All free groups of finite or infinite countable rank are subgroups of the free non-abelian group $F\_2$, which is linear. However, a free group of infinite uncountable rank will not be a subgroup of $F\_2$. Is it linear, too ?
This might easily follow from model theory, but I could not find a proof in the literature so far.



Are all free groups linear, i.e., admit a faithful representation to GL(n,K) for some field K ?
Free group of rank $c$ embeds in $Sl(2, F(t))$ where $F$ is a field of cardinality $c$. 


Edit: Here is the detailed argument which, as Yves noted in his comment, proves a stronger result. 


Theorem. Let $L$ be a field which is not an algebraic extension of a finite field and let $c$ be the cardinality of $L$. Then the free group of rank $c$ embeds in $SL(2, L)$. 


Proof. Let $P$ be the prime field of $L$; then $L$ has the form
$$
P\subset E \subset L
$$
where $E$ is a purely transcendenetal extension of $P$ and $L$ is an algebraic extension of $E$. Under our assumptions, $E$ and $L$ have the same cardinality, thus, it suffices to consider the case when $L=E$. Then $L$ is isomorphic to the functional field $L=F(t)$, where $F$ is a subfield of $L$. I will consider the case when $F$ is infinite since otherwise $L$ is countable and everything is clear (as the question reduces to the case of free groups of finite rank). Thus, $F$ has the same cardinality $c$ as $L$. 


Let $T$ be the Bruhat-Tits building associated with $G=SL(2, L)$: This building is a simplicial tree with the path-metric $d$, where every edge has unit length. The group $G$ acts on $T$ by simplicial automorphisms with the kernel $\pm 1$. Detailed description and properties of $T$ and the action of $G$ are in Serre's book "Trees." 


Let $v\in T$ be the vertex stabilized by $K=SL(2, O)$, where $O=F[t]$ is the ring of polynomial functions in $t$. Then the link $L\_v$ of $v$ in $T$ is naturally identified with the projective line over $F$ (so that $K$ acts on $L\_v$ by linear-fractional transformations). In particular, the group $K$ acts transitively on pairs of distinct points in $L\_v$. Let $g\in G\setminus K$ be a diagonal matrix with the axis $\gamma\subset T$. Then $\gamma$ contains $v$ and $g$ acts on $\gamma$ as a translation by some even integer distance $\ge 2$. In view of transitivity of the action of $K$ on pairs noted above, there exists a subset $K\_o\subset K$ of cardinality $c$ so that the elements $g\_k=kgk^{-1}$, $k\in K\_o$ have axes $k(\gamma)$ with the property that the 2-point sets 
$$
k(\gamma) \cap L\_v, k\in K\_o,
$$
are pairwise disjoint. (Call this property D.) Now, I claim that the elements $g\_k, k\in K\_o$, are free generators of a free subgroup of $G$. The proof is rather standard. For each $k\in K\_o$ let $D\_k\subset T$ denote the Dirichlet fundamental domain for the cyclic group $\langle g\_k \rangle$:
$$
D\_k=\{ x\in T: d(x, g\_k^m(v))> d(x, v), \forall m\in {\mathbb Z} \setminus 0\}. 
$$
 Since each $g\_v$ translates $v$ at least by $2$, and in view of Property D above, the domains $D\_k$ have pairwise disjoint complements. Thus, Tits' ping-pong argument (from his proof of the Tits alternative) applies in this setting and the subgroup of $G$ generated by the elements $g\_k$ is indeed free with free generators $g\_k$. qed. 


Note that one has to exclude fields $L$ with are algebraic extensions of finite fields, since in this case the group $GL(n, L)$ is torsion (for every finite $n$) and, hence, cannot contain a free subgroup. 


