Are matrices with determinant zero a manifold?

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 $\mathbb C$?

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Do you want a differentiable structure, or is a topological manifold good enough? – Potato Feb 22 '13 at 3:48
Any scaling-invariant smooth submanifold $M\subset \mathbb R^m$ that contains $0$ is a linear subspace. So much for differentiable structure. – user53153 Feb 22 '13 at 3:49
@Potato I can't imagine that the question was about producing differentiable structures on a non-smooth subset of a Euclidean space. – user53153 Feb 22 '13 at 3:59
– cats Feb 22 '13 at 4:03
The set of matrices with a fixed rank $k$ in $M_{m\times n}(\mathbb{R})$ is an embedded sub-manifold of $M_{m\times n}(\mathbb{R})$ with co-dimension $(m-k)(n-k)$. (For a proof, see example 8.14 of John M.Lee's Introduction to Smooth Manifolds). So the zero set of determinant of $M_{n}(\mathbb{R})$ is a disjoint union of $n$ embedded sub-manifold, one for each $0 \le k < n$. – achille hui Feb 22 '13 at 5:05

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.)

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Okay, here's what I have so far: Pick $n-1$ linearly independent vectors in $\mathbb{R}^n$: $v_1,v_2,\ldots,v_{n-1}$. Then consider a small $U$ neighborhood of:

$$M_0 = \left(\begin{array}{ccccc} & & & & \\ v_1 & v_2 & \cdots & v_{n-1} & 0 \\ & & & & \end{array}\right)$$

And consider $U \cap X$, where $X$ is our set of matrices that have determinant zero. If $U$ is small enough, for every matrix in $U \cap X$ the first $n-1$ vectors will be linearly independent. That means for $M \in U \cap X$ there exist unique real numbers $a_1(M),a_2(M),\ldots,a_{k-1}(M)$ such that the nth column of $M \in U \cap X$ (let's write it as $v_n(M)$) can be written:

$$v_n(M) = a_1(M)v_1(M) + a_2(M)v_2(M) + \cdots + a_{n-1}(M)v_{n-1}(M)$$

And furthermore it is clear that the $a_i(M)$ depend continuously on $M$ (In fact, the $a_i$ should just be rational functions of the elements of $M$, so they should be smooth as well). Adding in variations of the first $n-1$ vectors, each of which looks like an open subset of $\mathbb{R}^n$, we get that $U \cap X$ is homeomorphic to an open subset of $\mathbb{R}^{n^2 - 1}$

Now, by looking at a neighborhood of another point (the origin?) we should be able to determine that $X$ is not a manifold... though I cannot figure out a proof.

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