Definition of a vector space
I’m going to be working through Sheldon Axler’s Linear Algebra Done Right for the next little while, so I believe it will be useful to document my understanding of the proofs and definitions here.
Let $V$ be any set, and let $v_i\in V$ for any given natural number $i$; similarly, let $c_i\in\mathbb C$ for any given natural number $i$. Define functions $+$ and $\cdot$ as follows:
$+:V\times V\rightarrow V$ by $+(v_1,v_2)=v_1+v_2$
$\cdot~:\mathbb C\times V\rightarrow V$ by $\cdot(c_1,v_1)=c_1v_1$
The following definitions will be useful:
If $v_1+v_2=v_2+v_1$, we’ll say $+$ commutes in $V$. Similarly, if $c_1v_1=v_1c_1$, we’ll say $\cdot$ commutes in $V$.
If $(v_1+v_2)+v_3=v_1+(v_2+v_3)$, say $+$ associates in $V$. If $(c_1c_2)v_1=c_1(c_2v_1)$, say $\cdot$ associates in $V$.
If $c_1(v_1+v_2)=c_1v_1+c_1v_2$, say $\cdot$ distributes over $+$ in $V$.
Suppose $\cdot$ commutes in $V$. Then if $(c_1+c_2)v_1=c_1v_1+c_2v_1$, say $+$ distributes over $\cdot$ in $V$.
Now, consider the following conditions on the sets $V$ and $\mathbb C$ and the functions $+$ and $\cdot~$:
$+$ and $\cdot$ commute in $V$.
$+$ and $\cdot$ associate in $V$.
There is some element named 0 in $V$ with the property that $v+0=v$ for any given $v\in V$.
Given any $v\in V$, there is some element in $V$ named $-v$ with the property that $v+(-v)=0$.
There is an element named $1$ in $\mathbb C$ with the property that $1v=v$ for any given $v\in V$. Indeed, this is just the number $1$.
$\cdot$ distributes over $+$ in $V$ and $+$ distributes over $\cdot$ in $V$.
If the sets $V$ and $\mathbb C$ together with the functions $+$ and $\cdot$ satisfy each of the six conditions above, we say that the set $V$ is a vector space over $\mathbb C$ under $+$ and $\cdot$; further, any element of a vector space $V$ will be called a vector.
