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

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.

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