Algebraic and definable closure of a vector space is a span of its subset I am trying to show that for every infinite $K$-vector space $V$ (we use the first order language of vector spaces over $K$) and each $A \subseteq V$, $acl(A)$ is the $K$-linear subspace of $V$ spanned by $A$. The general notion I got is that it follows from strong minimality, yet I cannot see a connection between the two. How should I start thinking about it?
Is it also true for a definable closure?
EDIT [as suggested by @Berci in the comments]:
Definitions

*

*The first-order language of vector spaces is: a constant $0$, a binary function "$+$", and a unary function symbol $F_a$ for each $a\in K$. Given a $K$-vector space $V$, $0$ is interpreted by the identity element of $V$, "$+$" is interpreted by the addition of $V$, and each $F_a$ is interpreted by the operation of scalar multiplication by $a$.


*$acl(A)$ is the set of elements of $M$ that are algebraic over $A$ in an $L$-structure $\mathcal{M}$.


*An element $a$ of $M$ is algebraic over $A$ in $\mathcal{M}$ if there is an $L$-formula $\phi(x,y_1,\ldots,y_n)$ and elements $e_1,\ldots , e_n$ of $A$ such that:
(i) $\mathcal{M}\models \phi [a,e_1,\ldots ,e_n]$
AND
(ii) $\{c\in M|\mathcal{M}\models[c, e_1, \ldots ,e_n]\}$ is finite.
 A: First, recall that the theory of vector spaces over an infinite field $K$ admits quantifier elimination.  That is, for all formulae $\psi$ there is a quantifier free formula $\psi'$ such that:
$$\psi \leftrightarrow \psi'.$$
Now, for any atomic formula $\varphi(x,y_1,y_2,...,y_n)$, the language can only express terms that are linear combinations, so either:

*

*$\varphi(M) = \{\, m \in M \mid M \vDash \varphi(m,e_1,e_2,...,e_n)\,\}=M$ (occurs when $x$ appears with zero coefficient),

*Or, $M \vDash \varphi(x,e_1,e_2,...,e_n)$ implies that $x$ is in the linear span of the $e_i$'s, and therefore unique.

If $v$ is a linear span of elements in $A$, say $v = \lambda_1 a_1 + ... \lambda_n a_n$, consider the atomic formula:
$$\varphi(x,y_1,...,y_n) \equiv x = F_{\lambda_1}y_1 + ... + F_{\lambda_n}y_n.$$
Clearly $M \vDash \varphi(v,a_1,...,a_n)$ and $\varphi(M) =\{\,v\,\}$.  So $v \in acl(A)$ and $v \in dcl(A)$.
Conversely, if $v \in acl(A)$, then there exists a formula $\psi$ such that:

*

*$M \vDash \psi(v,a_1,...,a_n)$ ($a_i \in A$),

*And $\psi(M)$ is finite.

By quantifier elimination, we can assume $\psi$ is quantifier free.  By the above remarks, we can thus deduce that $v$ also satisfies an atomic formula of type 2. and so $v$ is in the linear span of $A$.
The definable and algebraic closures coincide since $span(A) \subseteq dcl(A) \subseteq acl(A) = span(A)$.
