How to establish the rank inequalities $$ \mbox{rank}(A) + \mbox{rank}(B) - n \leq \mbox{rank}(A B) \leq \min\{ \mbox{rank}(A), \mbox{rank}(B) \} \tag{1} $$ where $A$ and $B$ are real matrices of sizes $m \times n$, $n \times p$, respectively.
I checked the archives of this forum and the second inequality has been proved.
To recapitulate the arguments, first we note that $AB$ consists of vectors of the form $$ A b_1, A b_2, \ldots, A b_p $$ where $b_1, b_2, \ldots, b_p$ are the column vectors of $B$.
Thus, the column span of $A B$ is contained in the column span of $A$.
This shows that $$ \mbox{rank}(A B) \leq \mbox{rank}(A). $$
Next, we define the matrices $$ C = A^T, D = B^T $$ and note that $$ \mbox{rank}(D C) \leq \mbox{rank}(D) $$
Since a matrix $P$ and its transpose ($P^T$) have the same rank, it follows that $$ \mbox{rank}(C^T D^T) \leq \mbox{rank}(D^T) $$ or $$ \mbox{rank}(A B) \leq \mbox{rank}(B) $$
Hence, we showed that $$ \mbox{rank}(A B) \leq \min\{ \mbox{rank}(A), \mbox{rank}(B) \} $$
(This is not any new proof - I only recounted ideas presented in this forum.)
My query is how to prove the lower bound given in (1) for $\mbox{rank}(A B)$.
In other words, how to prove: $$ \mbox{rank}(A) + \mbox{rank}(B) - n \leq \mbox{rank}(A B) \tag{2} $$
I tried some examples, but I am not getting any intuition for it.
Your suggestions are welcome.
