Elementary properties of absolute value operator
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$|T|$ can be shown to be bounded this way, too: $$\||T|x\|^2 = \langle |T|x,|T|x\rangle = \langle |T|^2x,x\rangle = \langle T^*Tx,x\rangle = \langle Tx,Tx\rangle = \|Tx\|^2. $$
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stackedtritones
Updated on August 01, 2022Comments
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stackedtritones over 1 year
Let $T$ be a bounded linear operator on a Hilbert space. Define $|T| = (T^*T)^{1/2}$.
I know $|T|$ is well-defined because $T^*T$ is a positive operator, so the positive square root exists.
This is probably a trivial question, but why is $|T|$ bounded and linear? I don't see how to manipulate it because it is defined as a square root.
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Frederik vom Ende almost 6 yearsOne possibility is to use the polar decomposition $T=U|T|$ where $U$ is partial isometry so $|T|=U^\dagger T$ and thus $T$ can be written as a composition of bounded linear maps.
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Jake almost 6 years$T^*T$ is positive, so by the continuous functional calculus, it has a square root. The gelfand map from $C(\sigma(T^*T))\to C^*(1,T^*T)\subset B(H)$, so $|T|$ is indeed in B(H), i.e. bounded.
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Ramanujan over 3 yearsWhy isn't the answer accepted?
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