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I'd like to know how to prove that $|T^{*}T|=|T|^2$. I know that is $\leq$ since $|T|=|T^{*}|$ but I don't know how to prove the reverse inequality. Thanks.

Julian Kuelshammer
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Algge
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    Please, try to make the title of your question more informative. E.g., Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. – Julian Kuelshammer Nov 23 '12 at 04:41
  • Here's a pointer to Dana Williams' $C^*$ script: https://math.dartmouth.edu/~dana/bookspapers/cstar.pdf#page=36 – Hanno Jan 13 '23 at 14:24

2 Answers2

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We can prove equality directly by noticing that $T^*T$ is a self adjoint operator, and thus

$$ \|T^*T\|=\sup_{\|x\|=1}\vert\langle x,T^*Tx\rangle\vert=\sup_{\|x\|=1}\vert \langle Tx,Tx\rangle\vert=\|T\|^2 $$

Norbert
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icurays1
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As an alternative to icurays1's solution, where you need to know that the first equality in his proof holds, here is a direct way:

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

Then $$ \|T\|^2=\sup\{\|Tx\|^2:\ \|x\|=1\}\leq\|T^*T\| $$

Martin Argerami
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