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‎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.

  • 1
    In (a), you probably mean $(T^*T)^{-1/2}$ instead of $+1/2$. – 2017-01-13

1 Answers 1

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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: $$ \|\,|T|x\|^2=\langle |T|x,|T|x\rangle=\langle |T|^2x,x\rangle=\langle T^*Tx,x\rangle=\langle Tx,Tx\rangle=\|Tx\|^2. $$