In the language of vector spaces $\mathcal{L}=\{0,+,f_c(c\in\mathbb{F})\}$, where $+$ is usual vector addition and $f_c$ is scalar multiplication for each element $c\in \mathbb{F}$. Is it possible to write sentence(s) which say $V$ has dimension $n$?
I want to write a sentence which says every element is a linear combination of $n$ distinct elements. $\psi(x_1,...,x_n):=(x_1\neq x_2 \wedge x_2\neq x_3\wedge x_3\neq x_1\wedge ... \wedge x_{n-1}\neq x_n)$
Which should give me that the $x_1,..,x_n$ are distinct.
$\exists y_1,...,y_n\forall x(\psi(y_1,...,y_n)\wedge x = f_{c_1}(y_1)+f_{c_2}(y_2)+...+f_{c_n}(y_n))$
Can I then just take a collection of sentences for each combination of $c_i$?
My problem is that the dimension of $V$ can be finite but the field doesn't have to be, so I'm not sure if my sentences will work for every vector space or if it may only be possible to get a set of sentences for vector spaces over finite fields.
If your field is finite, then there are finitely many linear combinations of $n$ vectors possible. So there are $g_1(x_1, \ldots, x_n), \ldots, g_k(x_1, \ldots, x_n)$ where each $g_i(x_1, \ldots, x_n)$ is a linear combination of $x_1, \ldots, x_n$, and let $g_1$ be $0x_1 + \ldots + 0x_n$ (so $g_1$ always gives $0$). Then we can write down a formula saying the vector space has dimension $n$: $$ \exists x_1 \ldots x_n \left( \bigwedge_{1 \leq i < j \leq n} x_i \neq x_j \wedge \bigwedge_{1 < i \leq k} g_i(x_1, \ldots, x_n) \neq 0 \wedge \forall y \bigvee_{1 \leq i \leq k} g_i(x_1, \ldots, x_k) = y \right). $$ The first conjunction says that all the $x_i$ are distinct. The second conjunction says that the $x_i$ are linearly independent. The last disjunction says that every element is a linear combination of the $x_i$.
If your field is infinite, then you cannot write down such a formula. For this you can use the same argument as in this answer: if $n$-dimensional vector spaces (over an infinite field) do no respect the upwards Löwenheim-Skolem theorem.