IF $||T||\leq 1$ Then, $T(I-T^{*}T)^{1/2}$ = $(I-TT^{*})^{1/2}T$

Question

Let $T$ be a bounded linear operator on a hilbert space $H$ over complex numbers with $||T||\leq 1$. Then I need to show that $T(I-T^{*}T)^{1/2}$ = $(I-TT^{*})^{1/2}T$. I was able to show that equality holds when $h$ $\in$ $Ker((I-T^{*}T)^{1/2}$ and $h \in$ $Ker(I-TT^{*})^{1/2}$. Any ideas how to go ahead and prove this in general?

Answer 1

The original version of the question asked to show that $T(I-T^{*}T)^{1/2}=T(I-TT^{*})^{1/2}$. There's a counterexample to that below.

For the current version: Note first that by continuity we may assume that $||T||<1$. Now the function $(1-z)^{1/2}$ is holomorphic in the unit disk, so it has a power series $$(1-z)^{1/2}=\sum c_nz^n\quad(|z|<1).$$Hence $$(I-T^*T)^{1/2}=\sum c_n(T^*T)^n$$and $$(I-TT^*)^{1/2}=\sum c_n(TT^*)^n.$$The result follows since $$T(T^*T)^n=(TT^*)^nT.$$

Having already written up the counterexample to the original version we may as well include it: Let $T$ be the right shift on $\ell_2(\Bbb N)$: $$Tx=(0,x_1,x_2,\dots).$$Then $$T^*x=(x_2,x_3,\dots).$$So $T^*T=I$, hence $$T(I-T^*T)^{1/2}=0.$$But $$TT^*x=(0,x_2,x_3,\dots),$$so $$(I-TT^*)x=(x_1,0,0,\dots),$$so$$(I-TT^*)^{1/2}x=(x_1,0,0,\dots),$$hence $$T(I-TT^*)^{1/2}x=(0,x_1,0,0,\dots).$$