Let T be a normal operator. Prove that $\|T\|^{2n}=\|TT^*\|^n$
Has it got something to do with $\|T\|=\|T^*\|$?
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Actually, you don't need $T$ to be normal. Your equality holds for any $T$.
Note first that $\|T\|^2=\|T^*T\|$: this follows from $$ \langle \|Tx\|^2=\langle Tx,Tx\rangle=\langle T^*Tx,x\rangle\leq\|T^*T\|\,\|x\|^2, $$ so $\|T\|^2\leq\|T^*T\|\leq\|T^*\|\,\|T\|$. We immediately get $\|T\|\leq\|T^*\|$. As this reasoning applies also to $T^*$, we get $\|T\|=\|T^*\|$ and $\|T\|^2=\|T^*T\|$. In particular $$ \|T\|^2=\|T^*\|^2=\|(T^*)^*T^*\|=\|TT^*\|. $$ Taking $n^{\rm th}$ powers, we get $$ \|T\|^{2n}=\|TT^*\|^n,\ \ \ n\in\mathbb N. $$
Martin Argerami
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Hint: For a normal operator use the fact that
$$\|T^{*}T\|=\|T^2\|.$$
(Since $\langle T^*Tx,x\rangle=\|Tx\|^2=\langle TT^*x,x\rangle$).
Then use induction.
voldemort
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Why is that true? – Feb 09 '14 at 18:09
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@user108605: ok- let me provide a proof by editing my answer. – voldemort Feb 09 '14 at 18:12
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1Please: MathJax is not the work of primitive cave men. I changed $< T^Tx,x>=|Tx|^2=< TT^x,x>$ to $\langle T^Tx,x\rangle=|Tx|^2=\langle TT^x,x\rangle$. ${}\qquad{}$ – Michael Hardy Feb 09 '14 at 18:19