Difference between revisions of "Addition of vector spaces"

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(No, u_i is not a scalar (unless V_i is one-dimensional))
 
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==Notes==
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* See [[Notes:Vector space operations]]
  
 
==Definitions==
 
==Definitions==
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Often written: <math>V=V_1\boxplus V_2\boxplus\cdots\boxplus V_n</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><br/>
 
| 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><br/>
'''Operations:''' (given {{M|u,v\in V}} where {{M|u_i}} and {{M|c}} is a scalar in {{M|F}})
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'''Operations:''' (given {{M|u_i,v_i\in V_i}} and {{M|c}} is a scalar in {{M|F}})
 
* <math>(u_1,\cdots,u_n)+(v_1,\cdots,v_n)=(u_1+v_1,\cdots,u_n+v_n)</math>
 
* <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>
 
* <math>c(v_1,\cdots,v_n)=(cv_1,\cdots,cv_n)</math>

Latest revision as of 18:02, 18 March 2016

Notes

Definitions

All of this comes from the same reference[1]

Name Expression Notes
Finite
External direct sum Given V1,,Vn which are vector spaces over the same field F:

V=
Often written: V=V_1\boxplus V_2\boxplus\cdots\boxplus V_n

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

Operations: (given u_i,v_i\in V_i and c is a scalar in F)

  • (u_1,\cdots,u_n)+(v_1,\cdots,v_n)=(u_1+v_1,\cdots,u_n+v_n)
  • c(v_1,\cdots,v_n)=(cv_1,\cdots,cv_n)
Alternative form
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\} Consider the association:

(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]
That is, that maps a vector to a function which takes a number from 1 to n to the i^\text{th} component, and:
Given a function f:\{1,\cdots,n\}\rightarrow\cup_{i=1}^nV_i where f(i)\in V_i\ \forall i we can define the following association:
f\mapsto(f(1),\cdots,f(n))
Thus:

  • 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\}
  • V=\mathop{\boxplus}^n_{i=1}V_i=\left\{(v_1,\cdots,v_n)|v_i\in V_i,\ \forall i\right\}

Are isomorphic

Sum of vector spaces Given V_1,\cdots,V_n which are vector subspaces of V

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

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

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

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

References

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