# Difference between revisions of "Locally algebraic operator"

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A [[Linear operator|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. | A [[Linear operator|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 | + | Let $X$ be a [[Linear space|linear space]] over a [[Field|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 | + | 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|Algebraically closed field]]; [[Characteristic of a field|Characteristic of a field]]). |

− | Thus, a linear operator | + | 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. [[#References|[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|Algebraic operator]]). Thus, an algebraic operator is locally algebraic, but not conversely. |

− | A continuous locally algebraic operator acting in a complete linear [[Metric space|metric space]] | + | A continuous locally algebraic operator acting in a complete linear [[Metric space|metric space]] $X$ is algebraic (cf. [[#References|[a4]]]; for Banach spaces, see [[#References|[a1]]]). |

− | A locally algebraic operator | + | 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|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. [[#References|[a3]]]). The assumption about the completeness of | + | is not continuous (cf. [[#References|[a3]]]). The assumption about the completeness of $X$ is essential. |

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

− | i) | + | i) $\operatorname{dim}\operatorname {ker}p(T)=\operatorname{dim}\operatorname{ker}T.\operatorname{degree}p ( T ) $; |

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

− | iii) | + | 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. [[#References|[a2]]], [[#References|[a5]]]). This means that | + | (cf. [[#References|[a2]]], [[#References|[a5]]]). This means that $T$ is not an [[Algebraic operator|algebraic operator]]. |

====References==== | ====References==== | ||

− | <table>< | + | <table><tr><td valign="top">[a1]</td> <td valign="top"> I. Kaplansky, "Infinite Abelian groups" , Univ. Michigan Press (1954)</td></tr><tr><td valign="top">[a2]</td> <td valign="top"> J. Mikusiński, "Extension de l'espace linéaire avec dérivation" ''Studia Math.'' , '''16''' (1958) pp. 156–172</td></tr><tr><td valign="top">[a3]</td> <td valign="top"> D. Przeworska-Rolewicz, "Algebraic analysis" , PWN&Reidel (1988)</td></tr><tr><td valign="top">[a4]</td> <td valign="top"> D. Przeworska–Rolewicz, S. Rolewicz, "Equations in linear spaces" , PWN (1968)</td></tr><tr><td valign="top">[a5]</td> <td valign="top"> R. Sikorski, "On Mikusiński's algebraic theory of differential equations" ''Studia Math.'' , '''16''' (1958) pp. 230–236</td></tr></table> |

## Revision as of 17:03, 1 July 2020

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.\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.

#### References

[a1] | I. Kaplansky, "Infinite Abelian groups" , Univ. Michigan Press (1954) |

[a2] | J. Mikusiński, "Extension de l'espace linéaire avec dérivation" Studia Math. , 16 (1958) pp. 156–172 |

[a3] | D. Przeworska-Rolewicz, "Algebraic analysis" , PWN&Reidel (1988) |

[a4] | D. Przeworska–Rolewicz, S. Rolewicz, "Equations in linear spaces" , PWN (1968) |

[a5] | R. Sikorski, "On Mikusiński's algebraic theory of differential equations" Studia Math. , 16 (1958) pp. 230–236 |

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Locally algebraic operator.

*Encyclopedia of Mathematics.*URL: http://encyclopediaofmath.org/index.php?title=Locally_algebraic_operator&oldid=50482