What about $GL(n,\mathbb C)$? Is it open, dense in $M(n,\mathbb C)$?

Solution 1:

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!

Solution 2:

The amount of algebraic geometry needed to prove the denseness of $GL_n({\Bbb C})$ is indeed very minimal.

The topology of $M_n({\Bbb C})$ comes from its obvious identification with ${\Bbb C}^{n^2}$. The complement is the zero set of the determinant which is a polynomial in the $n^2$ entries of the matrix. The zero set of a polynomial cannot have inner points: if a polynomial vanishes on an open ball, its Taylor expansion centered at any point of the ball is identically 0, and by analicity the polynomial vanishes identically. Therefore there are no proper closed subsets of $M_n({\Bbb C})$ containing $GL_n({\Bbb C})$.

Solution 3:

Yes, it is both open and dense. One way to show it is open is by considering the determinant. Alternatively, the set of invertible elements of any Banach algebra is open (and $M(n,\mathbb{C})$ is a Banach algebra if you give it any submultiplicative norm), which follows from the fact that if $a$ is invertible and $\|a-b\|\lt \frac{1}{\|a^{-1}\|}$, then $b$ is invertible. One way to show it is dense is to consider the fact that $a\in M(n,\mathbb{C})$ has only a finite set of eigenvalues.

Solution 4:

For the denseness part, one way to look at is this: A matrix is in $GL_n({\mathbb C})$ iff its rows are linearly independent over ${\mathbb C}$. So what you need to show is that given any matrix $M$ and an $\epsilon > 0$, you can perturb each entry by less than $\epsilon$ (in magnitude) and end out with linearly independent rows.

Or you can appeal to algebraic geometry and say that $det(M) = 0$ defines an algebraic variety which therefore has measure zero. But of course this is pretty advanced :)