What conditions are needed to raise a linear operator to some power?

  • I
  • Thread starter fxdung
  • Start date
  • #1
323
15
Each operator has a domain, so for a power of an operator to exist, the domain of the operator must remain invariant under the operation.

Is that correct?

mentor note: edited for future clarity
 
Last edited by a moderator:

Answers and Replies

  • #3
14,756
12,138
Each operator has a domain, so for a power of an operator to exist, the domain of the operator must remain invariant under the operation.

Is that correct?

mentor note: edited for future clarity
The range must be included in the domain, simply to allow a consecutive application.
 
  • #4
S.G. Janssens
Science Advisor
Education Advisor
956
725
The range must be included in the domain, simply to allow a consecutive application.
This is the wrong approach if ##D(A)## is a proper subspace of a vector space ##X##.

For example, take ##X = C[0,1]## and ##D(A) = C^1[0,1]## and ##A : D(A) \to X## differentiation. Then the range of ##A## is not contained in its domain, but ##A^2## is well-defined with
$$
D(A^2) = C^2[0,1] = \{f \in D(A)\,:\,Af \in D(A)\}.
$$
(This also suggests the general definition.) This is important if one wants to talk about the generalized eigenspace of an unbounded operator, which is defined in terms of positive integer powers of ##\lambda I - A##, where ##I## is the identity on ##X## and ##\lambda## is an eigenvalue.
 
  • Informative
Likes Keith_McClary
  • #5
WWGD
Science Advisor
Gold Member
5,487
3,918
Look up the Borel Functional Calculus. It formalizes the general idea of applying general Mathematical operations to linear operators.
 
  • Informative
Likes Keith_McClary

Related Threads on What conditions are needed to raise a linear operator to some power?

Replies
2
Views
3K
Replies
2
Views
2K
Replies
7
Views
15K
Replies
15
Views
14K
Replies
1
Views
3K
  • Last Post
Replies
1
Views
536
Replies
3
Views
2K
Replies
2
Views
4K
Replies
2
Views
4K
Top