Vector space: Difference between revisions

From Wikibase
Jump to navigation Jump to search
Heal Q1862: create the classic page (the Q1862 fix backfill)
Tags: Replaced Reverted
Restore rev 8309 (my heal edit wrongly replaced the page content)
Tag: Undo
Line 1: Line 1:
== Overview ==
== Mathematical definition ==


mathematical concept
Set $\mathbf{V}$ is a '''vector space''' over $\mathbb{F}$ if:
 
1. Are defined operations
** of addition: $\mathbf{V} \times \mathbf{V} \to \mathbf{V}$
** of scalar multiplication: $\mathbb{F} \times \mathbf{V} \to \mathbf{V}$
 
2. Are satisfied the properties
** Commutativity: $\forall u,v\in\mathbf{V},\; u+v=v+u$;
** Associativity: $\forall u,v,w\in\mathbf{V},\;\forall a,b\in\mathbb{F},\; (u+v)+w=u+(v+w)\;\land\;(ab)v=a(bv)$;
** Additive identity: $\exists \vec 0\in\mathbf{V},\;\forall v\in\mathbf{V},\; \vec 0+v=v$;
** Additive inverse: $\forall v\in\mathbf{V},\;\exists w\in\mathbf{V},\; v+w=0$;
** Multiplicative identity: $\forall v\in\mathbf{V},\; 1v=v$;
** Distributivity: $\forall u,v\in\mathbf{V},\;\forall a,b\in\mathbb{F},\; a(u+v)=au+av\;\land\;(a+b)u=au+bu$.
 
<ref>{{#cite:Q1866}}</ref>

Revision as of 10:51, 29 September 2026

Languages: English · français · Esperanto

Mathematical definition

Set $\mathbf{V}$ is a vector space over $\mathbb{F}$ if:

1. Are defined operations

    • of addition: $\mathbf{V} \times \mathbf{V} \to \mathbf{V}$
    • of scalar multiplication: $\mathbb{F} \times \mathbf{V} \to \mathbf{V}$

2. Are satisfied the properties

    • Commutativity: $\forall u,v\in\mathbf{V},\; u+v=v+u$;
    • Associativity: $\forall u,v,w\in\mathbf{V},\;\forall a,b\in\mathbb{F},\; (u+v)+w=u+(v+w)\;\land\;(ab)v=a(bv)$;
    • Additive identity: $\exists \vec 0\in\mathbf{V},\;\forall v\in\mathbf{V},\; \vec 0+v=v$;
    • Additive inverse: $\forall v\in\mathbf{V},\;\exists w\in\mathbf{V},\; v+w=0$;
    • Multiplicative identity: $\forall v\in\mathbf{V},\; 1v=v$;
    • Distributivity: $\forall u,v\in\mathbf{V},\;\forall a,b\in\mathbb{F},\; a(u+v)=au+av\;\land\;(a+b)u=au+bu$.

[1]