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Given a complex inner product space X, and an operator $T: X \rightarrow X$ is normal i.e. $T^*T=TT^*$ How can we show $\| T^2\| = \|T\|^2$?

By the definition of operator norm, it follows that ||T|| = sup $\frac{||Tx||}{||x||}$ and ||T$^2$|| = sup $\frac{||T^2x||}{||x||}$. Then I can express the numerator as a form of inner product. But I still am not able to make these two equal. Any good ideas?

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If $T$ is normal, then $\|Tx\|^2=\left<Tx,Tx\right>=\left<x,T^*Tx\right> =\left<x,TT^*x\right>=\|T^*x\|^2$, so $\|Tx\|=\|T^*x\|$ (and therefore $\|T\|=\|T^*\|$). Then (replacing $x$ by $Tx$) $\|T^2x\|=\|T^*Tx\|$ so that $\|T^2\|=\|T^*T\|$. But also $\|Tx\|^2=\left<x,T^*Tx\right>\le\|T^*T\|\|x\|^2$ so that $\|T\|^2\le\|T^*T\| =\|T^2\|$. But $\|T^2\|\le\|T\|^2$. We conclude that $\|T^2\|=\|T\|^2$ whenever $T$ is normal.

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  • $\begingroup$ Thanks! By the way, how can we get (i) ||Tx|| = ||T*x|| implies that ||T|| = ||T*||? $\endgroup$ – Jonny Apr 23 at 4:54
  • $\begingroup$ (ii) why can we have <x, T$^*$Tx> $\leq$ ||T$^*$T|| ||x||$^2$? $\endgroup$ – Jonny Apr 23 at 4:56
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    $\begingroup$ By the definition: $\|T\|=\sup_{\|x\|=1}\|Tx\|$. @Eric $\endgroup$ – Lord Shark the Unknown Apr 23 at 4:56
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    $\begingroup$ @eric $|\left<x,Ax\right>|\le\|x\|\|Ax\|\le\|A\|\|x\|^2$. $\endgroup$ – Lord Shark the Unknown Apr 23 at 4:58

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