1006: Free Vectors Space on Set

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definition of free vectors space on set

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Topics

About: vectors space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Note

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Starting Context

• The reader knows a definition of %field name% vectors space.

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Target Context

• The reader will have a definition of free vectors space on set.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$S$: $\in \{\text{ the sets }\}$
$F$: $\in \{\text{ the fields }\}$
$\ast F(S,F)$: $=\{f:S\to F\in Pow(S\times F)|f^{-1}(F\setminus \{0\})\in \{\text{ the finite sets }\}\}$ with the addition and the scalar multiplication specified below, $\in \{\text{ the }F\text{ vectors spaces }\}$
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Conditions:
$\mathrm{\forall }{f}_1,f_2\in F(S,F)(f_1+f_2:s\mapsto f_1(s)+f_2(s))$
$\wedge$
$\mathrm{\forall }r\in F,\mathrm{\forall }f\in F(S,F)(rf:s\mapsto rf(s))$
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2: Note

The 1st "$F$" in "$F(S,F)$" does not mean the field, $F$, but is the abbreviation of "Free". So, when the field is $\mathbb{R}$, it becomes $F(S,\mathbb{R})$.

An $f$ is often denoted as $r^1{s}_1+...+r^k{s}_k$ where $r^j\in F$ and $s_j\in S$, which means that $f(s_j)=r^j$ and $f(s)=0$ for any $s\notin \{s_1,...,s_k\}$. The expression is commutative: $r^1{s}_1+r^2{s}_2+r^3{s}_3=r^2{s}_2+r^1{s}_1+r^3{s}_3=...$, because that does not change $f$ at all.

The operations are indeed well-defined: $(f_1+f_2{)}^{-1}(F\setminus \{0\})\in \{\text{ the finite sets }\}$ and $(rf{)}^{-1}(F\setminus \{0\})\in \{\text{ the finite sets }\}$, obviously.

Let us see that $F(S,F)$ is indeed a $F$ vectors space.

1) for any elements, $v_1,v_2\in F(S,F)$, $v_1+v_2\in F(S,F)$ (closed-ness under addition): already seen.

2) for any elements, $v_1,v_2\in F(S,F)$, $v_1+v_2=v_2+v_1$ (commutativity of addition): for each $s\in S$, $(v_1+v_2)(s)=v_1(s)+v_2(s)=v_2(s)+v_1(s)=(v_2+v_1)(s)$.

3) for any elements, $v_1,v_2,v_3\in F(S,F)$, $(v_1+v_2)+v_3=v_1+(v_2+v_3)$ (associativity of additions): for each $s\in S$, $((v_1+v_2)+v_3)(s)=(v_1+v_2)(s)+v_3(s)=v_1(s)+v_2(s)+v_3(s)=v_1(s)+(v_2(s)+v_3(s))=v_1(s)+(v_2+v_3)(s)=(v_1+(v_2+v_3))(s)$.

4) there is a 0 element, $0\in F(S,F)$, such that for any $v\in F(S,F)$, $v+0=v$ (existence of 0 vector): the $0$ function, $f_0$, is $0$, because for each $s\in S$, $(v+f_0)(s)=v(s)+f_0(s)=v(s)+0=v(s)$.

5) for any element, $v\in F(S,F)$, there is an inverse element, $v^{\prime }\in F(S,F)$, such that $v^{\prime }+v=0$ (existence of inverse vector): $-v$ is a $v^{\prime }$, because for each $s\in S$, $(-v+v)(s)=-v(s)+v(s)=0=f_0(s)$.

6) for any element, $v\in F(S,F)$, and any scalar, $r\in F$, $r.v\in F(S,F)$ (closed-ness under scalar multiplication): already seen.

7) for any element, $v\in F(S,F)$, and any scalars, $r_1,r_2\in F$, $(r_1+r_2).v=r_1.v+r_2.v$ (scalar multiplication distributability for scalars addition): for each $s\in S$, $((r_1+r_2).v)(s)=(r_1+r_2)v(s)=r_{1}v(s)+r_{2}v(s)=(r_{1}v)(s)+(r_{2}v)(s)=(r_1.v+r_2.v)(s)$.

8) for any elements, $v_1,v_2\in F(S,F)$, and any scalar, $r\in F$, $r.(v_1+v_2)=r.v_1+r.v_2$ (scalar multiplication distributability for vectors addition): for each $s\in S$, $(r.(v_1+v_2))(s)=r(v_1+v_2)(s)=r(v_1(s)+v_2(s))=r{v}_1(s)+r{v}_2(s)=(r{v}_1)(s)+(r{v}_2)(s)=(r.v_1+r.v_2)(s)$.

9) for any element, $v\in V$, and any scalars, $r_1,r_2\in F$, $(r_1{r}_2).v=r_1.(r_2.v)$ (associativity of scalar multiplications): for each $s\in S$, $((r_1{r}_2).v)(s)=(r_1{r}_2)v(s)=r_1(r_{2}v(s))=r_1(r_{2}v)(s)=(r_1.(r_2.v))(s)$.

10) for any element, $v\in V$, $1.v=v$ (identity of 1 multiplication): for each $s\in S$, $(1.v)(s)=1v(s)=v(s)$.

For each $s\in S$, there is the function, $f_s\in F(S,F)$, such that $f_s(s)=1$ and $f_s(s^{\prime })=0$ for each $s^{\prime }\in S\setminus \{s\}$ ($f_s=1s$ by the aforementioned notation), and $B:=\{f_s\in F(S,F):s\in S\}$ is a basis of $F(S,F)$: for each $f\in F(S,F)$, $f=f^1{f}_{s_1}+...+f^k{f}_{s_k}$, which is $f^1{s}_1+...+f^k{s}_k$ by the aforementioned notation; it is linearly independent, because for each $c^1{f}_{s_1}+...+c^k{f}_{s_k}=0$, $(c^1{f}_{s_1}+...+c^k{f}_{s_k})(s_j)=0(s_j)=0$, but the left hand side is $c^1{f}_{s_1}(s_j)+...+c^k{f}_{s_k}(s_j)=c^j$, so, $c^j=0$.

We need to be careful when $S$ is an $F$ vectors space, $V$: for each $v\in V$ and $r\in F\setminus \{1\}$, $r{f}_v=r(1v)=r(v)\ne (rv)=1(rv)=1f_{rv}=f_{rv}$, because $r{f}_v$ is the function that maps $v$ to $r$ while $f_{rv}$ is the function that maps $rv$ to $1$ but maps $v$ to $0$. When $S$ is just a set, $r{f}_s=rs$ is not ambiguous, because as there is no such operation as $rs$ on $S$, $r$ is inevitably operating on $F(S,F)$, but when $S=V$, an $F$ vectors space, $rv$ is ambiguous whether $r$ is operating on $V$, which is $1(rv)=1f_{rv}$, or $r$ is operating on $F(V,F)$, which is $r(v)=r{f}_v$.

We will adopt the notation that for each $v\in V$, $v$ denotes $v\in V$ while $(v)$ denotes $(v)=f_v\in F(V,F)$.

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References

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