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 = × ⊂ R2, 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 L2(X, m) given by
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.
Read more about this topic: Nilpotent Operator
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