TITLE: Are matrices with determinant zero a manifold?
QUESTION [7 upvotes]: Consider the set of matrices with determinant zero in $M_n(\mathbb R)$, where $n > 1$. Is it a manifold? In fact, is it even a topological manifold?
I would suspect not; but I do not have a proof.
If it is indeed not a manifold, is it perhaps a union of finitely many manifolds? Is there a way to determine them? Does the problem become easier if $\mathbb R$ is replaced by $\mathbb C$?

REPLY [6 votes]: Let $S$ be the set of matrices with determinant $0$. Let $M_{ij}$ be the matrix whose $(i,j)^{\rm th}$ entry is $1$ and all of whose other entries are $0$. Then $\gamma_{ij}(t)=tM_{ij}$ is a curve in $S$ for all $(i,j)$. 
Suppose for the sake of contradiction that $S$ is a manifold. Then $T_0S$ (exists and) contains $\gamma_{ij}'(0)=M_{ij}$ for all $(i,j)$. But the $M_{ij}$ span $T_0M_n(\Bbb{R})$. It follows that $T_0M_n(\Bbb{R})=T_0S$. So $S$ has dimension $n^2$, and thus is open in $M_n(\Bbb{R})$. But this is absurd (e.g., the matrices $\varepsilon \cdot\mathrm{id}$ do not lie in $S$, and come arbitrarily close to $0$). So $S$ is not a manifold.
(Note that this proof is morally very similar to that given in @5PM's link in the comments.)