Prove that if rkAB = n then rkA = rkB = n I have a question that I know the answer to, but there are many different answers to this question:
Q: Suppose that A and B are $n\times n$ matrices. Prove that if rk(AB) = n then rk(A) = rk(B) = n. A: $$rk(AB) = dimC(AB) = dim im(AB) = n$$
where C(AB) is the column space of AB and AB : $R^n$ -> $R^n$ is the
linear function associated to AB. Since dim im(AB) = n, AB is onto.
But AB = A  B. This implies that A must also be onto, and so
rkA = dim im(A) = n. Since rk(A) = rk(AB) = n, both A and AB are
invertible. Then B = $A^-1(AB)$ is the product of invertible matrices, and
so is invertible. Hence rkB = n as well. Therefore rkA = rkB = n.
This is basically using the dimension theorem and onto to solve the proof, but I'd like to see some other ways that may be possible, and perhaps quicker/easier to do. Thanks
A: *

*$\det(AB)=\det(A)\det(B)$  for any two $n \times n$ square matrices $A,B$

*(I am not sure how correct this is, I just started learning group theory) The set of all invertible matrices is a group with the group operation defined as matrix multiplication. Thus $AB$ belongs to this group. Thus $A$ and $B$ should belong to this group. In other words, $(AB)^{-1}=B^{-1}A^{-1}$, thus $A$, $B$ should be invertible which otherwise would violate the group rule $(AB)^{-1}=B^{-1}A^{-1}$

A: For any two $(n \times n)$-matrices $A$ and $B$ with $\newcommand{\rk}{\operatorname{rk}} \rk AB = n$,
$$
\begin{align}
n = \rk AB &\le \rk A \le n\\
\text{and} \quad n = \rk AB &\le \rk B \le n.
\end{align}
$$
Thus, $\rk A = \rk B = n$.
