Does $ \mbox{rank} (A B) \leq \min \left( \mbox{rank} (A), \mbox{rank}(B) \right)$ hold with probability $1$ when matrices $A$ and $B$ are random? For two non-random matrices $A$ of size $n\times p$ and $B$ of size $p \times m$ we know that
$$ \mbox{rank} (A B) \leq \min \left( \mbox{rank} (A), \mbox{rank}(B) \right)$$
One can construct examples where there is a strict inequality (such as $A = (1,0), B=(0,1)^T$) but how likely are these to occur if every entry of matrices $A$ and $B$ is sampled independently from a zero-mean Gaussian distribution with finite variance?
To me, it seems that if the entries are random, then intuitively we should have that equality holds with probability $1$, i.e.,
$$ \mbox{rank} (A B) = \min \left( \mbox{rank} (A), \mbox{rank}(B) \right)$$
but I cannot find any proof of this. Does anyone have a proof of this or some reference to where I can read more about this?
 A: First, we can observe that with probability $1$, $\operatorname{rank}(A) = \min\{n,p\}$ and $\operatorname{rank}(B) = \min\{p,m\}$. This holds not just for independent Gaussian entries, but in a very general case.
To prove this, suppose that we generate the rows of $A$ one at a time. At any given step, if the previous rows do not yet form a basis of $\mathbb R^p$, then their span is at most a hyperplane in $\mathbb R^p$, and the probability that the next row is in that hyperplane has probability $0$. So if $n \le p$, then the rows of $A$ are linearly independent with probability $1$; the other facts needed to prove that the rank is what we want are similar.
So we want to show that with probability $1$, $\operatorname{rank}(AB) = \min\{n,p,m\}$. This goes similarly to the above proof.
The columns of $AB$ are $A\mathbf b^{(1)}, \dots, A \mathbf b^{(m)}$ where $\mathbf b^{(1)}, \dots, \mathbf b^{(m)}$ are the columns of $B$. We'll show that if these are generated one at a time, then up until they span the column space of $A$, they'll remain linearly independent. The idea is the same: the set of vectors $\mathbf x$ such that $A\mathbf x$ is in the span of $A\mathbf b^{(1)}, \dots, A \mathbf b^{(i)}$ is a subspace of $\mathbb R^p$. If $A\mathbf b^{(1)}, \dots, A \mathbf b^{(i)}$ do not yet span the entire column space of $A$, then the set is not all of $\mathbb R^p$, so it's a proper subspace. Therefore a random $\mathbf x$ lies in it with probability $0$; in particular, the probability that $A \mathbf b^{(i+1)}$ lands in the span of $A\mathbf b^{(1)}, \dots, A \mathbf b^{(i)}$ has probability $0$.
So the columns of $AB$ are either linearly independent or span the column space of $A$. In the first case, $\operatorname{rank}(AB) = m$; in the second case, $\operatorname{rank}(AB) = \operatorname{rank}(A) = \min\{n,p\}$. So we conclude that $\operatorname{rank}(AB) \ge \min\{n,p,m\}$, and since this is also an upper bound, we get what we wanted.
