If $P$ is a projection and $T$ is positive operator such that $P\leq T$, then why $P(H)\subset\sqrt{T}(H)$?

Question

Suppose that $H$ is a Hilbert space. If $P\colon H\to H$ is a projection (i.e. $P^{2}=P^{*}=P$) and $T\colon H\to H$ is positive operator (i.e. $T\geq0$) such that $P\leq T$, then why $P(H)\subset\sqrt{T}(H)$? Here $\leq$ denotes the usual partial order on self-adjoint operators. I know that this result is true if $T$ is a projection, but I don't see the general case. I see that $P=\sqrt{P}\leq\sqrt{T}$ though. Any help would be greatly appreciated!

Answer 1

It's easy to prove that if $0\leq A\leq B$, then there exists a contraction $C$ such that $A^{1/2}=CB^{1/2}$.

So in your case, as $P^{1/2}=P$, you have $P=CT^{1/2}$. Taking adjoints, $P=T^{1/2}C^*$.