# Nilpotent operator

In operator theory, a bounded operator *T* on a Hilbert space is said to be **nilpotent** if *T ^{n}* = 0 for some

*n*. It is said to be

**quasinilpotent**or

**topologically nilpotent**if its spectrum

*σ*(

*T*) = {0}.

## Examples[edit]

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

The Volterra operator is the corresponding integral operator *T* on the Hilbert space *L*^{2}(0,1) given by

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.