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$\newcommand{\rank}{\operatorname{rank}}\renewcommand\Im{\operatorname{Im}}$ 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.

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  • $\begingroup$ From what you have, it's enough to show that $$\dim\ker(AB) - \dim\ker B = \dim(\Im B \cap \ker A). $$ $\endgroup$ – leo Sep 23 '14 at 17:21
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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.

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  • $\begingroup$ could you explain the last step please?$ \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))}$ $\endgroup$ – Lucas Sep 23 '14 at 22:18
  • $\begingroup$ @Lucas note that $(AB)x = A(B(x))$. Note that $B(x)$ is in the image of $B$. So, the rank of $AB$ is the dimension of the image of the image of $B$ under $A$. Furthermore, in order for $A(B(x))$ to be $0$, $B(x)$ (which is in the image of $B$) needs to be in the kernel of $A$. $\endgroup$ – Omnomnomnom Sep 23 '14 at 22:47

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