Vector Space

In mathematics, vector space is a set of objects that can be operated(added, multipled and so on) together with number

A vector space is a triple(V, +, .) and it must obey the following 9 rules:

(1) Closure by addition: $$ \forall u, w \in V, \quad u+w \in V $$
(2) Additive associality: $$ \forall u, v, w \in V, \quad (u+v)+w=u+(v+w) $$
(3) Additive commutativity: $$ \forall u, w \in V, \quad u+w = w+u $$
(4) Additive identity: $$ \exists 0 \in V, \quad u+0 = 0+u $$
(5) Additive inverse: $$ \forall u \in V, \quad -u \in V \quad s.t. \quad (-u)+u=u+(-u)=0 $$
(6) Closure by scalar multiplication: $$ \forall k \in R \quad and \quad u \in V, \quad ku \in V $$
(7) Multiplicative associality: $$ \forall k,l \in R \quad and \quad u \in V, \quad (kl)u=k(lu) $$
(8) Multiplicative identity: $$ \exists 1 \in V, \quad 1u=u$$
(9) Distributive property: $$ \forall k \in R \quad and \quad u,w \in V, \quad k*(u+w)=ku+kw $$ $$ \forall k,l \in R \quad and \quad u \in V, \quad (k+l)u=ku+l*u $$
*notice: the two operation in a triple is not necessary to be add and multiply

Example:

Let $ V=R_{>0} $ be the set of strictly positive real numbers. Let $u \boxplus v = u*v$ for all $u,v \in V$ and $k \odot u=u^k$ for all $u \in V$ and $k \in R$. Show that $(V, \boxplus, \odot)$ is a vector space.

Let $u,v \in V$, then $u \boxplus v = u*v$
$\because u$ and $v$ both are positive real number
$\therefore u * v$ is also a positive real number, so V is closure under $\boxplus$.

Let $v,1 \in V$, then $v \boxplus 1 = v*1 = v$
$\therefore$ V has $\boxplus$ identity.

Let $v \in V$ and its $\boxplus$ identity $=1$, then $v \boxplus (1/v)=v*(1/v)=1$,
$\therefore$ V has $\boxplus$ inverse.

Let $u,v,x \in V$, then $u \boxplus (v \boxplus x)=u \boxplus (v*x)=u*v*x=(u \boxplus v)*x=(u \boxplus v) \boxplus x$
$\therefore$ V has $\boxplus$ associality.

Let $u,v \in V$, then $u \boxplus v=u*v=v*u=v \boxplus u$
$\therefore$ V has $\boxplus$ commutativity.

Let $v \in V$, $k \in R$, then $k \odot v = v^k$
$\because v$ is a positive real number and $k$ is a real number
$\therefore$ $v^k $ is also a positive real number, so V is closure under $\odot$.

Let $v \in V$, then $1 \odot v=v^1=v$
$\therefore$ V has $\odot$ identity.

Let $v \in V$ and $a,b \in R$, then $(b \odot a) \odot v=(v^a)^b=b \odot v^a=b \odot (a \odot v)$
$\therefore$ V has $\odot$ associality.

Let $u,v \in V$ and $a,b \in R$,
then $a \odot (u \boxplus v)=a \odot (u*v)=(u*v)^a=u^a * v^a=(a \odot u) \boxplus (a \odot v)$
and $(a \boxplus b) \odot u=(a*b) \odot u=u^{a*b}=u^a * u^b=(a \odot u) \boxplus (b \odot u)$
$\therefore$ V has distributive property.

Therefore V fits all 9 rules and it’s a vector space.