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2 votes
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Infinite hermitian matrix

Not an answer really, but a collection of several comments. The "skew-symmetric" condition is not really natural for an operator on a complex Hilbert space, since it isn't preserved by unitary trans …
Nate Eldredge's user avatar
7 votes
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On the domains and extensions of unbounded operators

Yes, you've got it right. Given an unbounded self-adjoint operator $A$ with domain $D(A) \subset H$, using Zorn's lemma you can produce an everywhere defined operator $A'$ on $H$ which extends $A$. …
Nate Eldredge's user avatar