In other words, there exists a family of linear subspaces of , such that we have the following:
Each is finite-dimensional.
An equivalent condition only requires to be the spanned by finite-dimensional -invariant subspaces.[3][4] If is also a
Hilbert space, sometimes an operator is called locally finite when the sum of the is only
dense in .[2]: 78–79
Examples
Every linear operator on a finite-dimensional space is trivially locally finite.
Every
diagonalizable (i.e. there exists a
basis of whose elements are all
eigenvectors of ) linear operator is locally finite, because it is the union of subspaces spanned by finitely many eigenvectors of .
The operator on , the space of polynomials with complex coefficients, defined by , is not locally finite; any -invariant subspace is of the form for some , and so has infinite dimension.
The operator on defined by is locally finite; for any , the polynomials of
degree at most form a -invariant subspace.[5]
^Radford, David E. (Feb 1977). "Operators on Hopf Algebras". American Journal of Mathematics. 99 (1). Johns Hopkins University Press: 139–158.
doi:
10.2307/2374012.
JSTOR2374012.
^Scherpen, Jacquelien; Verhaegen, Michel (September 1995). On the Riccati Equations of the H∞ Control Problem for Discrete Time-Varying Systems. 3rd European Control Conference (Rome, Italy).
CiteSeerX10.1.1.867.5629.