Nilpotent operator

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] ⊂ R², with the Lebesgue measure m. On X, define the (kernel) function K by

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

The Volterra operator is the corresponding integral operator T on the Hilbert space L²(0,1) given by

${\displaystyle Tf(x)=\int _{0}^{1}K(x,y)f(y)dy.}$

The operator T is not nilpotent: take f to be the function that is 1 everywhere and direct calculation shows that Tⁿ f ≠ 0 (in the sense of L²) for all n. However, T is quasinilpotent. First notice that K is in L²(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.