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standard, and even an obvious algebraic fact following from (real or complex) Hilbert space axioms. (The argument that it doesn't appear in the
Japanese WP carries little force for me; WP in any language is imperfect and ongoing as you know.) As for its being peculiar to Hilbert spaces: sure, but that's the context for the present discussion of normal operators, and anyway I note that the proof of the trick you gave above seems to depend on complex Hilbert space axioms as well (and in basic conception is not far removed from the proof of the complex polarization identity). If you'd like the last word on this matter, be my guest, but I think I'm done. Peace.
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especially since there are some nice classes of non-normal operators with interesting algebraic structure and perhaps special numerical routines in the case of non-normal matrices. It would be nice to create a separate page for non-normal operators, or at minimum a new section in this article to signpost readers to other articles on special classes of non-normal operators so they can gain some intuition.
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standardly known than numerical radius.) I understand the point that the extra generality is not of earth-shaking importance , but sometimes it's good to know in what generality statements hold true. Conceptually clear proofs often make such generality manifest. :-) Thanks for the advice on places to discuss.
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an argument in favor of mine? What's wrong with the polarization identity is that it holds only for a
Hilbert space. It is preferable to work with more general concepts like norms. Also, I disagree that the identity is standardly known. It probably depends on your educational background, but I have
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The assertion that your proof is conceptually simpler is, as I see it, mere opinion, and I disagree with it, especially since both proofs are so short. (What is this "doesn't even have to" -- what's wrong with invoking the polarization identity? That's also used all the time, and arguably even more
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redirects to this page on normal operators. However, the text "non-normal" does not appear anywhere on this page. Readers are left to re-derive the properties of non-normal operators in the negative space left by the discussion of normal operators. This is not good, from a didactic standpoint ---
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Okay, thanks. Yes, I made the observation over at my talk page that it fails for real
Hilbert spaces. I view this as an argument in favor of the proof I gave, which I thought was conceptually clear and works either in the real or complex setting. But if you won't revert back, then I don't plan to
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is a fairly standard fact that is used for all the time. Yes, your proof does work for a real
Hilbert space, but is such a generality really important? It seems normal operators are usually discussed in the context of a complex Hilbert space, probably because many things could go wrong in a real
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It's probably silly to continue arguing about this, since I've already agreed not to revert. But FWIW, I agree that taste in these matters depends on educational background, and the fact that you consider the polarization identity "obscure" tells me something about yours. To me it's completely
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Please excuse my contacting you in this way, but I am a newcomer to
Knowledge editing and am not yet familiar with proper Knowledge protocol. I am trying to understand why you felt the need to undo my (first and so far only edit) at
776:
Sorry, I missed your comment at your talkpage. It's usually not a good idea to put a comment at your talkpage since other editors are probably not paying attention to it. As for the issue at hand, actually I thought my proof was
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In full generality it's false, otherwise any nilpotent operator would be normal. Besides, I think any statement involving paranormal operators should appear in the article about paranormal operators rather than this one.
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normal" implies "N normal"? I think I heard that the answer is affirmative when N is assumed to be, say, paranormal or something. I want to know the precise statement (so we can put in the article). --
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Is this the place to be having this type of conversation in your opinion? I thought since it's a side issue, we could iron it out away from the discussion page, but I wasn't sure of the proper venue.
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long before learning about the identity. (I don't remember my teacher in my
Hilbert space course ever invoking the identity. Also, it is telling that there is no Japanese version of the
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Hilbert space. If I remember correctly, the analogous fact fails for a real
Hilbert space. Finally, please don't get discouraged. In Knowledge, edits get reverted all the time. --
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Perhaps I don't have to respond any more, but about this "Conceptually clear proofs often make such generality manifest." Right, but isn't that
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Finally, about a place to discuss. You probably don't have to worry, but this talkpage is probably a right place. --
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on
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article.) So, I think now you can imagine my reaction to your change. I thought: of course, you
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Hilbert space like a numerical radius, as you pointed out, fails to be a norm.
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It's actually not an invalid argument. It follows from the fact that a
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don't know. nice to have someone interested in operator theory around.
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use the identity, but why wouldya? Why invoke an obscure identity? --
721:. As you can see, what is essential is that we're working with a
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is a norm. In fact, let me show you the proof quickly. Suppose
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The following was moved from the talkpage of User:TakuyaMurata.
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593:{\displaystyle 0=(Tx+iy\mid x+iy)=-i(Tx\mid y)+i(Ty\mid x)}
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