Let $m,n \in \mathbb{N}$ and $A,B \in \mathbb{R}^{m\times n}$ with $\text{rank}(B) = 1$. Show that $\text{rank}(A-B) = \text{rank}(A) - 1$ if and only if there exist vectors $x \in \mathbb{R}^n$ and $y \in \mathbb{R}^m$ such that $y^{T}Ax \neq 0$ and $B = \frac{Axy^{T}A}{y^{T}Ax}$.
I don't really know how to start this problem. I know that $y^{T}Ax$ is a scalar and that $B$ may be written as an outer product of two vectors $u^{T}v$. Other than that, I am pretty lost. I imagine that alternative characterizations of rank and the rank-nullity theorem might come in handy at some point.
Since $\operatorname{rank}(A-B)<\operatorname{rank}(A)$, we have $\operatorname{nullity}(A-B)>\operatorname{nullity}(A)$. Hence there exists some vector $x\in\ker(A-B)\setminus\ker(A)$, meaning that $0\ne Ax=Bx$. If we consider the left null spaces of $A-B$ and $A$ instead, there also exists some $y$ such that $0\ne y^TA=y^TB$.
So, if we write $B=uv^T$, we have $0\ne Ax=Bx=uv^Tx=(v^Tx)u$ and likewise $0\ne y^TA=(y^Tu)v^T$. Thus $y^TAx=(y^Tu)(v^Tx)\ne0$ and $$ B=uv^T=\frac{Ax}{v^Tx}\frac{y^TA}{y^Tu} =\frac{Axy^TA}{(y^Tu)(v^Tx)} =\frac{Axy^TA}{y^TAx}. $$ The converse should be easy. Note that $\ker(A)\subsetneqq\ker(A-B)$ when $B=\frac{Axy^TA}{y^TAx}\ne0$.