$\newcommand{\rank}{\operatorname{rank}}\renewcommand\Im{\operatorname{Im}}$ Let $A$ and $B$ be real matrices of sizes $m\times n$ and $n\times p$, respectively.
I have to prove that $\rank(AB)= \rank(B)- \dim(\Im B \cap \ker A)$
I haven't got much idea... but I started like this:
Using the first isomorphism theorem, we get the following relations:
$p= \rank(AB)+ \dim(\ker(AB))$
$p= \rank(B)+\dim(\ker B)$ and
$n= \rank(A)+\dim(\ker A)$
From the first and second relation we get that: $$\rank(AB)+ \dim(\ker(AB)) = \rank(B) + \dim(\ker B)$$
I don't know how to continue or if I am on the right way to prove it.
Thank you for your time and help. And sorry for my poor English.
Hint: we can be a bit more specific with the first isomorphism theorem. Certainly, we have$\newcommand{\rank}{\operatorname{rank}}$ $$ p= \rank(AB)+ \dim(\ker(AB)) $$ However, we can also think of $A$ and $B$ as maps $$ T_B : \Bbb R^p \to \text{Im}(B)\\ T_A: \text{Im}(B) \to \text{Im}(AB) $$ Where $T_{AB}(x) = ABx = T_A \circ T_B$. Now we have $$ p= \rank(B)+\dim(\ker B)\\ \rank(B) = \underbrace{\dim(\text{Im}(A\mid_{\text{Im}(B)}))}_{\rank(AB)} +\underbrace{\dim(\ker(A \mid_{\text{Im}(B)}))}_{\dim(\ker A \cap \text{Im}(B))} $$ This should be enough to get what you're looking for.