Rank of the product of two full rank matrices

Question

I have searched for the above topic and found some results, but the answer I am looking for is not found anywhere. Here is my question:

Given $A_{m \times n}$ matrix with rank $m$, and $B_{n \times p}$ matrix with rank $p$, where $n > p \geq m$. I know that $$ \operatorname{rank}(AB) \leq \min\left(\operatorname{rank}(A),\operatorname{rank}(B)\right) $$ What I want to know is if this expression holds for equality. I.e is this expression $$ \operatorname{rank}(AB) = \min\left(\operatorname{rank}(A),\operatorname{rank}(B)\right) = m $$ correct? If yes, how can it be proved?

Answer 1

if we take A and B be 2 non-zero matrices s.t AB= zero matrix then rank A,rank B >0 but rank of AB = 0 so rank AB is not equal to min{rank A, rank B}

Answer 2

(This is a reply to an earlier version of the question with $rank(A)=n$, $rank(B)=p$).

Yes, it holds. The assuptions imply $m\geq n\geq p$ and so you can find a submatrix $A'$ of $A$ that contains $n$ rows of $A$ and hence has size $n\times n$ and is regular. The columns of $B$ are independent and so are the columns of $A'B$. So indeed, $$\mathrm{rank} AB=\mathrm{rank} A'B=p=\min\{n,p\}.$$

But as Rupsa pointed out, if the matrices $A,B$ have smaller then the full column rank, then equality doesn't hold in general, such as in $(1,0){0\choose 1}=(0)$.