Invertibility of a matrix A=BC Let $A\in K^{m\times n}$ be a matrix such that $A=BC$, where $B\in K^{m\times l}, C\in K^{l\times n}$ and $K$ is a field.
I know that $A$ is invertible iff $B$ and $C$ are invertible (of course it depends on $m$,$n$ and $l$). Are surjectivity of $B$ and injectivity of $C$ sufficient conditions for the invertibility of $A$? In other words, is $\ker(A)=\ker(BC)$ and $\text{im}(A)=\text{im}(BC)$ true in general?
I would say so, because if $A=BC$ there aren't any other possibilities, but I wanted to be sure.
 A: An invertible matrix is necessarily square, so you must have $m=n$; the necessary and sufficient condition for invertibility of $A$ is that $\operatorname{rank}A=m$. Now we have three cases


*

*$l<m$

*$l=m$

*$l>m$


The first case is easily dismissed, because $\operatorname{rank}BC\le\min\{\operatorname{rank}B,\operatorname{rank}C\}$ and both ranks cannot exceed $l$, so $\operatorname{rank}A<m$ and $A$ is not invertible.
In the second case the above formula for the rank of a product and the fact that $A$ is invertible, tells us that $\operatorname{rank}B=\operatorname{rank}C=m$, so both matrices are invertible. Conversely, if both $B$ and $C$ are invertible, then $A$ is invertible.
For the third case the condition that both matrices have rank $m$ is necessary, but not sufficient. For instance, with $m=n=1$ and $l=2$, we can take
$$
B=\begin{bmatrix}1 & 1\end{bmatrix},
\qquad
C=\begin{bmatrix}1 \\ -1\end{bmatrix}
$$
so both matrices have rank $1$, but their product is the $1\times 1$ null matrix.
Conversely, if we consider
$$
B=\begin{bmatrix}1 & 0\end{bmatrix},\qquad
C=\begin{bmatrix}1\\0\end{bmatrix}
$$
we get that $BC$ is invertible. More generally if you have two invertible square $m\times m$ matrices $B_1$ and $C_1$, add to $B_1$ $l-m$ null columns at the right getting the $m\times l$ matrix $B$; similarly add to $C_1$ $l-m$ null rows at the bottom, getting the $l\times m$ matrix $C$. Then $BC=B_1C_1$ is again invertible.
