Locally finite operator
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In other words, there exists a family of linear subspaces of , such that we have the following:
- Each is finite-dimensional.
- 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 .
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