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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 <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/l/l120/l120140/l1201401.png" /> be a [[Linear space|linear space]] over a [[Field|field]] <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/l/l120/l120140/l1201402.png" />. Let <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/l/l120/l120140/l1201403.png" /> be the set of all linear operators with domains and ranges in <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/l/l120/l120140/l1201404.png" /> and let
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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
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/l/l120/l120140/l1201405.png" /></td> </tr></table>
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\begin{equation*} L _ { 0 } ( X ) = \{ A \in L ( X ) : \operatorname { dom } A = X \}. \end{equation*}
  
Denote by <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/l/l120/l120140/l1201406.png" /> the algebra of all polynomials in the variable <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/l/l120/l120140/l1201407.png" /> and with coefficients in <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/l/l120/l120140/l1201408.png" />. Usually, in applications it is assumed that the field <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/l/l120/l120140/l1201409.png" /> is of characteristic zero and algebraically closed (cf. also [[Algebraically closed field|Algebraically closed field]]; [[Characteristic of a field|Characteristic of a field]]).
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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 <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/l/l120/l120140/l12014010.png" /> is said to be locally algebraic if for any <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/l/l120/l120140/l12014011.png" /> there exists a non-zero polynomial <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/l/l120/l120140/l12014012.png" /> such that <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/l/l120/l120140/l12014013.png" /> (cf. [[#References|[a1]]]). If there exists a non-zero polynomial <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/l/l120/l120140/l12014014.png" /> such that <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/l/l120/l120140/l12014015.png" /> for every <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/l/l120/l120140/l12014016.png" />, then <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/l/l120/l120140/l12014017.png" /> is said to be algebraic (cf. [[Algebraic operator|Algebraic operator]]). Thus, an algebraic operator is locally algebraic, but not conversely.
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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]] <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/l/l120/l120140/l12014018.png" /> is algebraic (cf. [[#References|[a4]]]; for Banach spaces, see [[#References|[a1]]]).
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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 <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/l/l120/l120140/l12014019.png" /> acting in a complete linear metric space <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/l/l120/l120140/l12014020.png" /> (over the field <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/l/l120/l120140/l12014021.png" /> of complex numbers) and that is right invertible (cf. [[Algebraic analysis|Algebraic analysis]]) but not invertible, i.e.
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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.
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/l/l120/l120140/l12014022.png" /></td> </tr></table>
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\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 <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/l/l120/l120140/l12014023.png" /> is essential.
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is not continuous (cf. [[#References|[a3]]]). The assumption about the completeness of $X$ is essential.
  
If <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/l/l120/l120140/l12014024.png" /> satisfies, for any <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/l/l120/l120140/l12014025.png" />, the conditions:
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If $T \in L _ { 0 } ( X )$ satisfies, for any $p ( t ) , q ( t ) \in \mathbf{F} [ t ]$, the conditions:
  
i) <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/l/l120/l120140/l12014026.png" />;
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i) $\operatorname{dim}\operatorname {ker}p(T)=\operatorname{dim}\operatorname{ker}T.\operatorname{degree}p ( T ) $;
  
ii) <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/l/l120/l120140/l12014027.png" />;
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ii) $\operatorname { dim } \operatorname { ker }q ( T ) p ( T ) \leq \operatorname { dim } \operatorname { ker } q ( T ) + \operatorname { dim } \operatorname { ker } p ( T )$;
  
iii) <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/l/l120/l120140/l12014028.png" /> is locally algebraic; then there exists an operator <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/l/l120/l120140/l12014029.png" /> such that
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iii) $T$ is locally algebraic; then there exists an operator $A \in L _ { 0 } ( X )$ such that
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/l/l120/l120140/l12014030.png" /></td> </tr></table>
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\begin{equation*} T A - A T = I \end{equation*}
  
(cf. [[#References|[a2]]], [[#References|[a5]]]). This means that <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/l/l120/l120140/l12014031.png" /> is not an [[Algebraic operator|algebraic operator]].
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(cf. [[#References|[a2]]], [[#References|[a5]]]). This means that $T$ is not an [[Algebraic operator|algebraic operator]].
  
 
====References====
 
====References====
<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&amp;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>
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<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&amp;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
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