How to prove $\text{Rank}(AB)\leq \min(\text{Rank}(A), \text{Rank}(B))$? How to prove $\text{Rank}(AB)\leq \min(\text{Rank}(A), \text{Rank}(B))$?
 A: Surely vectors that are in the kernel of $B$ are also in the kernel of $AB$. Vectors that are in the kernel of $A^T$ are also in the kernel of $(AB)^T=B^TA^T$ therefore with the fact that Rank($A$)=Rank($A^T$) and the knowledge that the rank gives you the size of the kernel of a matrix you are done.
A: Hint: Show that rows of $AB$ are linear combinations of rows of $B$. Transpose this hint.
A: I used a way to prove this, which I thought may not be the most concise way but it feels very intuitive to me.
The matrix $AB$ is actually a matrix that consist the linear combination of $A$ with $B$ the multipliers. So it looks like...
$$\boldsymbol{AB}=\begin{bmatrix}
 &  &  & \\ 
a_1 & a_2 & ... & a_n\\ 
 &  &  & 
\end{bmatrix}
\begin{bmatrix}
 &  &  & \\ 
b_1 & b_2 & ... & b_n\\ 
 &  &  & 
\end{bmatrix}
=
\begin{bmatrix}
 &  &  & \\ 
\boldsymbol{A}b_1 & \boldsymbol{A}b_2 & ... & \boldsymbol{A}b_n\\ 
 &  &  & 
\end{bmatrix}$$
Suppose if $B$ is singular, then when $B$, being the multipliers of $A$, will naturally obtain another singular matrix of $AB$. Similarly, if $B$ is non-singular, then $AB$ will be non-singular. Therefore, the $rank(AB) \leq rank(B)$.
Then now if $A$ is singular, then clearly, no matter what $B$ is, the $rank(AB)\leq rank(A)$. The $rank(AB)$ is immediately capped by the rank of $A$ unless the the rank of $B$ is even smaller.
Put these two ideas together, the rank of $AB$ must have been capped the rank of $A$ or $B$, which ever is smaller. Therefore, $rank(AB) \leq min(rank(A), rank(B))$.
Hope this helps you!
A: Since we have $\text{Col }AB \subseteq \text{Col }A$ and $\text{Row }AB \subseteq \text{Row }B$, therefore $\text{Rank }AB \leq \text{Rank }A$ and $\text{Rank }AB \leq \text{Rank }B$, then the result follows.
A: You know that a linear transformation cannot increase the dimension of its domain; i.e. If $T: V\rightarrow W$ is a linear transformation, 
$$\dim(T(V))\le \dim(V).$$
