# Surjective (orthogonal) Projection

Let T be a compact bounded operator on a Hilbert space H. Suppose that P is a (orthogonal) projection with $$P \in B(H)$$ and $$P(H) \subseteq T(H)$$. I am trying to show that $$PT: H \to P(H)$$ is surjective. Clearly $$Image(PT) \subseteq P(H)$$. So all we have to show is the reverse inclusion.

So far I have this: let $$y \in P(H)$$ be given. Then there exists an $$x \in H$$ so that $$Px = y$$. Since P is a (orthogonal) projection, $$PPx = Px = y$$. What I am inclined to try to show is that $$PTx = PPx$$, however I am not sure how to show this. So, am I on the right track with wanting to show $$PTx = PPx$$ or do I need to approach this another way? Any direction or help will be appreciated.

We know that $$P(H) \subseteq T(H)$$. Thus, for every $$z\in P(H)$$ there exists $$x,y\in H$$ such that $$Py=z= Tx$$. Applying $$P$$ on both sides (and using that $$P$$ is a projection, i.e. $$P^2 =P$$) yiels $$z= Py = PPy =PTx \in PT(H).$$ As $$z\in P(H)$$ was arbitrary, we get $$P(H) \subseteq PT(H)$$.