Let $V\subset M_{m,n}(\mathbb{C})$ be a $n$-dimensional subspace ($m\geq 2$). Is there $X\in V$ such that $1\leq \operatorname{rank} X\leq m-1$?
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$\begingroup$ Not always. For example, let the three $3\times 3$ matrices with a single $1$ (and zeros otherwise) somewhere in the first row span a three-dimensional subspace. $\endgroup$– Christian RemlingCommented Mar 9 at 0:51
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2$\begingroup$ I believe your question would have been better suited for math.stackexchange.com . $\endgroup$– Christian RemlingCommented Mar 9 at 0:54
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$\begingroup$ As for your edited version, how about $X=0$ ? $\endgroup$– Christian RemlingCommented Mar 9 at 0:59
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$\begingroup$ I made a little change of the problem. $\endgroup$– user509119Commented Mar 9 at 1:03
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1 Answer
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Yes. If $n<m$ this is true since all nonzero $m \times n$ matrices have rank between $1$ and $m-1$ (uh unless $n=0$ I guess since $V$ then contains no matrices with positive rank).
If $n \geq m$, then matrices of rank $\leq m-1$ are a closed subset of $M_{n,m}$ containing zero of codimension $n+1-m$. The intersection of this set with $V$ contains the zero point, and thus is nonempty, and being the nonempty intersection of a dimension $n$ variety and a codimension $n+1-m$ variety has dimension $n- (n+1-m) \geq m-1 >0$.
Thus this intersection contains some nonzero point, which must have rank between $1$ and $m-1$, as desired.
The optimal lower bound is $\max( n+2-m, 1)$. Below this lower bound, a generic linear subspace does the trick.
