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Suppose two integers $n< m$, and one is given a matrix subspace of $n\times m$ complex matrices, says $S$.

I am asking for algorithms or conditions which can answer the following question:

Whether every non-zero element of $S$ has rank $n$?

Notice that any $n\times m$ complex matrix has rank less or equal to $n$, here we want the rank of the space is exactly $n$.

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  • $\begingroup$ My first thought was that this looks awfully similar to the Flanders theorem and its generalizations and I was about to ask whether you can allow rank $\leq n$ instead of precisely $n$. Can you? (I guess not). $\endgroup$ Commented Nov 1, 2012 at 11:19
  • $\begingroup$ @Felix, I guess not, since all $n\times m$ matrix has rank less or equal to $n$. $\endgroup$
    – gondolf
    Commented Nov 1, 2012 at 11:23

2 Answers 2

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Your problem is known as the MinRank problem : given a linear subspace $V$ of matrices and an integer $r$, determine the locus of matrices of $V$ of rank less than $r$, or decide if it is empty or not. Over finite fields, the decision problem is known to be NP-Hard.

You may find this paper interesting : Jean-Charles Faugère, Mohab Safey El Din, Pierre-Jean Spaenlehauer, « Computing loci of rank defects of linear matrices using Gröbner bases and applications to cryptology » ( http://dx.doi.org/10.1145%2f1837934.1837984 )

They explore to different formulations of the problem in term of polynomial systems, the first one being the formulation of the previous answer by Dima Pasechnik.

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  • $\begingroup$ @Lierre, thank you Lierre, notice here we consider the rank is $n$ over $n\times m$ matrices space, isn't this constraint helpful? $\endgroup$
    – gondolf
    Commented Nov 1, 2012 at 11:15
  • $\begingroup$ @gondolf — That is certainly a simplifying constraint, but I think the difficulty remains essentially the same. $\endgroup$
    – Lierre
    Commented Nov 2, 2012 at 8:38
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An obvious answer is to look at the $n\times n$ minors. At least one of them must be non-zero for a nonzero element of $S$. Assuming that your subspace is given parametrically, i.e. you have $S=${$\sum t_{ij} S_{ij}$}, for $S_{ij}\in \mathbb{C}^{n\times m}$, $1\leq i\leq n$, $1\leq j\leq m$. Then you have to check that the only solution of the system of equations $$ \det (\sum t_{ij} S_{ij})_J, \quad J\in \binom{[1..m]}{n},$$ where $A_J$ denotes the $n\times n$-minor corresponding to the columns in the set $J$, in $t_{ij}$, which take values in $\mathbb{C}^{n\times m}$, is $(t_{ij})=0$

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  • $\begingroup$ And this one can do (in principle!) using Groebner bases. $\endgroup$ Commented Nov 1, 2012 at 4:16
  • $\begingroup$ @Dima @Mariano, thank you for your answer, this seems very complicated. Is there any simpler characterization? $\endgroup$
    – gondolf
    Commented Nov 1, 2012 at 5:45
  • $\begingroup$ no, that's basically as simple as it gets in general. You can do some ad hoc things like finding invertible $n\times n$ (resp. $m\times m$) matrices $U$ and $V$ so that $U\dot S\dot V$ looks simpler than $S$, and so that the system of polynomial equations gets sparser. $\endgroup$ Commented Nov 1, 2012 at 6:04

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