What does $\text{rank}(AB) = \text{rank}(A)$ imply? Suppose now I have two matrices $A$ and $B$ which are of size $m\times n$ and $n\times l$ respectively. For simplicity, assume $n&lt;m&lt;l$.

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?
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 $$A = \begin{pmatrix} 1 &amp; 0 &amp; 0\end{pmatrix}, B = \begin{pmatrix} 1 &amp; 0 &amp; 0 \\ 0 &amp; 1 &amp; 0 \\ 0 &amp; 0 &amp; 0 \end{pmatrix}$$

Then $$AB = \begin{pmatrix} 1 &amp; 0 &amp; 0 \end{pmatrix}.$$

Hence $AB = A$, in particular $Rank(AB) = Rank(A)$. But $Rank(B) = 2 &lt;3$.

Edit, another example:

For some reason, we want to consider $A$ $m \times n$ and $B$ $n \times \ell$ will $n &lt; m &lt; \ell$.

Let $$A = \begin{pmatrix} 0 &amp; 0 \\ 0 &amp; 0 \\ 0&amp; 0 \end{pmatrix}, B = \begin{pmatrix} 1 &amp; 0 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 0 &amp;0 \end{pmatrix}$$ 

Then $A$ is rank 0, $AB$ rank 0, $B$ rank 1 &lt; 2.
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 $$A = \begin{pmatrix} 1 &amp; 0 &amp; 0\end{pmatrix}, B = \begin{pmatrix} 1 &amp; 0 &amp; 0 \\ 0 &amp; 1 &amp; 0 \\ 0 &amp; 0 &amp; 0 \end{pmatrix}$$

Then $$AB = \begin{pmatrix} 1 &amp; 0 &amp; 0 \end{pmatrix}.$$

Hence $AB = A$, in particular $Rank(AB) = Rank(A)$. But $Rank(B) = 2 &lt;3$.

Edit, another example:

For some reason, we want to consider $A$ $m \times n$ and $B$ $n \times \ell$ will $n &lt; m &lt; \ell$.

Let $$A = \begin{pmatrix} 0 &amp; 0 \\ 0 &amp; 0 \\ 0&amp; 0 \end{pmatrix}, B = \begin{pmatrix} 1 &amp; 0 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 0 &amp;0 \end{pmatrix}$$ 

Then $A$ is rank 0, $AB$ rank 0, $B$ rank 1 &lt; 2.