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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.

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    Please, try to make the title of your question more informative. E.g., *Why does a imply a+c?* is much more useful for other users than *A question about inequality.* From [How can I ask a good question?](http://meta.math.stackexchange.com/a/589/): *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.*2012-11-23

2 Answers 2

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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\| $

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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 $