# How to prove a property of ranks: $\operatorname{rank}(AB)= \operatorname{rank}(B)- \dim(\operatorname{Im} B \cap \ker A)$

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

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

## 1 Answer

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.

• could you explain the last step please?$\rank(B) = \underbrace{\dim(\text{Im}(A\mid_{\text{Im}(B)}))}_{\rank(AB)} +\underbrace{\dim(\ker(A \mid_{\text{Im}(B)}))}_{\dim(\ker A \cap \text{Im}(B))}$ – Lucas Sep 23 '14 at 22:18
• @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 $A$. – Omnomnomnom Sep 23 '14 at 22:47