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Assume $M(n,m)$ denote the vector space of $n\times m$ matrices. Consider this as an affine space. Now consider the subset of matrices with rank exactly $r$. Is this subset an irreducible subvariety? If it is, can we find its dimension?

The only thing I can recall about rank is that we can use minors of a matrix to determine the rank. In this way I guess I can prove the set is locally closed (in the Zarisky topology). So this should be a subvariety. About irreducibility, I guess I should express this subset as the image of some map, but I am not sure. Can I decompose a matrix with rank r by two other matrices?

Of course, the hardest part is the dimension. Totally no idea.

Rodrigo de Azevedo
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  • Here is a similar question but using differential methods: https://math.stackexchange.com/q/518202/16490 – ziggurism Mar 30 '20 at 18:12

2 Answers2

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Here is a proof of the irreducibility of the variety $V_r$ of matrices of rank $r$.
Consider the action of $GL(n)\times GL(m)$ on $M(n,m)$ given by $$(G,H)\bullet A=GAH^{-1}$$ Two matrices matrices $A,B\in M(n,m)$ have the same rank if and only if they are in the same orbit.
Hence if one fixes a matrix $A_r$ of rank $r$ , the variety $V_r$ of matrices of rank $r$ is the orbit of $A_r$ under this action , which means that we have a regular surjective morphism $$GL(n)\times GL(m)\twoheadrightarrow V_r:(G,H)\mapsto GA_rH^{-1} $$ Since $GL(n)\times GL(m)$ is irreducible, so is $V_r$.

Georges Elencwajg
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To find the dimension of our set $V_r$ is not difficult. It is $r(m+n-r)$; cf. for example How to calculate the degrees of freedom of an $r$-ranked matrix with the size being $n\times n$?

EDIT 1. Let $W_r=\{A\mid \det(U_{r+1})=0$ for every $(r+1) \times(r+1) $ submatrix $U_{r+1}$ of $A\}$, be the affine algebraic set of matrices with rank at most $r$. Note that $W_r$ is irreducible in $K^{mn}$ (and $W_{r-1}$ is the singular locus of $W_r$ when $r<n$); consequently, $W_r$ is said to be an affine variety (in fact, the definitions are not really fixed).

$V_r$ is a Zariski open dense subset of $W_r$ and therefore is a quasi-affine variety (or a constructible set or simply a variety).

A Zariski topological space is called irreducible if it cannot be written as a union of two proper closed sets. For this definition, the Zariski open $V_r$ is irreducible. Georges gave a direct proof of this fact.

EDIT 2. Answer to ziggurism. As a simple example, consider $W_{n-1}=\{A=[a_{i,j}]\in M_n;\det(A)=0\}$. If $A$ is a singular point of $W_{n-1}$, then, for every $i,j$,

$\dfrac{\partial \det(A)}{\partial a_{i,j}}=cofactor(a_{i,j})=0$, that is, $A\in W_{n-2}$.

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    So how to prove it is irreducible? –  Oct 20 '15 at 02:44
  • Can you clarify why $W_{r-1}$ is the singular locus of $W_r$? I am under the impression that the variety of rank $r$ matrices is non-singular everywhere. – ziggurism Mar 30 '20 at 19:17
  • @ziggurism , cf. my edit. $W_r$ is the set of matrices with rank $\leq r$. –  Apr 03 '20 at 03:45
  • Yeah, that makes sense. But that description fails in the $r=n$ case, right? the space of rank $n$ matrices is an affine open in the space of all $n\times n$ matrices, and the space of rank at most $n$ matrices is the whole space. So I guess it's only true for $r<n$ that the space of rank at most $r-1$ matrices is the singular locus of the space of rank at most $r$ matrices? – ziggurism Apr 03 '20 at 16:19
  • @ziggurism , yes you are right. –  Apr 03 '20 at 16:22
  • thanks for clarifying – ziggurism Apr 03 '20 at 16:23
  • I tried to write it down precisely, but i got confused again. If $r=n$, then there are no $n+1$ minors, so the map sending each matrix to its $n+1$ degree cofactor matrix is the constant map at the empty matrix, whose derivative vanishes everywhere (I think). More simply, is all of $\mathbb{A}^n$ singular since it is the zero locus of the zero function whose derivative vanishes everywhere? I'm confused but maybe I should write a new question... – ziggurism Apr 04 '20 at 14:15
  • Ok nevermind. $\mathbb{A}^n$ is the zero locus of $0$, but it's not enough for derivative to vanish. Vanishing derivative has to guarantee decreased rank, which doesn't happen in this case. – ziggurism Apr 04 '20 at 16:50
  • @ziggurism , $W_n=M_n$ (the whole set of matrices) is defined neither by an implicit function $f(A)=0$ nor by a parameterization $A=f(u)$; then its singular locus is empty. 1. In the case $f(A)=0$, the singular locus is the set of $A$ s.t. $\nabla(f)(A)=0$. 2. In the case $A=f(u)$, the singular locus is the image of $u$ s.t. $Df_u=0$. and we must add the self intersections ($f(u_1)=f(u_2)$). –  Apr 04 '20 at 17:34
  • For #1, but $M_n$ is defined by an implicit function $f(A) = 0$, it's the locus of the constant function. But it's fine because $\nabla(f)(A)=0$ is only a valid criterion for critical point if $f$ is irreducible, which $f\equiv 0$ is not. For #2, I guess $M_n$ is parametrized by itself, so $f$ is the identity function and $Df_u$ is never zero. Right? – ziggurism Apr 04 '20 at 18:26