$R(A)\subset R(B)$ is equivalent to $A=BB^\dagger A$.

Question

Let $A,B\in \mathcal B(\mathcal H,\mathcal K),$ the space of linear maps from a Hilbert space $\mathcal H$ to $\mathcal K$. Let $A^\dagger$ and $B^\dagger$ be the moore-Penrose inverse of $A$. Then,

$R(A)\subset R(B)$ is equivalent to $A=BB^\dagger A$.

$(\Leftarrow)$ is easy by observing that,

$$R(A)=R(BB^\dagger A)\subset R(BB^\dagger)\subset R(B)$$

But for the forward implication, I dont know how to prove.

Thanks in advance!

Answer 1

$(\implies)$

Let $x\in \mathcal H$,

Now, \begin{align} BB^\dagger A x & = BB^\dagger By\quad (\because Ax\in R(A)\subset R(B), \text{ there exists } y\in \mathcal H\, s.t\, Ax=By)\\ & =B y \quad (\because B^\dagger \text{ is moore-Penrose inverse of } A)\\ & =Ax \end{align}

Hence, $A=BB^\dagger A$.