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.
Why does $|T^{*}T|\geq |T|^2$ hold for the operator norm?
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functional-analysis
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1Please, 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
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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 $