Ancillary statistic - Revision history

en>Yobot: /* Example */WP:CHECKWIKI error fixes using AWB (10093)

2014-05-05T11:15:39Z

Example: WP:CHECKWIKI error fixes using AWB (10093)

← Older revision		Revision as of 13:15, 5 May 2014
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	~~{{Unreferenced\|date=July 2009}}~~		Not much to say about me at all.<br>Finally a member of this site.<br>I really wish I am useful at all<br><br>Feel free to visit my site ... [http://michaelherlache.com/index.php?do=/profile-5147/info/ 4Inkjets Promotion]
	~~In [[operator theory]], a bounded operator ''T'' on a [[Hilbert space]] is said~~ to ~~be '''[[nilpotent]]''' if ''T~~<~~sup~~>n<~~/sup~~>~~'' = 0 for some ''n''. It is said to be '''quasinilpotent''' or '''topological nilpotent''' if its [[spectrum (functional analysis)\|spectrum]] ''σ''(''T'') = {0}.~~

	~~==Examples==~~
	~~In the finite dimensional case, i.e. when ''T'' is a square matrix with complex entries, ''σ''(''T'') = {0} if and only if~~
	~~''T'' is similar to a matrix whose only nonzero entries are on the superdiagonal, by the [[Jordan canonical form]]. In turn this is equivalent to ''T~~<~~sup~~>n<~~/sup~~>~~'' = 0 for some ''n''~~. ~~Therefore, for matrices, quasinilpotency coincides with nilpotency~~.

	~~This is not true when ''H'' is infinite dimensional~~. ~~Consider the [~~[~~Volterra operator]], defined as follows~~: ~~consider the unit square ''X'' = [0,1] × [0,1] ⊂ '''R'''<sup>2<~~/~~sup>, with the Lebesgue measure ''m''~~. ~~On ''X'', define the (kernel) function ''K'' by~~

	~~:<math>K(x,y) =~~
	~~\left\{~~
	~~\begin{matrix}~~
	~~1, & \mbox{if} \; x \geq y\\~~
	~~0, & \mbox{otherwise}~~.
	~~\end{matrix}~~
	~~\right.~~
	~~</math>~~

	~~The Volterra operator is the corresponding [[integral operator]] ''T'' on the Hilbert space ''L''<sup>2</sup>(''X'', ''m'') given by~~

	~~:<math>T f(x)~~ = ~~\int_0 ^1 K(x,y) f(y) dy.<~~/~~math>~~

	~~The operator ''T'' is not nilpotent: take ''f'' to be the function that is 1 everywhere and direct calculation shows that~~
	~~''T<sup>n<~~/~~sup> f'' ≠ 0 (in the sense of ''L''<sup>2<~~/sup>) for all ''n''. However, ''T'' is quasinilpotent. First notice that ''K'' is in ''L''<sup>2</sup>(''X'', ''m''), therefore ''T'' is [[compact operator on Hilbert space\|compact]]. By the spectral properties of compact operators, any nonzero ''λ'' in ''σ''(''T'') is an eigenvalue. But it can be shown that ''T'' has no nonzero eigenvalues, therefore ''T'' is quasinilpotent.

	~~{{DEFAULTSORT:Nilpotent Operator}}~~
	~~[[Category:Operator theory]~~]

89.246.209.244: /* Ancillary complement */ deleted confusing qualifier "to T"

2013-07-25T14:24:43Z

Ancillary complement: deleted confusing qualifier "to T"

← Older revision		Revision as of 16:24, 25 July 2013
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			{{Unreferenced\|date=July 2009}}
			In [[operator theory]], a bounded operator ''T'' on a [[Hilbert space]] is said to be '''[[nilpotent]]''' if ''T<sup>n</sup>'' = 0 for some ''n''. It is said to be '''quasinilpotent''' or '''topological nilpotent''' if its [[spectrum (functional analysis)\|spectrum]] ''σ''(''T'') = {0}.

			==Examples==
			In the finite dimensional case, i.e. when ''T'' is a square matrix with complex entries, ''σ''(''T'') = {0} if and only if
			''T'' is similar to a matrix whose only nonzero entries are on the superdiagonal, by the [[Jordan canonical form]]. In turn this is equivalent to ''T<sup>n</sup>'' = 0 for some ''n''. Therefore, for matrices, quasinilpotency coincides with nilpotency.

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			:<math>K(x,y) =
			\left\{
			\begin{matrix}
			1, & \mbox{if} \; x \geq y\\
			0, & \mbox{otherwise}.
			\end{matrix}
			\right.
			</math>

			The Volterra operator is the corresponding [[integral operator]] ''T'' on the Hilbert space ''L''<sup>2</sup>(''X'', ''m'') given by

			:<math>T f(x) = \int_0 ^1 K(x,y) f(y) dy.</math>

			The operator ''T'' is not nilpotent: take ''f'' to be the function that is 1 everywhere and direct calculation shows that
			''T<sup>n</sup> f'' ≠ 0 (in the sense of ''L''<sup>2</sup>) for all ''n''. However, ''T'' is quasinilpotent. First notice that ''K'' is in ''L''<sup>2</sup>(''X'', ''m''), therefore ''T'' is [[compact operator on Hilbert space\|compact]]. By the spectral properties of compact operators, any nonzero ''λ'' in ''σ''(''T'') is an eigenvalue. But it can be shown that ''T'' has no nonzero eigenvalues, therefore ''T'' is quasinilpotent.

			{{DEFAULTSORT:Nilpotent Operator}}
			[[Category:Operator theory]]

en>Michael Hardy: /* Example */ Don't italicize digits in this context; see WP:MOS

2012-08-24T22:52:45Z

Example: Don't italicize digits in this context; see WP:MOS

New page

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