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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.2012-05-13
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    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 matrices2012-05-13
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    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 .2012-05-13
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    @GeorgesElencwajg I deleted because I considered post a comment and discuss the problematic points on the argument.2012-05-13
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    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 dimension2012-05-13

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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 $n>2$?2012-05-13
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    Dear @Robert, I have written an Edit addressing your excellent question.2012-05-13
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    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.)2012-05-13
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    Dear @Marc, I have modified the typography according to your preference. I hope you find the text more legible now.2012-05-13
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    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.2012-05-13
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    Dear @Robert, in algebraic geometry the zero-set of $x^2$ is a non-reduced scheme and thus not smooth. This explanation probably just proves in your eyes that algebraic geometers are swindlers and that they don't answer questions in good faith! However in the present case the determinant is a reduced polynomial, i.e. it has no repeated factor (it is actually even irreducible). Hence the smoothness of its zero locus for algebraic geometers coincides with that for differential geometers and you can apply the jacobian criterion to prove non-smoothness.2012-05-13
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    Hmm. $x^8 - y^3$ is an irreducible polynomial whose derivative at $(0,0)$ is $0$, and its zero locus (in ${\mathbb R}^2$) is a differentiable manifold.2012-05-13
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    Dear @Robert, "differentiable manifold" or "smooth manifold" without qualification generally means $C^\infty $ manifold. Your example $x^8−y^3=0$ is only a $C^2$ manifold. I have added an edit to my answer, because I am only sure that a singularity in the algebraic (or analytic) geometry sense implies $C^\infty$ non-smoothness when the base field is $\mathbb C$. So,to be on the safe side, assume the base field is $\mathbb C$ in my preceding comment .2012-05-13
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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 ${\mathbb R}^3$.