Pseudoinverse of pseudoinverse of matrix A equals A: ${(A^{+})}^{+}=A$

Question

As you know, a Matrix $A^{+}\in \mathbb{R}^{m\times n}$ is called a a pseudoinverse of $A\in \mathbb{R}^{n\times m}$ if $\Vert b-A A^{+} b \Vert_2\leq\Vert b- A y\Vert_2 \forall b\in \mathbb{R}^{n} \forall y\in \mathbb{R}^m$ and $A^{+}b$ is the smallest vector by norm to do so, which is equally characterized by $\langle b- A A^{+}b,Ay\rangle=0$, since $A A^{+}$ is the orthogonal projection of $\mathbb{R}^{n}$ onto $R(A)$.

Because of it being a orthogonal projection the penrose axioms ${AA^{+}}^T=AA^{+}$ and $A A^{+}A=A$ follow. What I would like to do now is proof the other two axioms $A^{+}A A^{+}=A^{+}$ and ${A^{+} A}^T= A^{+}A$ given a Matrix and its pseudoinverse, by showing that ${(A^{+})}^{+}=A$ or in other words, that $A^{+}A$ is a projection onto $R(A^{+})=N(A)^{\perp}$. On the other hand, if those two penrose-axioms could be proven another way ${(A^{+})}^{+}=A$ would follow by the penrose-axioms.

I would appreciate any proof - especially one that is not using SVD.

Answer 1

$A(b-A^{+}Ab)=0\Rightarrow b-A^{+}Ab\in $N(A)$\Rightarrow \langle b-A^{+}Ab,A^{+}y\rangle=0$ since $A^{+}y \in R(A^{+})={N(A)}^{\perp}$