Difference between revisions of "Addition of vector spaces"

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==Definitions==
 
==Definitions==
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All of this comes from the same reference<ref>Advanced Linear Algebra - Third Edition - Steven Roman - Graduate Texts in Mathematics</ref>
 
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| [[Internal direct sum]]
 
| [[Internal direct sum]]
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| <math>V=\bigoplus\mathcal{F}</math> or <math>V=\bigoplus_{i\in I}</math> where the following hold:
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| Given a family of subspaces of {{M|(V,F)}}, <math>\mathcal{F}=\{V_i|i\in I\}</math>, the internal direct sum is defined as follows:<br/>
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# A
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<math>V=\bigoplus\mathcal{F}</math> or <math>V=\bigoplus_{i\in I}</math> where the following hold:
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# B
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# <math>V=\sum_{i\in I}V_i</math> - that is that {{M|V}} is the sum (or join) of the family {{M|\mathcal{F} }}
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# <math>\forall i\in I</math> we have <math>V_i\cap\left(\sum_{j\ne i}V_j\right)=\{0\}</math>
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*For the second condition each {{M|V_j}} is called a ''direct summand'' of {{M|V}}
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* If {{M|\mathcal{F} }} is finite, that is <math>\mathcal{F}=\{V_1,\cdots,V_n\}</math> then we often write:
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*: <math>V=V_1\oplus\cdots\oplus V_n</math>
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* If {{M|1=V=S\oplus T}} then we call {{M|T}} a '''complement of {{M|S}} in {{M|V}}'''
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* The {{M|2^\text{nd} }} condition is stronger than saying the members of {{M|\mathcal{F} }} are pairwise disjoint - the book makes this clear although I see it as obvious. (Even though they're not quite pairwise disjoint!)
 
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Revision as of 11:24, 24 April 2015

Definitions

All of this comes from the same reference[1]

Name Expression Notes
Finite
External direct sum Given [math]V_1,\cdots,V_n[/math] which are vector spaces over the same field [ilmath]F[/ilmath]:

[math]V=\mathop{\boxplus}^n_{i=1}V_i=\left\{(v_1,\cdots,v_n)|v_i\in V_i,\ i=1,2,\cdots,n\right\}[/math]
Often written: [math]V=V_1\boxplus V_2\boxplus\cdots\boxplus V_n[/math]

This is the easiest definition, for example [math]\mathbb{R}^n=\mathop{\boxplus}^n_{i=1}\mathbb{R}=\underbrace{\mathbb{R}\boxplus\cdots\boxplus\mathbb{R}}_{n\text{ times}}[/math]

Operations: (given [ilmath]u,v\in V[/ilmath] where [ilmath]u_i[/ilmath] and [ilmath]c[/ilmath] is a scalar in [ilmath]F[/ilmath])

  • [math](u_1,\cdots,u_n)+(v_1,\cdots,v_n)=(u_1+v_1,\cdots,u_n+v_n)[/math]
  • [math]c(v_1,\cdots,v_n)=(cv_1,\cdots,cv_n)[/math]
Alternative form
[math]V=\mathop{\boxplus}^n_{i=1}V_i=\left\{\left.f:\{1,\cdots,n\}\rightarrow\bigcup_{i=1}^nV_i\right|f(i)\in V_i\ \forall i\in\{1,\cdots,n\}\right\}[/math] Consider the association:

[math](v_1,\cdots,v_n)\mapsto\left[\left.f:\{1,\cdots,n\}\rightarrow\bigcup_{i=1}^nV_i\right|f(i)=v_i\ \forall i\right][/math]
That is, that maps a vector to a function which takes a number from 1 to [ilmath]n[/ilmath] to the [ilmath]i^\text{th} [/ilmath] component, and:
Given a function [math]f:\{1,\cdots,n\}\rightarrow\cup_{i=1}^nV_i[/math] where [math]f(i)\in V_i\ \forall i[/math] we can define the following association:
[math]f\mapsto(f(1),\cdots,f(n))[/math]
Thus:

  • [math]V=\mathop{\boxplus}^n_{i=1}V_i=\left\{\left.f:\{1,\cdots,n\}\rightarrow\bigcup_{i=1}^nV_i\right|f(i)\in V_i\ \forall i\right\}[/math]
  • [math]V=\mathop{\boxplus}^n_{i=1}V_i=\left\{(v_1,\cdots,v_n)|v_i\in V_i,\ \forall i\right\}[/math]

Are isomorphic

Sum of vector spaces Given [ilmath]V_1,\cdots,V_n[/ilmath] which are vector subspaces of [ilmath]V[/ilmath]

[math]\sum^n_{i=1}V_i=\left\{v_1+\cdots+v_n|v_i\in V_i,\ i=1,2,\cdots,n\right\}[/math]
Sometimes this is written: [math]V_1+V_2+\cdots+V_n[/math]

For any family of vectors (here [ilmath]K[/ilmath] will denote an indexing set and [math]\mathcal{F}=\left\{V_i|i\in K\right\}[/math] (a family of vector spaces over [ilmath]F[/ilmath]))
Direct product [math]V=\prod_{i\in K}V_i=\left\{\left.f:K\rightarrow\bigcup_{i\in K}V_i\right|f(i)\in V_i\ \forall i\in K\right\}[/math] Generalisation of the external direct sum
External direct sum [math]V=\mathop{\boxplus}_{i\in K}V_i=\left\{\left.f:K\rightarrow\bigcup_{i\in K}V_i\right|f(i)\in V_i\ \forall i\in K,\ f\text{ has finite support}\right\}[/math] Note:
  • The alternative notation [math]\bigoplus_{i\in K}^\text{ext}[/math] is sometimes used
Finite support:
A function [ilmath]f[/ilmath] has finite support if [ilmath]f(i)=0[/ilmath] for all but finitely many [ilmath]i\in K[/ilmath] So it is "zero almost everywhere" - the set [math]\{f(i)|f(i)\ne 0\}[/math] is finite.
Internal direct sum Given a family of subspaces of [ilmath](V,F)[/ilmath], [math]\mathcal{F}=\{V_i|i\in I\}[/math], the internal direct sum is defined as follows:

[math]V=\bigoplus\mathcal{F}[/math] or [math]V=\bigoplus_{i\in I}[/math] where the following hold:

  1. [math]V=\sum_{i\in I}V_i[/math] - that is that [ilmath]V[/ilmath] is the sum (or join) of the family [ilmath]\mathcal{F} [/ilmath]
  2. [math]\forall i\in I[/math] we have [math]V_i\cap\left(\sum_{j\ne i}V_j\right)=\{0\}[/math]
  • For the second condition each [ilmath]V_j[/ilmath] is called a direct summand of [ilmath]V[/ilmath]
  • If [ilmath]\mathcal{F} [/ilmath] is finite, that is [math]\mathcal{F}=\{V_1,\cdots,V_n\}[/math] then we often write:
    [math]V=V_1\oplus\cdots\oplus V_n[/math]
  • If [ilmath]V=S\oplus T[/ilmath] then we call [ilmath]T[/ilmath] a complement of [ilmath]S[/ilmath] in [ilmath]V[/ilmath]
  • The [ilmath]2^\text{nd} [/ilmath] condition is stronger than saying the members of [ilmath]\mathcal{F} [/ilmath] are pairwise disjoint - the book makes this clear although I see it as obvious. (Even though they're not quite pairwise disjoint!)

References

  1. ↑ Advanced Linear Algebra - Third Edition - Steven Roman - Graduate Texts in Mathematics
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