# Locally algebraic operator

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A linear operator such that for each element of the space under consideration there exists a polynomial in this operator (with scalar coefficients) annihilating this element.

Let be a linear space over a field . Let be the set of all linear operators with domains and ranges in and let Denote by the algebra of all polynomials in the variable and with coefficients in . Usually, in applications it is assumed that the field is of characteristic zero and algebraically closed (cf. also Algebraically closed field; Characteristic of a field).

Thus, a linear operator is said to be locally algebraic if for any there exists a non-zero polynomial such that (cf. [a1]). If there exists a non-zero polynomial such that for every , then is said to be algebraic (cf. Algebraic operator). Thus, an algebraic operator is locally algebraic, but not conversely.

A continuous locally algebraic operator acting in a complete linear metric space is algebraic (cf. [a4]; for Banach spaces, see [a1]).

A locally algebraic operator acting in a complete linear metric space (over the field of complex numbers) and that is right invertible (cf. Algebraic analysis) but not invertible, i.e. is not continuous (cf. [a3]). The assumption about the completeness of is essential.

If satisfies, for any , the conditions:

i) ;

ii) ;

iii) is locally algebraic; then there exists an operator such that (cf. [a2], [a5]). This means that is not an algebraic operator.

How to Cite This Entry:
Locally algebraic operator. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Locally_algebraic_operator&oldid=13689
This article was adapted from an original article by Danuta Przeworska-Rolewicz (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article