TITLE: Given $BA$, find $AB$.
QUESTION [11 upvotes]: If $A, B$ are $4\times 3, 3\times 4$ real matrices, respectively,
$$BA=\begin{bmatrix} -9 & -20 & -35 \\ 2 & 5 & 7 \\ 2 & 4 &8 \end{bmatrix}$$
$$AB=\begin{bmatrix} -14 & 0 & -15&-32 \\ 2x-9 & 1 & 3x-9&4x-19\\ 2 & 0 & 3&4\\6&0&6&14 \end{bmatrix}$$
What is $x$ ?
I try some ways, for example, $\det AB=0$, but it didn't work

REPLY [5 votes]: For any square matrix $S$, let us denote its minimal polynomial by $m_S$  and denote  its characteristic polynomial by $p_S$.
Facts: 

A square matrix is diagonalizable (over $\Bbb C$) iff its minimal polynomial has no multiple root.
If $S$ is a diagonalizable square matrix of order $n$ and if $\lambda$ is an eigenvalue of $S$ with multiplicity $k$, then the rank of $\lambda I_n-S$ is $n-k$.
Let $A$ and $B$ be matrices of size $m\times n$ and $n\times m$ respectively. If $q(BA)=0$ for some polynomial $q$, then $AB\cdot q(AB)=A\cdot q(BA)\cdot B=0$.
Moreover, if $m\ge n$ additionally, then $p_{AB}(\lambda)=\lambda^{m-n}p_{BA}(\lambda)$.


In your question, as implicitly shown in user1551's answer, 
$$m_{BA}(\lambda)=(\lambda-1)(\lambda-2),\quad p_{BA}(\lambda)=(\lambda-1)^2(\lambda-2).$$
Then from 3. and 4. (and the fact that eigenvalues are roots of minimal polynomial) we know
$$m_{AB}(\lambda)=\lambda(\lambda-1)(\lambda-2),\quad p_{AB}(\lambda)=\lambda(\lambda-1)^2(\lambda-2).$$
Then from 1. and 2. we know $AB$ is diagonalizable and the rank of $I_4-AB$ is $2$. 
As a result,  the determinant of any $3\times 3$ submatrix of $I_4-AB$ is $0$, which implies that $x=0$.