Ancillary statistic

2013-07-25T14:24:43Z

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

{{Unreferenced|date=July 2009}}
In [[operator theory]], a bounded operator ''T'' on a [[Hilbert space]] is said to be '''[[nilpotent]]''' if ''Tn'' = 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 ''Tn'' = 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'''2, 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''2(''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
''Tn f'' ≠ 0 (in the sense of ''L''2) for all ''n''. However, ''T'' is quasinilpotent. First notice that ''K'' is in ''L''2(''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]]

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