For a finite group $G$, set $\mu(G):=\min\{|S|: G=\langle S\rangle\}$.
Given an infinite field $F$ and an integer $n\ge 1$, set
$$\mu_{n,F}:=\sup\{\mu(G): G \le GL(n,F), |G| \text{ is finite}\}.$$
$$\tau_{n,F}:=\sup\{\mu(G): G \le GL(n,F) \text{ is a finite abelian subgroup }\}.$$
Clearly, $\tau_{n,F}\le \mu_{n,F}$.
(1) Is it known when $\mu_{n,F}<\infty?$
(This is clearly true if $n=1$, since $GL(1,F)$ is just the multiplicative subgroup of $F$, and it is known that all it's finite subgroups are cyclic).
(2) Is it known when $\tau_{n,F}$ is finite?
Please help.
UPDATE: As shown by Eric Wofsey in the comments, $\tau_{n,F}$ is infinite whenever $F$ is an infinite field of finite positive characteristic and $n>1$.
The case of when $F$ has characteristic zero remains open.
Since we can extend scalars, to show finiteness of $\mu_{n,F}$ in characteristic zero $F$, it suffices to check for $F=\mathbb{C}$, since any finite subgroup only needs finitely many coefficients of the matrices, and we can pick an embedding into $\mathbb{C}$.
So now we quote the following result, due to Jordan (Theorem 14.2 in Isaac's Character Theory of finite groups), that there exists a function $f:\mathbb{N}\rightarrow \mathbb{N}$ such that any finite subgroup of $GL_n(\mathbb{C})$ has an abelian normal subgroup $A$ of index less than $f(n)$. So knowing this, given any finite subgroup $G$ of $GL_n(\mathbb{C})$, we have this normal abelian subgroup, and we can restrict our map to get an $n$ dimensional faithful representation of $A$, so we see that $A$ requires at most $n$ generators, since our representation splits into a sum of one dimensional representations, each with cyclic image.
So then since $A$ has index less than $f(n)$, we can pick lifts of each element in $G/A$, and together these generate $G$, so we have generated $G$ with at most $n+f(n)$ elements, so $\mu_{n,F}$ is finite for $F$ of characteristic zero.
One can also read about the theorem here https://en.wikipedia.org/wiki/Jordan%E2%80%93Schur_theorem, which gives some explicit upper bounds for $f(n)$.