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 $\$ of linear subspaces of $V$, such that we have the following:
* $\bigcup_ V_i=V$
* (\forall i\in I) f _isubseteq 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 $\$ 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 /math>, 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 _0(x) for some f_0(x)\in\mathbb /math>, and so has infinite dimension.
* The operator on \mathbb /math> defined by $T(f(x))=\frac$ is locally finite; for any $n$, the polynomials of degree at most $n$ form a $T$-invariant subspace.Joppy
(Apr 28, 2018)
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Locally Finite Operator
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References

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