How can I show rank $( AB)$ = rank $( A)$? 
Can you help me how to show rank $( AB)$ = rank $( A)$ iff null$(A)$ $\cap $ range$(B) = \{0\} $?

I can understand that rank $(AB)$ would be no greater than rank A. But not sure how to show this by using null(A) $\cap $ range$(B) = \{0\}$. 
 A: The statement is not true as given. In fact, both implications are false.
($\!\implies\!$) Counterexample: take $A=0$ and $B=I$. Then $\mathrm{rank}(AB)=0=\mathrm{rank}(A)$, but $\mathrm{null}(A)=\mathrm{range}(B)$ (with both equalling the whole vector space).
($\!\impliedby\!$) Counterexample: take $A=I$ and $B=0$. Then $\mathrm{null}(A)=\mathrm{range}(B)=\{\vec{0}\}$, but $\mathrm{rank}(A)\neq\mathrm{rank}(AB)$ (the former has full rank while the latter has rank $0$).
A: Think of a matrix $M$ as (the datum of) a linear map $f_M$. The rank is then the dimension of the image.
Since the product of matrices correspond to the composition of functions, in order to have the dimension of the image of $f_{AB}=f_A\circ f_B$ equal to the dimension of the image of $f_B$ you need that $f_A$ restricted to the image of $f_B$ be injective, i.e. that no element in the kernel of $f_A$ belongs to the image of $f_B$.
In order to have the rank of $AB$ equal to rank of $A$ the condition is similar but somewhat more complicated.
