)\leq \text{rank}\big((AB)^{n}\big)\implies \dim W_1 = \dim W_2$
Now for the refinement:
for $\mathbf x \in W_2$ we have $A\mathbf x = A(BA)^n\mathbf x' = (AB)^n (A\mathbf x')\implies \big(A\mathbf x\big)\in W_1$ and by nearly identical argument, $\mathbf y \in W_1\implies \big(B\mathbf y\big)\in W_2$
so $A\mathbf B_2 = \mathbf B_1 M_A $ and $B\mathbf B_1 =\mathbf B_2 M_B$
$\implies (AB)\mathbf B_1 = A\mathbf B_2M_B =\mathbf B_1 M_A M_B$
$\implies M_A, M_B \in GL_r(\mathbb F)$, for avoidance of doubt: this occurs since $\ker AB\subseteq \text{ker }(AB)^n$ and $\text{ker }(AB)^n \cap W_1 = \big\{\mathbf 0\big\}$ and thus $M_B$ is injective hence invertible since square, which implies $M_A$ is as well. Similarly $(BA)\mathbf B_2 = B\mathbf B_1 M_A =\mathbf B_2 M_B M_A$.
Thus after choice of bases $\mathbf B_1$ and $\mathbf B_2$ we see $(AB)_{\vert W_1}$ has matrix representation $M_A M_B$, while $(BA)_{\vert W_2}$ has matrix representation $M_B M_A$ and $M_A^{-1}\big(M_A M_B\big)M_A = M_B M_A$ so these two matrices have the same Jordan structure i.e. $(AB)_{\vert W_1}$ and $(BA)_{\vert W_2}$ have the same Jordan structure.
