1

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!

Calculix
  • 3,486

1 Answers1

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, $P=T^{1/2}C^*$.

Martin Argerami
  • 217,281
  • Thanks for your reply! I see your point! If $T$ is compact, do you know why this implies that $P$ has finite rank? I can only conclude that $\sqrt{T}$ is compact. – Calculix Apr 24 '20 at 19:42
  • 1
    Yes. If $P$ has infinite rank, then its (closed) image is infinite-dimensional. So $T^{1/2}$ applies to the unit ball cannot be compact (it contains the closed unit ball of the image of $P$). Thus $T^{1/2}$ is not compact, and neither is $T$. – Martin Argerami Apr 24 '20 at 20:05
  • So I assume one cannot take a shortcut by proving that $P(H)\subset T(H)$ and conclude from this that $P$ has finite rank. In other words, does one really need to use the square root of $T$ to conclude that $P$ has finite rank? The square root feels like a detour. – Calculix Apr 24 '20 at 20:13
  • 1
    At first sight, I wouldn't know how to prove that. Note that in general, the inclusion $\operatorname{ran}T\subset \operatorname{ran}T^{1/2}$ is proper. – Martin Argerami Apr 24 '20 at 20:45
  • First of all, thanks a lot for your help! To be honest, I don't understand your argument on why $T^{1/2}$ is not compact. Why can $T^{1/2}(\text{ball}(H))$ not be compact if $P(H)$ is finite dimensional? And why does this imply that $T^{1/2}(\text{ball}(H))$ cannot be pre-compact (since this is the definition of compact operators)? – Calculix Apr 25 '20 at 14:30
  • 1
    I assume "finite" was a typo. If $P$ has infinite-dimensional range, its range is an infinite-dimensional closed subspace $PH$. Take an orthonormal basis${e_j}$ for $PH$; this is inside the unit ball of $H$. An orthonormal basis does not have a convergent subsequence. As $e_j=T^{1/2}C^*e_j$ and $C$ is a contraction, you get that ${e_j}\subset T^{1/2}(\operatorname{ball}H)$. – Martin Argerami Apr 25 '20 at 16:02