TITLE: What about $GL(n,\mathbb C)$? Is it open, dense in $M(n,\mathbb C)$?
QUESTION [18 upvotes]: What about $GL(n,\mathbb C)$? Is it open, dense in $M(n,\mathbb C)$?

REPLY [28 votes]: It seems like we agree that the openness of $\operatorname{GL}_n(\mathbb{C})$ is most easily proved by observing that it is the preimage of the open set $\mathbb{C} \setminus \{0\}$ under the continuous function $\operatorname{det}: M_n(\mathbb{C}) \rightarrow \mathbb{C}$.
We are giving different answers for the density, and I have not yet seen the simple linear algebra answer I was expecting to see.  
Here it is: let $M \in M_n(\mathbb{C})$.  It is sufficient to show that there exists $\epsilon > 0$ such that for all nonzero complex numbers $t$ with $|t| < \epsilon$, the matrix 
$M + t I_n$ is nonsingular (because then $M$ is a limit point of this set of nonsingular matrices).  But viewing $t$ as an indeterminate, the determinant of $M+t I_n$ is nothing else than the characteristic polynomial $P(t)$ of $-M$.  Being a monic polynomial of degree $n$, it has $n$ roots $\alpha_1,\ldots,\alpha_n$ (possibly with multiplicity), and we may take $\epsilon = \min_{i \ | \ \alpha_i \neq 0} |\alpha_i|$.  
Note that this argument works verbatim for matrices over $\mathbb{R}$ or any nondiscrete normed field.  In fact, for an arbitrary field $k$, if one works instead with the Zariski topology on $M_n(k)$, then what we really need is for any cofinite subset of $k$ (i.e., the complement in $k$ of some finite set) to be dense in the Zariski topology.  This is true iff $k$ is infinite.  If $k$ is finite, the Zariski topology on $M_n(k)$ is discrete, and the result is false.
Added: A fun application of this "linear algebraic perturbation perspective" is to show that $\operatorname{GL}_n(\mathbb{C})$ is connected (whereas $\operatorname{GL}_n(\mathbb{R})$ is not).  Hint: by general nonsense, it is equivalent to show path-connectedness, so try that instead.  If I am trying to find a path from point $A$ to point $B$ in the complex plane, requiring that my path avoid a finite set of points is not nearly as problematic as it would be on the real line!