Locally finite operator

In mathematics, a linear operator ${\displaystyle f:V\to V}${\displaystyle f:V\to V} is called locally finite if the space ${\displaystyle V}${\displaystyle V} is the union of a family of finite-dimensional ${\displaystyle f}${\displaystyle f}-invariant subspaces.^{[1] [2] :40}

In other words, there exists a family ${\displaystyle \{V_{i}\vert i\in I\}}${\displaystyle \{V_{i}\vert i\in I\}} of linear subspaces of ${\displaystyle V}${\displaystyle V}, such that we have the following:

• ${\displaystyle \bigcup _{i\in I}V_{i}=V}${\displaystyle \bigcup _{i\in I}V_{i}=V}
• ${\displaystyle (\forall i\in I)f[V_{i}]\subseteq V_{i}}${\displaystyle (\forall i\in I)f[V_{i}]\subseteq V_{i}}
• Each ${\displaystyle V_{i}}${\displaystyle V_{i}} is finite-dimensional.

An equivalent condition only requires ${\displaystyle V}${\displaystyle V} to be spanned by finite-dimensional ${\displaystyle f}${\displaystyle f}-invariant subspaces.^{[3] [4]} If ${\displaystyle V}${\displaystyle V} is also a Hilbert space, sometimes an operator is called locally finite when the sum of the ${\displaystyle \{V_{i}\vert i\in I\}}${\displaystyle \{V_{i}\vert i\in I\}} is only dense in ${\displaystyle V}${\displaystyle V}.^{[2] :78–79}

• Every linear operator on a finite-dimensional space is trivially locally finite.
• Every diagonalizable (i.e. there exists a basis of ${\displaystyle V}${\displaystyle V} whose elements are all eigenvectors of ${\displaystyle f}${\displaystyle f}) linear operator is locally finite, because it is the union of subspaces spanned by finitely many eigenvectors of ${\displaystyle f}${\displaystyle f}.
• The operator on ${\displaystyle \mathbb {C} [x]}${\displaystyle \mathbb {C} [x]}, the space of polynomials with complex coefficients, defined by ${\displaystyle T(f(x))=xf(x)}${\displaystyle T(f(x))=xf(x)}, is not locally finite; any ${\displaystyle T}${\displaystyle T}-invariant subspace is of the form ${\displaystyle \mathbb {C} [x]f_{0}(x)}${\displaystyle \mathbb {C} [x]f_{0}(x)} for some ${\displaystyle f_{0}(x)\in \mathbb {C} [x]}${\displaystyle f_{0}(x)\in \mathbb {C} [x]}, and so has infinite (or zero) dimension.
• The operator on ${\displaystyle \mathbb {C} [x]}${\displaystyle \mathbb {C} [x]} defined by ${\displaystyle T(f(x))={\frac {f(x)-f(0)}{x}}}${\displaystyle T(f(x))={\frac {f(x)-f(0)}{x}}} is locally finite; for any ${\displaystyle n}${\displaystyle n}, the polynomials of degree at most ${\displaystyle n}${\displaystyle n} form a ${\displaystyle T}${\displaystyle T}-invariant subspace.^[5]