Talk:Hermitian adjoint

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Shouldn't one use inner product notation here rather than bra-ket notation?

I.e. \langle Ax,y\rangle rather than \langle Ax | y\rangle.

PJ.de.Bruin 00:49, 5 Jul 2004 (UTC)

at least one concrete example would be nice! otherwise it seems A*=A^{-1} as in A*A=AA*=I... anyone?

[edit] Hermiticity & Self-Adjointness: Distinction

The distinction is not made clear. A is self-adjoint if:

\displaystyle\begin{array}{rl} A^* &= A, \text{ and}\\ \text{dom}(A^*) &= \text{dom}(A). \end{array}

Hermiticity does not necessarily guarantee the latter statement.

Dirc 20:52, 14 February 2007 (UTC)

Hermiticity and self-adjoitness are synonymous. (the notation A* = A implicitly implies Dom(A) = Dom(A*), usually.) so doesn't quite make sense to say self-adjointness is not guaranteed by Hermiticity. on the other hand, being symmetric does not imply an operator is self-adjoint/Hermitian in general. Mct mht 23:48, 14 February 2007 (UTC)

[edit] Proof?

Hmm, the article says:

Moreover, \| A^* A \| _{op} = \| A \| _{op}^2

I don't really see how this follows from the properties above it. Can someone provide a simple proof for this?

Never mind, think I found it. How dumb can you be?
A * A is Hermitian, as one can easily show. Therefore
\displaystyle\begin{array}{rl} ||T^* T|| & = \sup_{||x|| = 1} |\langle T^* T x, x \rangle| = \sup_{||x|| = 1} |\langle T x, T x \rangle| \\ & = \sup_{||x|| = 1} ||T x||^2 = \left( \sup_{||x|| = 1} ||T x|| \right)^2 = ||T||^2 \end{array}
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