TITLE: Are all free groups linear, i.e., admit a faithful representation to GL(n,K) for some field K ?
QUESTION [22 upvotes]: 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.

REPLY [10 votes]: Linearity of uncountable free groups 
can be easily deduced from linearity of free groups of countable rank using basic set theory and algebra only, without using compactness theorem, or ultraproducts, or non-trivial arguments of Misha Kapovich. Here is the simple proof.
Proposition. Let $K$ be a field, $n\ge 2$, and $S$ a subset of ${GL}_n(K)$. Suppose all elements of all matrices from $S$ are distinct and algebraically independent in $K$. Then $S$ is a free subset of ${GL}_n(K)$.
Proof.  Suppose not. Then there exist  distinct matrices $A_1,\dots,A_m$ from $S$ and a non-trivial  reduced group word $w(x_1,\dots,x_m)$ such that 
$w(A_1,\dots,A_m)=E$.  
For any field $F$ and  matrix $C=(c_{ij})$ in $GL_n(F)$,
the inverse of $C$ can be found using a formula of the form
$C^{-1}=(f_{ij}(c_{11},\dots,c_{nn})),$
where all $f_{ij}$ are certain rational functions over $\mathbb Z$ in $n^2$ variables,
which do not depend on the choice of $F$ and $C$.
Therefore,  for any field $F$ and matrices $C_1,\dots,C_m$
 in $GL_n(F)$, where $C_k=(c^k_{ij})$, we have
$$w(C_1,\dots,C_m)=
(g_{ij}(c^1_{11},\dots,c^1_{nn},\dots,c^m_{11},\dots,c^m_{nn})),
$$
for some  rational functions $g_{ij}$  over $\mathbb Z$ 
in $mn^2$ variables; these $g_{ij}$ do not depend on the choice of
$F$ and  $C_1,\dots,C_m$.
Let $A_k=(a^k_{ij})$. Since $w(A_1,\dots,A_m)=E$, 
the algebraic independence of all $a^k_{ij}$ 
implies the $g_{ij}\equiv 0$ for 
$i\ne j$, and $g_{ij}\equiv 1$ for $i=j$, for any field of characteristic $p$,
where $p={\rm char}(K)$ (zero or prime).
Hence $w(C_1,\dots,C_m)=E$
for any field $F$ of characteristic $p$ and matrices $C_1,\dots,C_m$  in $GL_n(F)$.
Hence, for any field $F$ of characteristic $p$, in  $GL_n(F)$ there is no free subgroup of rank $\ge m$. 
This contradicts to the known fact that
free subgroups of countable rank embed into
$GL_n(\mathbb Q)$ and $GL_n({\bf F}_q(t))$, for any prime $q$. $\square$
Corollary. If $\lvert K\rvert=\kappa>\aleph_0$
then the free group of rank $\kappa$ embeds
into ${GL}_n(K)$.
Proof.  Since $\kappa>\aleph_0$, 
any transcendence basis $B$ of $K$
is of power $\kappa$. Then there is
a subset $S$ of $M_n(K)$ of power $\kappa$ such that all elements of matrices from $S$ belong to $B$ and are distinct.
Since the elements of $B$ are algebraically independent,
for $A\in S$ we have ${\rm det}(A)\ne 0$ 
and so $A\in {GL}_n(K)$. By Proposition, $S$ is free.  $\square$
EDITED VERSION. In his answer
Misha Kapovich proved, using his 
non-trivial geometric arguments, that
for any field $K$ of uncountable power $\kappa$, the free group of rank $\kappa$ embeds into $SL_n(K)$, for any $n\ge 2$. In my answer I gave an elementary proof of a weaker result, with $GL$ instead of $SL$. However, my proof can be easily modified to the case of $SL$. Here is this modified proof.
For a non-singular square matrix $A$ denote by $\bar A$ the
matrix  obtained from $A$ by replacing any element $a$
in the first row of $A$ with $a/{\rm det}(A)$. Clearly, 
${\rm det}(\bar A)=1$, and $\bar A=A$ if ${\rm det}(A)=1$.
Proposition. Let $K$ be a field, $n\ge 2$,
and $S$ a subset of ${GL}_n(K)$. 
Suppose all elements of 
all matrices from $S$ are distinct and algebraically independent in $K$. 
Then ${\bar S} =\{{\bar A}: A\in S\}$ is a free subset of ${SL}_n(K)$.
Proof. Suppose not. Then there exist  distinct matrices 
$A_1,\dots,A_m$ from $S$ such that 
$w({\bar A}_1,\dots,{\bar A}_m)=E$
for some  non-trivial  reduced group word $w(x_1,\dots,x_m)$.
As above, for any field $F$ and matrices $C_1,\dots,C_m$
 in $GL_n(F)$, where $C_k=(c^k_{ij})$, we have
$$w({\bar C}_1,\dots,{\bar C}_m)=
(g_{ij}(c^1_{11},\dots,c^1_{nn},\dots,c^m_{11},\dots,c^m_{nn})),
$$
for some  rational functions $g_{ij}$  over $\mathbb Z$ 
in $mn^2$ variables; these $g_{ij}$ do not depend on the choice of
$F$ and  $C_1,\dots,C_m$.
Let $A_k=(a^k_{ij})$. Since $w({\bar A}_1,\dots,{\bar A}_m)=E$, 
the algebraic independence of all $a^k_{ij}$ 
implies the $g_{ij}\equiv 0$ for 
$i\ne j$, and $g_{ij}\equiv 1$ for $i=j$,
for any field of characteristic $p$ , where $p={\rm char}(K)$  (zero or prime). 
Hence 
$w({\bar C}_1,\dots,{\bar C}_m)=E$
for any field $F$ of characteristic $p$ and matrices $C_1,\dots,C_m$  in $GL_n(F)$. In particular, 
$w(C_1,\dots, C_m)=E$
for any field $F$ of characteristic $p$ and matrices $C_1,\dots,C_m$  in $SL_n(F)$.
Hence, for any field $F$ of characteristic $p$, in  $SL_n(F)$ there is no free subgroup of rank $\ge m$. This contradicts to the fact that
free subgroups of countable rank embed into $SL_n(\mathbb Q)$ and $SL_n({\bf F}_q(t))$, for any prime $q$.
 $\square$
Corollary. If $\lvert K\rvert=\kappa>\aleph_0$
then the free group of rank $\kappa$ embeds
into ${SL}_n(K)$. 
Proof.  Since $\kappa>\aleph_0$, 
any transcendence basis $B$ of $K$
is of power $\kappa$. Then there is
a subset $S$ of $M_n(K)$ of power $\kappa$ such that all elements of matrices from $S$ belong to $B$ and are distinct.
 Since the elements of $B$ are algebraically independent,
for $A\in S$ we have ${\rm det}(A)\ne 0$ 
and so $A\in {GL}_n(K)$. 
Clearly, $\lvert \bar S\rvert=\kappa$.
By Proposition, $\bar S$ is 
free in $SL_n(K)$.  $\square$