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.