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Does "${\displaystyle N^{2}}$ 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). -- Taku ( talk) 23:52, 11 February 2009 (UTC) reply

don't know. nice to have someone interested in operator theory around. Mct mht ( talk) 03:18, 12 February 2009 (UTC) reply

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. Gamesou ( talk) 11:07, 3 April 2009 (UTC) reply

The following was moved from the talkpage of User:TakuyaMurata.

Please excuse my contacting you in this way, but I am a newcomer to Wikipedia editing and am not yet familiar with proper Wikipedia protocol. I am trying to understand why you felt the need to undo my (first and so far only edit) at Normal operator, which was made in response to what I believed was an invalid argument, claiming that ${\displaystyle \langle Sx,x\rangle =\langle Tx,x\rangle }$ implies ${\displaystyle S=T}$. I can explain further at my talk page. Thanks. Duffifi ( talk) 14:38, 7 August 2009 (UTC)Duffifi reply

It's actually not an invalid argument. It follows from the fact that a numerical radius is a norm. In fact, let me show you the proof quickly. Suppose ${\displaystyle (Tx\mid x)=0}$ for all x. In particular,

${\displaystyle 0=(Tx+y\mid x+y)=(Tx\mid y)+(Ty\mid x)}$

${\displaystyle 0=(Tx+iy\mid x+iy)=-i(Tx\mid y)+i(Ty\mid x)}$

Combing the two we get:

${\displaystyle (Tx\mid y)=0}$

for all x and y. Take ${\displaystyle y=Tx}$ and we get ${\displaystyle Tx=0}$ for all x. As you can see, what is essential is that we're working with a complex Hilbert space. If I remember correctly, the analogous fact fails for a real Hilbert space. Finally, please don't get discouraged. In Wikipedia, edits get reverted all the time. -- Taku ( talk) 18:06, 7 August 2009 (UTC) reply

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 press the point.

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. Duffifi ( talk) 22:51, 7 August 2009 (UTC)Duffifi reply

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 conceptually simpler. It doesn't even have to use the polarization identity, and I thought ${\displaystyle \langle Tx,x\rangle =0}$ implies ${\displaystyle T=0}$ 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 Hilbert space like a numerical radius, as you pointed out, fails to be a norm.

Finally, about a place to discuss. You probably don't have to worry, but this talkpage is probably a right place. -- Taku ( talk) 01:31, 8 August 2009 (UTC) reply

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 standardly known than numerical radius.) I understand the point that the extra generality is not of earth-shaking importance [which is why I won't insist on reverting back], 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. Duffifi ( talk) 14:50, 8 August 2009 (UTC)Duffifi reply

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 exactly 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 been using a trick that shows ${\displaystyle (Tx|x)=(Sx|x)}$ to show ${\displaystyle T=S}$ 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 polarization identity article.) So, I think now you can imagine my reaction to your change. I thought: of course, you can use the identity, but why wouldya? Why invoke an obscure identity? -- Taku ( talk) 22:53, 8 August 2009 (UTC) reply

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 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. Duffifi ( talk) 20:02, 15 August 2009 (UTC)Duffifi reply