Yesterday, I asked if that property was true :
Rangespace of Moore-Penrose pseudoinverse : $\mathcal{R}(A^+)=\mathcal{N}(A)^\perp$?
Someone found a demonstration, but he used SVD, and I wanted to see if it can be avoided, using only the definition of the MP pinverse : http://en.wikipedia.org/wiki/Moore%E2%80%93Penrose_pseudoinverse#Definition
Is the following demonstration correct ?
Let $A:E\rightarrow F$, let $y\in F$ and $x\in \mathcal{N}(A)$ : $$(Ax)^Ty=0 \\ \Rightarrow x^TA^Ty=0 \\ \Rightarrow x^T(AA^+A)^Ty=0 \\ \Rightarrow x^T(A^+A)^TA^Ty=0 \\ \Rightarrow x^T(A^+AA^Ty)=0 \\ $$
So we have $\mathcal{R}(A^+AA^T)\subset\mathcal{N}(A)^\perp$
Let's prove that $\mathcal{R}(A^+AA^T)=\mathcal{R}(A^+)$. Immediately we have $\mathcal{R}(A^+AA^T)\subset\mathcal{R}(A^+)$
Besides, $rank(A^+AA^T)=rank((AA^+A)^T)=rank(A^T)=rank(A)=rank(A^+)$, thus we prove $\mathcal{R}(A^+AA^T)=\mathcal{R}(A^+)$ and $\mathcal{R}(A^+)\subset\mathcal{N}(A)^\perp$
We get equality by using dimensionality once again since $dim(\mathcal{N}(A)^\perp)=rank(A^+)$
Is this demonstration correct ?