Suppose now I have two matrices $A$ and $B$ which are of size $m\times n$ and $n\times l$ respectively. For simplicity, assume $n<m<l$.
Assume that I have $\text{rank}(AB) = \text{rank}(A) \ne \text{rank}(B)$. Then can I conclude that the matrix $B$ will have full row rank $n$?
I know there is a special case that when $\text{rank}(A) = \text{rank}(B)$, for example the simplest case that the matrices are zero matrices, then the conclusion is not correct. So I would exclude this case. I am wondering that if I exclude the special case, is the conclusion above correct?
Best Answer
It implies $rank(B)$ is at least $rank(A)$. Since the rank of the matrix is the dimension of its image, then $\dim A(B(\mathbb{R}^\ell)) \leq \dim B(\mathbb{R}^\ell)$, hence $rank(AB) \leq rank(B)$.
As noted in the comments, you cannot deduce that $B$ is full rank.
Consider $$A = \begin{pmatrix} 1 & 0 & 0\end{pmatrix}, B = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix}$$
Then $$AB = \begin{pmatrix} 1 & 0 & 0 \end{pmatrix}.$$
Hence $AB = A$, in particular $Rank(AB) = Rank(A)$. But $Rank(B) = 2 <3$.
Edit, another example:
For some reason, we want to consider $A$ $m \times n$ and $B$ $n \times \ell$ will $n < m < \ell$.
Let $$A = \begin{pmatrix} 0 & 0 \\ 0 & 0 \\ 0& 0 \end{pmatrix}, B = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 &0 \end{pmatrix}$$
Then $A$ is rank 0, $AB$ rank 0, $B$ rank 1 < 2.