let $T \in B ( H )$ and $ T =U \mid T \mid$ is Polar decomposition of $ T$.
How can be proved the following conditions?
a: if $ T^{*} T$ is invertible, then $ U $ is a isometric and $ U= T ( T^{*} T ) ^{1/2}$.
b: if $ T^{*} T$ is invertible, then $ U $ is a unitary operator.
c: if $T \in B ( H )$,then for all $ x \in H $, $ \parallel \mid T \mid x \parallel = \parallel T x \parallel $.
thank for your attention.
If $T^*T$ is invertible, its inverse it positive and so it makes sense to get $(T^*T)^{-1/2}$, so $|T|$ is also invertible. Since $T=U|T|$, by multiplying on the right by $|T|^{-1}$ we get $$ U=T\,|T|^{-1}=T(T^*T)^{-1/2}. $$ Now $$ U^*U=(T^*T)^{-1/2}T^*T(T^*T)^{-1/2}=I, $$ so $U$ is an isometry. A similar computation shows that $UU^*=I$, so $U$ is a unitary.
The last equality is a straightforward computation: $$ \|\,|T|x\|^2=\langle |T|x,|T|x\rangle=\langle |T|^2x,x\rangle=\langle T^*Tx,x\rangle=\langle Tx,Tx\rangle=\|Tx\|^2. $$