Let $A$ a m-by-n matrix. $A^+$ its MP pseudoinverse.
Is $\mathcal{R}(A^+)$ the orthogonal complement of $\mathcal{N}(A)$ ?
If yes, how do you prove it ?
I don't think the $AA^+A=A$ and $A^+AA^+=A^+$ properties are enough, since you can find a application mapping $\mathcal{R}(A)$ to a non-orthogonal complement of $\mathcal{N}(A)$ which also have those properties.
So if it's true, I guess the fact that $AA^+$ is hermitian must be used, but I'm not sure to see how.
$$ x=A^+Ax+(x-A^+Ax) $$ where $$A(x-A^+Ax)=0 $$ so the space is the sum of $\mathcal{R}(A^+)$ and $\mathcal{N}(A)$.
The ortogonality of the two spaces is also easy.
If $A=UDV$ is his SVD decomposition, then $$v\in \mathcal{N}(A)\iff UDVv=0\iff D(Vv)=0\iff Vv=\begin{pmatrix} 0\\*\end{pmatrix}$$ We have also $A^+=V^tD^+U^t$, so, given any $v\in \mathcal{N}(A)$, and any $w=A^+z\in \mathcal{R}(A^+)$, we have $$ w^tv=z^t(A^+)^tv=z^tUD^+Vv=0$$ thus proving the ortogonality