I am attempting to show that C(I − A+A) = N(A)
The column space of the quantity above equals the null space of A, where I is the identity matrix, A is an n x m matrix and A+ is the Moore-Penrose g-inverse of A.
I have been attempting to change the left side around using identities and the conditions of a generalized inverse. Can I maybe use A+A = I? If I use that then I just have C(0) = N(A). I mean, I know the zero vector is always in the null space but that seems trivially easy.
Is there some other direction I should be going with this?
Claim 1: $A^+A$ is an orthogonal projection
Proof: It suffices to note that $P = A^+A$ is symmetric (which is given as a property of the MP inverse) and show that $P^2 = P$. Indeed, we see that $$ P^2 = A^+AA^+A = A^+(AA^+A) = A^+A = P. \qquad \square $$
Claim 2: The row space of $P = A^+A$ is equal to the row space of $A$ (or equivalently, the null space of $P$ is equal to the null space of $A$)
Proof: The row space of $PQ$ is always a subspace of the row space of $Q$. Thus, the row space of $P = A^+A$ is a subspace of the row space of $A$ and the row space of $A = AP$ is a subspace of the row space of $P$. So, the two row spaces are equal. $\hspace{1 cm}\square$
From there, it suffices to note that the null space is the orthogonal complement of the row space. Thus, $I - P$ must be the projection onto the null space of $A$.