About the relation of rank(AB), rank(A), rank(B) and the zero matrix Let $A$ be a $2 \times 4$ matrix and $B$ be a $4 \times 4$ matrix, prove that if $rank(A) = 2$ and $rank(B)=3$ then $AB \neq 0$.
I got stuck at $rank(AB) \leq 2 $
How do I continue from here?
 A: As pointed out in the comments your matrix $A$ is the wrong way round... As, for our purposes, is the inequality! (It's right, just not helpful since what we want is $\mathrm{rank}(AB)\ge1$.)
Here's an equivalent way of saying $AB=0$.

The image of $B$ is entirely contained within the kernel of $A$.

Can you see why this can't happen?
A: We know that,
$rank (AB)=rank (B)-\dim(Img(B)\cap Ker A)$
Reason: Take the Vector Space $Img(B)$ .Let T be a linear transformation on $Img(B)$ represented by the matrix $A$
Then by rank -nullity theorem we have,
$rank (T)+dim(Ker(T))=m$ where m is the dimension of $Img (B)$ and this is equal to $rank (B)$.
Now we have $rank (AB)=rank (T)$ and $Ker (T)=\{v\in Img (B)|Av=0\}=\{v\in Img (B)|v\in Ker (A)\}=Img (B)\cap Ker (A)$.
From this $rank (AB)=rank (B)-\dim(Img(B)\cap Ker A)$ follows.
Now we have in case of this problem 
$\dim(Ker (A))=2$(Reason :$\dim(Ker (A))+rank (A)=4$ and as $rank (A)=2$)
$\Rightarrow \dim(Img(B)\cap Ker A) \leq \dim(Ker (A))=2$
$rank (AB)=rank (B)-\dim(Img(B)\cap Ker A) \geq 3-2 = 1$
So $AB\ne 0$
