Locally finite operator

In mathematics, a linear operator $f:V\to V$ is called locally finite if the space $V$ is the union of a family of finite-dimensional $f$-invariant subspaces.

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

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

An equivalent condition only requires $V$ to be the spanned by finite-dimensional $f$-invariant subspaces. If $V$ is also a Hilbert space, sometimes an operator is called locally finite when the sum of the $\{V_{i}\vert i\in I\}$ is only dense in $V$.

Examples

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