I have difficulty proving the following rank equality , which becomes the Wedderburn’s rank-one reduction formula when $k=1$ .
Let $A\in M_{m,n}(F)$ . If $X\in M_{n,k}$ and $Y\in M_{m,k}(F)$ , and if $W=Y^TAX$ is non-singular , then we have rank equality $rk(A - AXW^{-1}Y^TA) = rk(A) - rk(AXW^{-1}Y^TA)$ .
I could only obtain some information :
by full rank factorization , $rk(W)=k \iff W=Y^T(AX) \iff W=(Y^TA)X $ such that $rk(Y^T)=rk(AX)=rk(Y^TA) = rk(X) =k$ . [ For the first "iff" : This is because $C(W)\subseteq C(Y^T) \implies k \le rk(Y^T) $ but $ rk(Y^T) \le k$ . Similarly , $R(W)\subseteq R(AX)\implies k \le rk(AX)$ but $ rk(AX) \le k$ . ]
$W^{-1}$ is non-singular . Since multiplication with non-singular matrices will not change rank , apply full rank factorization again to get $rk(AX(W^{-1}Y^TA)) = k$ .
Obviously in general $rk(U-V) \neq rk(U)-rk(V)$ for any compatible matrices $U,V$ . But I don't see why the equality holds in this case .
Let $\operatorname{rank}(A)=r$ and $A=UV^T$ be a rank factorisation, where $U\in M_{m,r}(F)$ and $V\in M_{n,r}(F)$ have full column ranks. Let $B=Y^TU$ and $C=V^TX$. Then $W=BC$. Since $U$ and $V$ have full column ranks, the ranks of $$ \begin{aligned} AXW^{-1}Y^TA&=UC(BC)^{-1}BV^T\\ \text{and}\quad A-AXW^{-1}Y^TA&=U\left(I_r-C(BC)^{-1}B\right)V^T\\ \end{aligned} $$ are those of $C(BC)^{-1}B$ and $I_r-C(BC)^{-1}B$ respectively. However, as $C(BC)^{-1}B$ and $I_r-C(BC)^{-1}B$ are mutually complementary projections, their ranks must add up to $r=\operatorname{rank}(A)$. Hence the result follows.