Let $A \in \mathbb{R}^{m \times n }; A=U-V$ be a splitting such that $R(A)=R(U), N(A)=N(U)$ where $R(A),N(A)$ denote the range space and null space $A.$
Definition of Moore-Penrose inverse of a matrix
$A^\dagger y=x$ if $Ax=y, x \in R(A^T)$ and $A^\dagger y=0$ if $y \in N(A^T).$
For what vectors does $A^+ A v = v$, where $A^+$ is the Moore-Penrose pseudoinverseof $A$?
https://en.wikipedia.org/wiki/Generalized_inverse
Doubt: Let $A=U-V$ be a splitting such that $R(A)=R(U), N(A)=N(U)$ Since $R(A)=R(U) \implies R(V) \subseteq R(U)$ Thus $UU^\dagger V=V$ Hence $A=U-UU^\dagger V=U(I-U^\dagger V).$ Claim: $(I-U^\dagger V)$ is non-singular
My approach: Assume $U^\dagger V x=x;$ for some $0 \ne x \in \mathbb{R}^n$ $\implies x\in R(U^\dagger)=R(U^T)=R(A^T)---(*)$ and $UU^\dagger Vx=Ux\implies Vx-Ux=0 \implies Ax=0 \implies x\in N(A)---(**)$
From (*) and (**) we can say that $x=0$ since $x\in R(A^T)$ and $x \in N(A).$
Hence $(I-U^\dagger V)$ is non-singular.
Is this proof is correct?
Thanks in advanced.