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. It is said to be quasinilpotent or topologically nilpotent if its spectrum ÃÂ(T) = {0}.
In the finite-dimensional case, i.e. when T is a square matrix (Nilpotent matrix) with complex entries, ÃÂ(T) = {0} if and only if T is similar to a matrix whose only nonzero entries are on the superdiagonal(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] × [0,1] â R<sup>2</sup>, 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<sup>2</sup>(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<sup>n</sup> f â 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. By the spectral properties of compact operators, any nonzero û in ÃÂ(T) is an eigenvalue. But it can be shown that T has no nonzero eigenvalues, therefore T is quasinilpotent.