# Is the set of singular matrices ever a differentiable manifold?

I can see that invertible matrices are a differentiable manifold however I don't know how to show that something is not a differentiable manifold so easily.

Is it ever the case that singular matrices form a differentiable manifold?

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The matrixes with rank at least k form an open manifold, but I don't know how it helps in there. –  checkmath May 13 '12 at 6:39
I interpret your question as asking whether the set of all singular matrices is a submanifold and I have given an answer below. Of course if you take the set of multiples of a singular nonzero matrix you will obtain a line, which is a submanifold, consisting only of singular matrices –  Georges Elencwajg May 13 '12 at 7:03
I want to congratulate the user who posted an answer and deleted it after some commenters pointed at a flaw in it: everybody makes mistakes but not everybody reacts so gracefully . –  Georges Elencwajg May 13 '12 at 7:22
@GeorgesElencwajg I deleted because I considered post a comment and discuss the problematic points on the argument. –  matgaio May 13 '12 at 7:24
I was thinking on this: if we take matirces of form $$\left( \begin{array}{ccc} 0 & \times & \times \\ 0 & \times & \times \\ 0 & \times & \times \end{array} \right)$$ and of form $$\left( \begin{array}{ccc} \times & 0 & 0 \\ \times & 0 & 0 \\ \times & 0 & 0 \end{array} \right)$$ the union of these two tipes would be like linear subspaces of different dimensions intersecting transversally, but I thought it could not be a manifold, because of the unconstancy of dimension –  matgaio May 13 '12 at 7:27

The set $Sing_n(\mathbb R)\subset M_n(\mathbb R)$ consisting of singular matrices of size $n$ is the zero set $Z(det)$ of the determinant $det: M_n(\mathbb R)\to \mathbb R$ .
It is an algebraic cone of degree $n$ in $M_n(\mathbb R)=\mathbb R^{n^2}$ since the determinant is a polynomial of degree $n$ in its variables.
It is thus not smooth at the origin if $n\gt 1$ : an algebraic cone is smooth if and only if it is a linear subspace.

NB The exact same analysis works if $\mathbb R$ is replaced by $\mathbb C$.

Edit Let me try to answer Robert's interesting question and determine in the neighbourhood of which matrices $A\in Sing_n(\mathbb R)$ the singular matrices form a locally closed submanifold of $M_n(\mathbb R)$.

Fix $A\in Sing_n(\mathbb R)$. We have for the total derivative (=Fréchet differential) $D\:\operatorname {det}_A: M_n(\mathbb R)\to \mathbb R$ the formula ( in which $H\in M_n(\mathbb R)$) $$D\:\operatorname {det}_A(H)=\operatorname {Tr}(A^{adj}\cdot H)$$ Since $$D\: \operatorname {det}_A(E_{ij})=\operatorname {Tr}(A^{adj}\cdot E_{ij})=(A^{adj})_{ji}$$ we see that the linear form $D\:\operatorname {det}_A$ is not zero , and thus that $det$ is a submersion, at every $A\in Sing_n(\mathbb R)$ such that $A^{adj}\neq0$ or equivalently such that $rank(A)=n-1$.
Thus $Sing_n(\mathbb R)$ is a locally closed submanifold of $M_n(\mathbb R)$ in the neighbourhood of a matrix such that $rank(A)=n-1$ .

On the other hand, if the matrix $A$ satisfies $rank(A)\leq n-2$, then $A^{adj}=0$ and thus $D\:\operatorname {det}_A:M_n(\mathbb R)\to \mathbb R$ is the zero linear form so that the cone $Sing_n(\mathbb R)$ has a singularity at $A$ as an algebraic or analytic subset.
I am not sure that this prevents $Sing_n(\mathbb R)$ from being a $C^\infty$ manifold at $A$, but I would be surprised if $A$ were a $C^\infty$- smooth point.
However I am sure that over $\mathbb C$ an algebraic singularity at $A$ prevents the variety $Sing_n(\mathbb C)$ from being $C^\infty$- smooth at $A$ (The whole edit is valid if $\mathbb R$ is replaced by $\mathbb C$)

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For $n=2$, $\det$ has nonzero gradient everywhere except $0$, so the nonzero singular matrices do form a submanifold of dimension $3$ in ${\mathbb R}^{4}$. What happens when $n>2$? –  Robert Israel May 13 '12 at 8:58
Dear @Robert, I have written an Edit addressing your excellent question. –  Georges Elencwajg May 13 '12 at 10:27
I had a bit of difficulty reading the answer. I suppose $Ddet$ means the total derivative of the determinant map, a linear map $M_n(\mathbb R)\to\mathbb R$. I think I would have had less difficulty if $det$ were typeset in roman, and this becomes $D\det$. (I think all multi-letter symbols like det, Sing, Tr, adj conventionally use roman type.) –  Marc van Leeuwen May 13 '12 at 10:57
Dear @Marc, I have modified the typography according to your preference. I hope you find the text more legible now. –  Georges Elencwajg May 13 '12 at 11:06
It's not clear to me how the fact that the derivative is $0$ implies a singularity. $f(x,y) = x^2$ has derivative $0$ on its zero-set, but that zero-set is a smooth manifold. –  Robert Israel May 13 '12 at 17:46

In the case $n=2$, consider $C = \{(x_{11},x_{12},x_{21},x_{22}) \in {\mathbb R}^4: x_{11} x_{22} - x_{12} x_{21} = 0\}$. With the change of variables $y=x_{11}+x_{22},z=x_{11}-x_{22},v=x_{12}+x_{21},w=x_{12}-x_{21}$, this becomes $\{(y,z,v,w) \in {\mathbb R}^4: y^2 - z^2 - v^2 + w^2 = 0\}$. The intersection of this with the unit sphere is $\{(y,z,v,w) \in {\mathbb R}^4: y^2 + w^2 = z^2 + v^2 = 1/2\}$, which is the Cartesian product of two circles, i.e. a $2$-torus. But in ${\mathbb R}^3$ a region whose boundary is a $2$-torus is not simply connected. So no neighbourhood of the origin in $C$ is homeomorphic to ${\mathbb R}^3$.

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