Prove that if $A^{t}A$ is idempotent then $A^{t}=A^{+}$

Question

Let $A_{m\times{n}}$
prove that if $A^{t}A$ is idempotent, then $A^{t}=A^{+}$.

I already proved the reciprocal, but I am having some troubles with this one. I think that I have to prove that the 4 conditions of the Moore-Penrose inverse holds. I would like to know if my prove is correct.

My attempt: Because $A^{t}A$ is idempotent, then $A^{t}AA^{t}A=A^{t}A$ $\hspace{1cm}$ associating we have

$(A^{t}AA^{t})A=A^{t}A$

Then $A^{t}AA^{t}=A^{t}$

Is this correct? To prove that $AA^{t}A=A$ is analogous. And to prove that $AA^{t}$ and $A^{t}A$ is symmetric is a property already proved and I can use it.

My question is: what property guarentees that if $(A^{t}AA^{t})A=A^{t}A$ then $A^{t}AA^{t}=A^{t}$ ?

Answer 1

Let $B=AA^TA$. Then $B^TB=A^TA=A^TB$. It follows that $\|B\|_F^2=\|A\|_F^2=\langle A,B\rangle$ and in turn, $\langle A,B\rangle=\|A\|_F\|B\|_F$. Hence $A=B=AA^TA$. Take transposes on both sides, we also get $A^T=A^TAA^T$.

Answer 2

Use the following facts

1. $\mathcal{R}(A^tA)=\mathcal{R}(A)$
2. Fact 1 and that $A^tA$ is idempotent implies that $A^tA$ is a projector onto $\mathcal{R}(A)$

Hence $(A^tA)A^t=A^t$.