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| 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}.
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| ==Examples==
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| In the finite dimensional case, i.e. when ''T'' is a square matrix with complex entries, ''σ''(''T'') = {0} if and only if
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| ''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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| 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
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| :<math>K(x,y) =
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| \left\{
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| \begin{matrix}
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| 1, & \mbox{if} \; x \geq y\\
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| 0, & \mbox{otherwise}.
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| \end{matrix}
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| \right.
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| </math>
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| The Volterra operator is the corresponding [[integral operator]] ''T'' on the Hilbert space ''L''<sup>2</sup>(''X'', ''m'') given by
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| :<math>T f(x) = \int_0 ^1 K(x,y) f(y) dy.</math>
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| The operator ''T'' is not nilpotent: take ''f'' to be the function that is 1 everywhere and direct calculation shows that
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| ''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.
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| {{DEFAULTSORT:Nilpotent Operator}}
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| [[Category:Operator theory]]
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Not much to say about me at all.
Finally a member of this site.
I really wish I am useful at all
Feel free to visit my site ... 4Inkjets Promotion