2
$\begingroup$

I'm not a professional geometer / topologist, so please thanks for your patience :)

Setup

The following questions are the first in a series of steps I'm undertaking in an attempt to break down a bigger question (MO link) into standalone chewable pieces.

So, let $F$ be a field (e.g $\mathbb R$, $\mathbb C$, $\mathbb F_2$, etc.), $n$, $p$, $q$, and $r$ be positive integers and let $F^{n \times p}$ and $Y \in F^{n \times r}$ be fixed matrices. Consider the system of equations in unknown matrices $U \in F^{p \times q}$ and $V \in F^{q \times r}$ $$ \begin{split} X^T(XUV - Y)V^T &= 0 \in F^{p \times q},\\ U^TX^T(XUV - Y) &= 0 \in F^{q \times r}. \end{split} $$

Note that this is a system of polynomial equations in the entries of the matrices $U$ and $V$. Thus the set of solutions to the above system of equations is an algebraic variety $\Omega$ over the field $F$. I'm interested in the geometry and topology of $\Omega$ (number and nature of isolated points, connected components, etc.).

Possible questions

  • What is a good upper bound on the number of isolated points of $\Omega$ ?
  • What is a good upper bound on the average number of isolated points of $\Omega$ when the rows of $X$ and $Y$ are random i.i.d ?

Observations

  • My wild guess is that some kind of [Bézout theorem][2] should be applicable here, but I don't have any experience with that theory.
  • The two matricial equations above have a common factor $X^T(XUV-Y)$.
  • Symmetries: It's easy to see that if $(U,V) \in \Omega$ and $A$ is an invertible $p$-by-$p$ matrix, then $(UA^{-1},AV) \in \Omega$. Thus the mapping $(A, (U, V)) \mapsto (UA^{-1},AV)$ defines a group-action on $\Omega$, of $\operatorname{GL}_p$, the linear algebraic group of invertible $p$-by-$p$ matrices. Furthermore this group-action is regular. See comment section of this related question (MO link).

Maybe these observations can be exploited somehow in the answer ?

Updates

  • A user has remarked that the group-action above implies that the only possible isolated point of $\Omega$ is the origin $(U=0,V=0)$. Consequently it has also been suggested that a better question is counting the number of isolated orbits. So

Further questions

(A) What's the structure / size of the quotient "$\Omega/\operatorname{GL}_p$" ?

(B) More "precisely",

  • How many isolated orbits does $\Omega$ have ?
    • Can we describe / characterize them ?
    • What are their dimensions ?
    • etc.
  • Same questions as above, but for connected components.
  • Same questions as above, but for irreducible components
  • etc.
$\endgroup$
  • 1
    $\begingroup$ Unless I'm misunderstanding, there seems to be a size mismatch. $X^T X W_0 W_1$ is $p_0 \times p_2$, while $X^T Y$ is $p_0 \times p_1$. Is $Y$ supposed to be $n \times p_2$? $\endgroup$ – Zach Teitler Nov 9 '18 at 17:39
  • 2
    $\begingroup$ Thanks for catching. Yes it was a typo. Fixed. $\endgroup$ – dohmatob Nov 9 '18 at 17:51
  • 2
    $\begingroup$ I don't see why the group action preserves the solution set. E.g. consider the term $X^T Y W_1$ in the first equation... But if you did have such a group action, the only possible isolated point is the origin, since that's the only fixed point of this action. $\endgroup$ – Joshua Grochow Nov 9 '18 at 21:25
  • 1
    $\begingroup$ Oh... If you put $W_1^T$ in the first equation your symmetries work, and then my previous comment applies. $\endgroup$ – Joshua Grochow Nov 9 '18 at 21:54
  • 1
    $\begingroup$ Sorry, I meant the second part of my comment applies: with such a symmetry group, the only possible isolated point is the origin. A more interesting question might thus be to ask about isolated orbits, but I suppose that depends on your motivation. $\endgroup$ – Joshua Grochow Nov 10 '18 at 0:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.