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Let $n>n'\gg m$ and $V$ be a subspace of $\mathbb{C}^n$ of dimension $n'$. I am trying to characterize the set $X$ of $m\times n$ matrices $A=(a_{ij})$ satisfying $\ker(A)\not\subseteq V$, that is, $A\mathbb{x}=0$ admitting a solution not in $V$.

I think $X$ is the image of a closed subscheme of $(\mathbb{P}^n\setminus \mathbb{P}^{n'})\times\mathbb{A}^{m\times n}$ (cut out by $A\mathbb{x}=0$) under the projection to the second factor $\mathbb{A}^{m\times n}$. As $n>n'\gg m$, a generic matrix $A$ lies in $X$. So $X$ is dense in $\mathbb{A}^{m\times n}$. And it is constructible by Chevalley's theorem on constructible sets (is it open in $\mathbb{A}^{m\times n}$?). I am wondering if it is possible to explicitly determine the system of polynomial equations cuting out the closure of $\mathbb{A}^{m\times n}\setminus X$. For simplicity just assume the subspace $V$ is spanned by the first $n'$ coordinate vectors, if it helps.

Thanks!

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    $\begingroup$ You wrote "ker$(A)\supsetneq V$, that is, $Ax=0$ admitting a solution not in $V$." These are not equivalent; did you mean ker$(A)\not\subseteq V$? Or did you really intend that the kernel should include all of $V$? $\endgroup$ Apr 30, 2013 at 12:39
  • $\begingroup$ Oops. Thanks for pointing it out. Fixed. $\endgroup$
    – Zeyu
    Apr 30, 2013 at 13:11

2 Answers 2

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Denote by $\newcommand{\bC}{\mathbb{C}}$ $L(\bC^m\to \bC^n)$ the space of linear operator $\bC^m\to\bC^n$. Then $X$ is the complement in $L(\bC^m\to\bC^n)$ of the vector subspace

$$ \bigl\lbrace A\in L(\bC^m\to\bC^n);\;\; Av=0,\;\;\forall v\in V\bigr\rbrace\subset L(\bC^m,\bC^n). $$

In particular, $X$ is open in $L(\bC^m,\bC^n)$ (as the complement of a closed set) and constructible.

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  • $\begingroup$ I apologize but as pointed out by Andreas in the comment the original question has a typo (fixed now). The complement of $X$ is $\bigl\lbrace A\in L(\bC^m\to\bC^n);\;\; Av\neq 0,\;\;\forall v\not\in V\bigr\rbrace\subset L(\bC^m,\bC^n)$. $\endgroup$
    – Zeyu
    Apr 30, 2013 at 13:14
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Write $\mathbb C^n = V \oplus W$ and $A=A|_V + A|_W$. Then $A\in X$ iff

  • $A|_W$ has kernel $\ne 0$ or $Im(A|_V)\cap Im(A|_W)\ne 0$.
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