I learnt of this idea 2 days ago while I was chatting with the MATH201 TA after a recitation, complaining about the absurdity of the course showing matrices before linear transformations.
TL;DR: Given an abelian group $V$ and a field $F$, if we can find a ring homomorphism $\phi:F\to\mathrm{End}(V)$, then $V$ forms a vector space over $F$ when we define $av:=\phi(a)(v)$.
Note: Everything mentioned here works for modules as well as vector spaces by just substituting a ring for the field.
The main idea
Let $V$ be a vector space over a field $F$ and let $u,v,w\in V$.
Consider what a scalar does. By the vector space axioms, scaling preserves the structure of the vector space, so if $u+v=w$ then $au+av=aw$ for all $\alpha\in F$. Because of this structure preservation, considering our vector space as an abelian group, this scaling operation is just an endomorphism on that group. So, every fixed $\alpha\in F$ would correspond to $\phi_\alpha:V\to V$ where $\phi_\alpha(v)=\alpha v$.
Conversely, $\phi_\alpha$ being an endomorphism would imply $\alpha u+\alpha v$ without requiring $V$ to be a vector space in the first place.
Obviously, the structure of scalars don’t only appear as the homomorphism property, they appear in the context of other scalars as well. I am of course talking about $\alpha(\beta v)=(\alpha\beta)v$ and $(\alpha+\beta)v=\alpha v+\beta v$ for $\alpha,\beta\in F$. Translating this to endomorphisms, we have $\phi_\alpha(\phi_\beta(v))=(\phi_\alpha\phi_\beta)(v)$ and $(\phi_\alpha+\phi_\beta)(v)=\phi_\alpha(v)+\phi_\beta(v)$.
The above suggests that multiplication of endomorphisms be defined as composition and addition be defined as pointwise addition. This is precisely the structure of $\mathrm{End}(V)$, the endomorphism ring of $V$!
All of the structure we define a vector space to have is already present in its endomorphism ring. This means that the only thing we have to do is map the field elements to the endomomorphisms in a way that preserves the desirable structure of the endomorphism ring. So we’re looking for a structure-preserving map, which is precisely a homomorphism!
In light of this, we can see that at the core of the vector space is a homomorphism $\phi:F\to\mathrm{End}(V)$, where $\alpha v=\phi(\alpha)(v)$ for all $\alpha\in F$ and $v\in V$.
Defining the vector space
Standard definition
Fraleigh’s A first course in abstract algebra defines a vector space over a field $F$ as an additive abelian group $V$ equipped with an operation of scalar multiplication satisfying the following properties for $u,v\in V$ and $\alpha,\beta\in F$:
- $\alpha v\in V$
- $\alpha(\beta v)=(\alpha\beta)v$
- $(\alpha+b)v=\alpha v+bv$
- $\alpha (u+v)=\alpha u+\alpha v$
- $1v=v$ where $1$ is the multiplicative identity of $F$.
Applying the idea
Let $V$ be an abelian group, $F$ a field, and $\phi:F\to\mathrm{End}(V)$ be a ring homomorphism from $F$ to the endomorphism ring of $V$. We claim that defining $\alpha v:=\phi(\alpha )(v)$ for $\alpha \in F$ and $v\in V$ makes $V$ a vector space over $F$.
$\alpha v\in V$
$\phi(\alpha )$ is an endomorphism of $V$, so $\alpha v\in V$.
$(\alpha\beta)v=\alpha(\beta v)$
$(\alpha\beta)v=\phi(\alpha\beta)(v)=(\phi(\alpha)\circ\phi(\beta))(v)=\phi(\alpha)(\phi(\beta)(v))=\phi(\alpha)(\beta v)=\alpha(\beta v)$. Here, $\phi(\alpha \beta)(v)=(\phi(\alpha )\circ \phi(\beta))(v)$ is because multiplication in $\mathrm{End}(V)$ is defined as composition.
$(\alpha+\beta)v=\alpha v+\beta v$
$(\alpha+\beta )v=\phi(\alpha+\beta )(v)=(\phi(\alpha)+\phi(\beta ))(v)=\phi(\alpha)v+\phi(\beta )v=\alpha v+\beta v$.
$\alpha(u+v)=\alpha u+\alpha v$
$\phi(\alpha)(u+v)=\phi(\alpha)(u)+\phi(\alpha)(v)=\alpha u+\alpha v$.
$1v=v$ where $1$ is the multiplicative identity of $F$
$1v=\phi(1)(v)=\iota(v)=v$ where $\iota$ is the multiplicative identity of $\mathrm{End}(V)$ (the identity endomorphism). We use the definition of ring homomorphism in which multiplicative identity is preseved.
And we are done. In order to show that these two definitions are equivalent it remains to prove that the standard definition implies this algebraic definition, which is trivial.