Addition of vector spaces

Notes

• See Notes:Vector space operations

Definitions

All of this comes from the same reference

Name	Expression	Notes
Finite
External direct sum	Given $$V_1,\cdots,V_n$$ which are vector spaces over the same field $F$: $$V=\mathop{\boxplus}^n_{i=1}V_i=\left\{(v_1,\cdots,v_n)|v_i\in V_i,\ i=1,2,\cdots,n\right\}$$ 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: $$V=\sum_{i\in I}V_i$$ - that is that $V$ is the sum (or join) of the family $\mathcal{F}$ $$\forall i\in I$$ 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!)