Editing Nilpotent operator

In [[operator theory]], a [[bounded operator]] ''T'' on a [[Banach space]] is said to be '''[[nilpotent]]''' if ''T<sup>n</sup>'' = 0 for some positive integer ''n''.<ref>{{cite book|title=Introductory Functional Analysis with Applications|last=Kreyszig|first=Erwin|publisher=Wiley|year=1989|pages=393|chapter=Spectral Theory in Normed Spaces 7.5 Use of Complex Analysis in Spectral Theory, Problem 1. (Nilpotent operator)}}</ref> It is said to be '''quasinilpotent'''  or '''topologically  nilpotent''' if its [[spectrum (functional analysis)|spectrum]] ''σ''(''T'') = {0}.

==Examples==
In the finite-dimensional case, i.e. when ''T'' is a square matrix ([[Nilpotent 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<ref>
{{cite book
| title = Linear Algebra Done Right
| first = Sheldon
| last = Axler
| author-link = Sheldon Axler
| section = Nilpotent Operator
| section-url = https://linear.axler.net/NilpotentOperators.pdf
}}
</ref>(this fact is used to prove the existence of [[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] &times; [0,1] ⊂ '''R'''<sup>2</sup>, with the [[Lebesgue measure]] ''m''. On ''X'', define the [[Integral kernel|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>(0,1) 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.

==References==
{{Reflist}}

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