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?
 A: 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 ${\mathbb R}^3$.  
A: 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 $\mathbb C$)
