Prove that $AA^+$ is the projection operator onto the column space of $A$
If $A$ has independent column vectors $A^TA$ is invertible and the projection operator onto the column space of $A$ is $P=A(A^TA)^{-1}A^T$.
It is stated that, If $A$ has linearly independent columns, then the pseudoinverse of $A$ becomes $A^+=(A^TA)^{-1}A^T$ and $AA^+=A(A^TA)^{-1}A^T$ is the projection operator on to the column space of $A$.
I know where the first statement comes from in the case for $A$ has independent column vectors, but how do I bring in the pseudoinverse into projection ?
Note: Similar question had been asked Show $AA^+$ is projection matrix onto the column space but the post failed to address the mentioned doubts.
$$ (AA^+)^2=AA^+AA^+=I\\ (AA^+)^T=(U\Sigma V^TV\Sigma^+U^T)^T=(U\Sigma \Sigma^+U^T)^T=U(\Sigma\Sigma^+)^TU^T=U(\Sigma\Sigma^+)U^T\\ =U\Sigma V^TV\Sigma^+U^T=AA^+ $$ $\implies AA^+$ is an orthogonal projector, but how do we show that it projects on to the column space of $A$ ?
Is it enough ?
If $y\in C(A)$, ie., $y=Ax$, for some $x$ $$PA=AA^+A=U\Sigma V^TV\Sigma^+ U^TU\Sigma V^T=U\Sigma\Sigma^+\Sigma V^T=U\Sigma V^T=A\\ Py=PAx=Ax=y$$
If $x\in N(A^T)$, ie., $A^Tx=0$ then $$(I-P)x=(I-AA^+)x=x-AA^+x=x-$$
A matrix $P$ is an orthogonal projection if and only if $P^2=P$ (idempotence) and $P^T=P$ (symmetry). The subspace that $P$ projects onto is its image/column space. So if we wish to show that $AA^+$ is projection onto the column space of $A$, we need to show 4 things:
You proved the first 3 of these. The fourth is a general fact: no matter what $B$ is (assuming the multiplication makes sense), $C(AB)\subset C(A)$.