Prove this homomorphism gives an equivalent definition of Linear Space Let $\mathbb V(\oplus)$ be an abelian group and $\mathbb K(+,\cdot)$ a field; of course $\mathbb K(\cdot)$ is the group under product. We represent the group of automorphisms of $\mathbb V(\oplus)$ with the notation $\{\mathbb Vf\mathbb V\}$, and let $*:\mathbb K(\cdot)\rightarrow\{\mathbb Vf\mathbb V\}$ be a homomorphism such that $*(a+b)(u)=*(a)(u)\oplus*(b)(u)$. Prove we have defined a linear space.
The elements of the abelian group are the vectors, while scalar multiplication is defined by the homomorphism. We need to prove the relations that hold for scalar product. Basically we are saying that a linear space exists when we have a homomorphism between a field and a group of automorphisms of an abelian group that satisfies one condition.
 A: The homomorphism $*$ maps elements of the fiels into automorphisms for the abelian group. The condition given on $*$ makes it so that $*(a+b)$ applied to $u$, gives the same result if we operate $*a(u)$ and $*b(u)$ in the group. Equivalently, $(a+b)*u=(a*u)\oplus(b*u)$.
Obviously, the scalar product is defined in terms of the homomorphism $*$ in such a way that $*(a)$ is the automorpphism that maps $u\mapsto a*u$. Since the objects in the image of $*$ are automorphisms, we have $*a(u\oplus v)=*a(u)\oplus*a(v)$, which is another way of saying $a*(u\oplus v)=(a*u)\oplus (a*v)$.
Secondly, the fact that $*$ is a homomorphism means that $1\in\mathbb K(\cdot)$ is the identity morphism for our abelian group. Also, $*(a\cdot b)=*(a)\circ*(b)$ because $*$ is an homomorphism with composition as operation in the image. Another way of expressing this is by $(a\cdot b)*u=a*(b*u)$. This gives a scalar producto of the objects in the field and the abelian group, turning it into a linear space over $\mathbb K$.
