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Consider the vector space of 2x2 matrices $M_{2x2}$ over the reals. There's a clear isomorphism from the space to $\mathbb{R}^4$. Take a singular matrix in this space, say $\left[\begin{array}{ccc} a & 0 \\ 0 & 0 \end{array}\right]$, $a \in \mathbb{R}$. Taking the set of all matrices of this form clearly forms a subspace, which can be thought of as a line in $\mathbb{R}^4$, yet the set of all singular matrices is not a subspace. My question is thus: in general, do sets of singular matrices of certain forms always have nice geometric intuitions like this? Furthermore, in this space of matrices, can we always form subspaces of singular matrices, and if so, are they always one or two dimensional? Clearly $\left[\begin{array}{ccc} a & b \\ 0 & 0 \end{array}\right]$, $a, b \in \mathbb{R}$ is a two-dimensional subspace consisting entirely of singular matrices in $M_{2x2}$, which can be interpreted as plane in $\mathbb{R}^4$, but $\left[\begin{array}{ccc} a & b \\ c & 0 \end{array}\right]$, $a, b, c \in \mathbb{R}$ is $not$ a subspace of singular matrices. Is there a general theory for subspaces of nxn singular matrices of certain forms in $M_{nxn}$?

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    $\begingroup$ The set of singular matrices is the solution of the equation $ad-bc=0$. It is not a subspace in the sense of vector subspaces, that is true. However, it is a solution of a polynomial equation in four variables. There is a general theory for solutions of polynomial equations called algebraic geometry. $\endgroup$ – Lee Mosher Apr 4 '18 at 15:25
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    $\begingroup$ The determinant is a polynomial in four variables, and the singular matrices form its zero set. It's probably better to think of this in terms of algebraic geometry than linear algebra, but sadly, I don't know anything about that. $\endgroup$ – saulspatz Apr 4 '18 at 15:25
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    $\begingroup$ Singular matrices do not form a vector subspace, but they do form a cone, meaning that any scalar multiple of a singular matrix is also singular. Something that is often done with cones is to go to projective space to study them: here each affine patch of the projective space looks like $\mathbb{R}^3$, so is visualisable; the singular matrices are the zeroes of an equation like $xy=z$, which looks like a saddle. This may or may not be enlightening. $\endgroup$ – Joppy Apr 4 '18 at 15:52
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So I thought about this a lot, and one of the things that I learned is that this set separates $\mathbb{R}^4$ into two connected components (individually corresponding to positive and negative determinant). Fixing $d \neq 0$ gives $a$ as a function of $b$ and $c$, which has a saddle-shaped critical point at the origin. Taking any of the four components to be zero implies that at least one of the other components is zero, so the "degenerate" cases are four planes, individually homeomorphic to $\mathbb{R}^2$. I'm sure one could think more about, say, taking $d$ as a "slider" that naturally induces a homotopy from different planes, and stuff like that, but it doesn't seem too interesting.

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