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In operator theory, a bounded operator T on a Banach space is said to be nilpotent if T = 0 for some positive integer n. It is said to be quasinilpotent or topologically nilpotent if its spectrum σ(T) = {0}.

Examples

In the finite-dimensional case, i.e. when T is a square matrix (Nilpotent 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(this fact is used to prove the existence of Jordan canonical form). In turn this is equivalent to T = 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 = × ⊂ R, 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 L(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 T f ≠ 0 (in the sense of L) for all n. However, T is quasinilpotent. First notice that K is in L(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.

References

1. Kreyszig, Erwin (1989). "Spectral Theory in Normed Spaces 7.5 Use of Complex Analysis in Spectral Theory, Problem 1. (Nilpotent operator)". Introductory Functional Analysis with Applications. Wiley. p. 393.
2. Axler, Sheldon. "Nilpotent Operator" (PDF). Linear Algebra Done Right.

• Operator theory