A and B are square nxn matrices and I'm asked to show that if rank(A)=rank(B)=n then rank(AB)=n
I'm aware this is likely quite simple, but I can't seem to figure how to start the proof for some reason. I would greatly appreciate any help. If you do find how, please try to do so only using very basic theorems about rank. This is a potential exam question and I'm not sure which theorems she'll allow us to just know or claim we have to show.
Also, I know it's very looked down upon to post a question that you haven't at least tried, but I genuinely don't know how to even start this. My first thought was to think about row equivalent matrices and showing that if this is true for A and B in reduced row echelon form, then it must be for the non-reduced A and B. But I'm not sure how to show that and it seems overly complex to begin with.
Note that I was taught the definition of rank to be the dimension of the row space. And also note that, through some trivial connections, rank(A)=n very clearly makes the invertible matrix theorem useful. This probably makes the problem even simpler.
The easiest way to go I to note that $\operatorname{rank} M=n\iff\det M\neq 0$ for and $n\times n$ matrix.
Now, let $M,N$ be of rank $n$. Then:
$\det(M.N)=\det(M).\det(N)\neq 0$ by hypothesis since $\operatorname{rank} M=\operatorname{rank} N=n$
Since $\det(M.N)\neq 0$ one has that $\operatorname{rank}(M.N)=n$
Hope it helps.