$A$ and $B$ are $n \times n$ matrix. rank($A$) = $n-1$. rank($B$) = $n$. I want to show $rank(AB) = n-1$.
A solution provided to me is:
We want to find the N($AB$), the null space of $AB$. To find the N($AB$), find $ABx=0$. We can write this as $Ay=0$, where $Bx = y$. So $x = B^{-1}y$. So N($AB$) = $B^{-1}N(A)$.
So I'm confused. First and foremost, how does this show that the rank of $AB$ = $n-1$? I'm trying to get someone to explain this proof.
On
(1) Since $A$ has rank $n-1$ so there exists $y\neq 0$ : $Ay=0$. And since $B$ has rank $n$ so there exists $x\neq 0$ s.t. $Bx=y$. So $$ ABx=Ay=0$$ So $AB$ has nontrivial kernel. So ${\rm rank}\ (AB)\leq n-1$
(2) Assume that $z\neq 0$ is another kernel of $AB$ s.t. $z\neq cx$ where $$x\neq 0\in {\rm ker} \ AB$$ Then $Bz\neq 0$ but $ABz=0$ Hence $A$ has a kernel $Bz$. Since $A$ has rank $n-1$ then $$ Bz=cy$$ for some $c$. So $z=cB^{-1}y=cx$. Contradiction.
Note that for every invertible matrix $P$ and any matrix $B$ we have $$\text{rank}(PB)=\text{rank}(B)$$
Proof $$\begin{align}\text{rank}(PB)=k &\iff \dim\left\{x|PBx=0\ \ x\in \mathbb{R}^n\right\}=n-k\\ & \iff \dim\left\{x|P^{-1}(PBx)=0\ \ x\in \mathbb{R}^n\right\}=n-k\\ & \iff \dim\left\{x|Bx=0\ \ x\in \mathbb{R}^n\right\}=n-k\\ &\iff \text{rank}(B)=k \end{align}$$
Note that this follows from the equivalences: $$(PBx)=0\iff P^{-1}(PBx)=0\iff Bx=0 $$