Question on positive square root of an operator

Question

Let $T$ is a bounded linear operator on a Hilbert Space $H$. Let there exists a unique positive square root of $T^{*}T$. Let us call it as $P$

Then how can I show that the dimension of the range of $T$ is the same as that of the range of $P$?

For this, I am trying to show that their null spaces are equal.

I started with $x \in KerT$ $\rightarrow$ $Tx=0$ $\rightarrow$ $\langle Tx,Tx\rangle=0$ $\rightarrow$ $\langle T^{*}Tx,x\rangle=0$

From here, I can't imply that $Px=0$.

Help, please

Answer 1

Hint: If $\ P\ $ is a square root of $\ T^*T\ $, then $\ T^*T=P^2\ $.

Answer 2

Here is an alternative method of proof. By the polar decomposition one may write $ T=UP, $ where $U$ is a partial isometry which also satisfies, $ U^*T=P. $

This implies that $U$ is an isometric isomorphism from $\text{Ran}(P)$ to $\text{Ran}(T)$ and, in particular, $$\text{dim}(\text{Ran}(P)) = \text{dim}(\text{Ran}(T)).$$

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PS: Thanks to user "lonza leggiera" for pointing out my initial misunderstanding of the method based on the fact that $\text{Ker}(T)=\text{Ker}(P)$. This implies that $$ \text{Ran}(T)\simeq H/\text{Ker}(T) = H/\text{Ker}(P) \simeq \text{Ran}(P), $$ thus solving the problem.

However this method only proves that the ranges are algebraically isomorphic and cannot decide, for instance, whether or not one range is closed in case the other one is known to be closed.