Vector space - Revision history

Rongzhou: /* Base */

2026-10-12T11:30:42Z

Base

@@ Line 71: / Line 71: @@
 <ref>{{#cite:Q1893}}</ref>
-== Base ==
+== Basis (plural: bases) ==
 If for $B=\{\vec u_1,\dots,\vec u_n\}$:
 {{#content:Q2144}}
 == See also ==

Rongzhou at 11:14, 12 October 2026

2026-10-12T11:14:13Z

← Older revision		Revision as of 11:14, 12 October 2026
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	{{#content:Q2144}}		{{#content:Q2144}}

			== See also ==

			* [[Inner product space]]
			* [[Vector subspace]]
			* [[Hyperplane]]

Rongzhou at 07:46, 12 October 2026

2026-10-12T07:46:46Z

← Older revision		Revision as of 07:46, 12 October 2026
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	<ref>{{#cite:Q1893}}</ref>		<ref>{{#cite:Q1893}}</ref>

			== Base ==

			If for $B=\{\vec u_1,\dots,\vec u_n\}$:

			{{#content:Q2144}}

Rongzhou: /* Elementary properties */

2026-10-05T06:59:10Z

Elementary properties

← Older revision		Revision as of 06:59, 5 October 2026
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	'''Proof:''' Let $\vec 0,\vec 0_1\in V$ satisfy $\forall v\in V,\;\vec 0+v=v$ and $\forall v\in V,\;\vec 0_1+v=v$. Then		'''Proof:''' Let $\vec 0,\vec 0_1\in V$ satisfy $\forall v\in V,\;\vec 0+v=v$ and $\forall v\in V,\;\vec 0_1+v=v$. Then
	$$\vec 0_1=\vec 0+\vec 0_1=\vec 0.$$
			$$
			\vec 0-\vec 0_1=\vec 0 \\
			\vec 0 - \vec 0 = \vec 0_1 \\
			\vec 0=\vec 0_1
			$$

	Hence $\vec 0=\vec 0_1$.		Hence $\vec 0=\vec 0_1$.

Rongzhou: /* Elementary properties */

2026-09-29T13:28:48Z

Elementary properties

← Older revision		Revision as of 13:28, 29 September 2026
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	<ref>{{#cite:Q1866}}</ref>		<ref>{{#cite:Q1866}}</ref>

			== Elementary properties ==

			For vector space $V$ over $\mathbb{F}$:

			=== The $\vec 0$ is unique ===

			$$\exists!\,\vec 0\in V,\;\forall v\in V,\;\vec 0+v=v.$$

			'''Proof:''' Let $\vec 0,\vec 0_1\in V$ satisfy $\forall v\in V,\;\vec 0+v=v$ and $\forall v\in V,\;\vec 0_1+v=v$. Then
			$$\vec 0_1=\vec 0+\vec 0_1=\vec 0.$$
			Hence $\vec 0=\vec 0_1$.

			=== The additive inverse for each vector is unique ===

			$$\forall v\in V,\;\exists!\,w\in V,\;v+w=\vec 0.$$

			'''Proof:''' Let $w,w'\in V$ satisfy $v+w=\vec 0$ and $v+w'=\vec 0$. Then
			$$w=w+\vec 0=w+(v+w')=(w+v)+w'=\vec 0+w'=w'.$$
			Hence $w=w'$.

			=== Scalar $0$ mutiplies by any vector results in $\vec 0$ (the vector zero) ===

			$$\forall v\in V,\;0v=\vec 0.$$

			'''Proof:''' For $v\in V$,
			$$0v=(0+0)v=0v+0v.$$
			Adding the additive inverse of $0v$, $-0v$, to both sides, we obtain
			$$\vec 0=0v.$$

			=== Any scalar multiples of $\vec 0$ is $\vec 0$ ===

			$$\forall a\in\mathbb{F},\;a\vec 0=\vec 0.$$

			'''Proof:''' For $a\in\mathbb{F}$,
			$$a\vec 0=a(\vec 0+\vec 0)=a\vec 0+a\vec 0.$$
			Adding the additive inverse of $a\vec 0$, $-a\vec 0$ to both sides, we obtain
			$$\vec 0=a\vec 0.$$

			=== Scalar $-1$ multiples by any vector results in the additive inverse of the vector ===

			$$\forall v\in V,\;(-1)v=-v.$$

			'''Proof:''' For $v\in V$,
			$$v+(-1)v=1v+(-1)v=(1+(-1))v=0v=\vec 0,$$
			which shows that $(-1)v$ is the additive inverse $-v$ of $v$.

			<ref>{{#cite:Q1893}}</ref>

SeedBot: Restore rev 8309 (my heal edit wrongly replaced the page content)

2026-09-29T10:51:02Z

Restore rev 8309 (my heal edit wrongly replaced the page content)

← Older revision		Revision as of 10:51, 29 September 2026
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	== ~~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>

SeedBot: Heal Q1862: create the classic page (the Q1862 fix backfill)

2026-09-29T10:50:13Z

Heal Q1862: create the classic page (the Q1862 fix backfill)

← Older revision		Revision as of 10:50, 29 September 2026
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	== ~~Mathematical definition~~ ==		== Overview ==

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

	~~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>~~

Rongzhou: /* Mathematical definition */

2026-09-29T10:32:08Z

Mathematical definition

← Older revision		Revision as of 10:32, 29 September 2026
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	** Multiplicative identity: $\forall v\in\mathbf{V},\; 1v=v$;		** Multiplicative identity: $\forall v\in\mathbf{V},\; 1v=v$;
	** Distributivity: $\forall u,v\in\mathbf{V},\;\forall a,b\in\mathbb{F},\; a(u+v)=au+av\;\land\;(a+b)u=au+bu$.		** 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>

Rongzhou: Created page with "== 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 i..."

2026-09-29T10:25:08Z

Created page with "== 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 i..."

New page

== 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: $\forall u,v\in\mathbf{V},\;\forall a,b\in\mathbb{F},\; a(u+v)=au+av\;\land\;(a+b)u=au+bu$.