Explicit example of a vector space over a finite field, and linear transformation of vector spaces over different fields I am trying to do two things. First, come up with an explicit example of a vector space over a finite field (checking the axioms for vector space would be helpful). Second, come up with an explicit example of a linear transformation from that domain to a vector space over an infinite field (I'm having trouble understanding if a linear transformation necessitates both the domain and codomain to be vector spaces over the same field).
Any help in stating these examples would be appreciated.
 A: One of the great things about linear algebra is that almost all the stuff you do with linear algebra over the reals extends to any field (including finite fields).
Let's start by considering the finite field $\mathbb{Z} / \mathbb{3Z}$, and consider the cartesian product of this with itself (i.e. vectors with two entries from the field of integers mod 3). We're looking at $(\mathbb{Z} / \mathbb{3Z})^2$.
Do the axioms hold?


*

*Is it an abelian group under addition? Yes. Addition commutes (and is associative) and is closed, there is an identity (zero vector), and by existence of additive inverses there exists a unique inverse for every vector.

*Left and right distributivity by scalar multiplication holds.

*For $u, v$ in the space, is $au+bv$ in the space, where $a, b \in \mathbb{Z} / \mathbb{3Z}$? You can verify that this holds as well. 
As user Randall said (I directly quote), you need to keep the same ground field for your linear transformations. Linearity is always with respect to a field, because the axioms for linearity rely on addition and scalar multiplication in that field. (Sometimes, people will even use the word K-linear to denote a linear map with respect to a field K, which should convince you that linearity is really tied to some notion of a base field.)
I would also like to make clear that the language of "vector subspaces" also makes perfect sense in a vector space over a field. Consider $(\mathbb{Z} / \mathbb{3Z})^3$ (vectors with 3 entries) and then consider the subspace consisting of all vectors with zero in the third entry. (In plain old $\mathbb{R}^3$, this is just a plane spanned by two axes.) You can verify that this subspace is a vector space.
A: Let $\mathbb{F}$ be any field.  Define $V^n$ to be the set of all $n$-component vectors with entries from $\mathbb{F}$.  Given slot-wise addition and scaling, this is an $\mathbb{F}$-vector space.  You won't be shocked that any f.d. $\mathbb{F}$-vector space of dimension $n$ is isomorphic to $V^n$.
