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# Nilpotent operator

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 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 ^{n}* = 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

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

The Volterra operator is the corresponding integral operator *T* on the Hilbert space *L*^{2}(*X*, *m*) given by

\( T f(x) = \int_0 ^1 K(x,y) f(y) dy. \)

The operator *T* is not nilpotent: take *f* to be the function that is 1 everywhere and direct calculation shows that *T ^{n} 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. 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.

The electromagnetic field of a plane wave without sources is nilpotent when it is expressed in terms of the algebra of physical space.[6]

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