Let $A,B\in M_{10}(\Bbb{R})$ such that $A,B$ have rank $3,2$ respectively. If Img$B\subset$Coimg$A$,then find rank of $AB$.

Question

By definition $\operatorname{Img}B$ denotes the column space of $B$ i.e. $\operatorname{Img}B=\{Bx\in\Bbb{R}^{10}\ |\ x\in\Bbb{R}^{10}\}$ and $\operatorname{Coimg}A$ denotes the row space of $A$ or equivalently the column space of $A^T$ i.e. $\operatorname{Coimg}A=\{A^Ty\in\Bbb{R}^{10}\ |\ y\in\Bbb{R}^{10}\}$
Now, $\operatorname{Img}B\subset \operatorname{coimg}A\implies $ for any $x\in\Bbb{R}^{10}$, $\exists y\in\Bbb{R}^{10}$ such that $ Bx=A^Ty\implies ABx=AA^Ty\in\operatorname{Img}AA^T$.
Hence, $\operatorname{Img}AB\subset \operatorname{Img}AA^T$. So, $\operatorname{rank}(AB)\le\operatorname{rank}(AA^T)=\operatorname{rank}(A)=3$, which gives me nothing.
Can anyone give the answer? Thanks for the assistance in advance.

Answer 1

Hint

rank$(A)=3$ and rank$(B)=2$, implies the nullity$(B)=8$ and nullity$(A)=7$. Then nullity$(AB) \geq \text{nullity}(B)=8$. This means rank$(AB) \leq 2$. Can you proceed from here?

Added note:

Ask yourself the following: Can $\exists \, \mathbf{x} \in \mathbb{R}^{10}$ such that $ \mathbf{x} \not\in \ker{B}$ but $ \mathbf{x} \in \ker{AB}$?

If it does, then $B\mathbf{x} \perp \text{coimage}(A)$. But we are given that $\text{Image}(B) \subset \text{Coimage}(A)$. Thus we can conclude that $B\mathbf{x}$ has to be the $\mathbf{0}$ vector. This means $ \mathbf{x} \in \ker{B}$, a contradiction. Thus nullity$(AB) \not> 8$. So rank$(AB) \geq 2$.