# 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 $X$ be a linear space over a field $\mathbf{F}$. Let $L ( X )$ be the set of all linear operators with domains and ranges in $X$ and let

\begin{equation*} L _ { 0 } ( X ) = \{ A \in L ( X ) : \operatorname { dom } A = X \}. \end{equation*}

Denote by $\mathbf{F} [ t ]$ the algebra of all polynomials in the variable $t$ and with coefficients in $\mathbf{F}$. Usually, in applications it is assumed that the field $\mathbf{F}$ is of characteristic zero and algebraically closed (cf. also Algebraically closed field; Characteristic of a field).

Thus, a linear operator $T \in L _ { 0 } ( X )$ is said to be locally algebraic if for any $x \in X$ there exists a non-zero polynomial $p ( t ) \in \mathbf{F} [ t ]$ such that $p ( T ) x = 0$ (cf. [a1]). If there exists a non-zero polynomial $p ( t ) \in \mathbf{F} [ t ]$ such that $p ( T ) x = 0$ for every $x \in X$, then $T$ 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 $X$ is algebraic (cf. [a4]; for Banach spaces, see [a1]).

A locally algebraic operator $T$ acting in a complete linear metric space $X$ (over the field $\mathbf{C}$ of complex numbers) and that is right invertible (cf. Algebraic analysis) but not invertible, i.e.

\begin{equation*} \ker T = \{ x \in X : T x = 0 \} \neq \{ 0 \}, \end{equation*}

is not continuous (cf. [a3]). The assumption about the completeness of $X$ is essential.

If $T \in L _ { 0 } ( X )$ satisfies, for any $p ( t ) , q ( t ) \in \mathbf{F} [ t ]$, the conditions:

i) $\operatorname{dim}\operatorname {ker}p(T)=\operatorname{dim}\operatorname{ker}T\cdot\operatorname{degree}p ( T )$;

ii) $\operatorname { dim } \operatorname { ker }q ( T ) p ( T ) \leq \operatorname { dim } \operatorname { ker } q ( T ) + \operatorname { dim } \operatorname { ker } p ( T )$;

iii) $T$ is locally algebraic; then there exists an operator $A \in L _ { 0 } ( X )$ such that

\begin{equation*} T A - A T = I \end{equation*}

(cf. [a2], [a5]). This means that $T$ 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=50856
This article was adapted from an original article by Danuta Przeworska-Rolewicz (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article