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As a by-product of a numerical linear algebra result on structured matrices, I can prove the following result:

For each $m$-dimensional subspace $\mathcal{U}$ of $\mathbb{C}^n$, one can find a $n\times n$ permutation matrix $\Pi$ such that $\mathcal{U}=\operatorname{span} \Pi\begin{bmatrix}I\\\\X\end{bmatrix}$ and all the entries of $X$ satisfy $\left\vert X_{ij}\right\vert\leq 1$.

A self-contained proof is not hard.

The maps $X\mapsto \Pi \begin{bmatrix}I\\\\X\end{bmatrix}$ are local charts that form perhaps the simplest atlas of the $(n,m)$ Grassmannian, so I imagine this could be a natural question for people studying Grassmannians. I'd be surprised if this result were not already known.

Do you have a reference for it? Seeing a new theory and a different approach that produces the same result could give me new insight on the original problem I am studying.

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You could try "Algebraic Geometry: A First Course" by Joe Harris –  Michael Bächtold Mar 17 '11 at 15:02
    
Maybe I should add that the chart you describe without the restriction on the entries are the usual charts on Grassmannians. –  Michael Bächtold Mar 17 '11 at 15:05

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I can't give you a reference, but I can give you a quick proof. There is nothing special about $Gn(n,2n)$, so I'll prove it for $Gr(k,m)$.

Consider the $\binom{m}{k}$ Plucker coordinates. One of them must be the largest; permute basis coordiantes such that $|p_{12\cdots k}| \geq |p_I|$ for every $I$. Work in the coordinate chart that looks like $\left( \mathrm{Id} \ X \right)$, for an $k \times (m-k)$ matrix $X$, we'll number the columns of $X$ from $k+1$ to $m$. Then $X_{ij} = (-1)^{k-i} p_{12 \cdots \hat{i} \cdots k j}/p_{12 \cdots k}$. So all the $|X_{ij}|$ are at most $1$.

By the way, the other ratios $p_I/p_{123\cdots k}$ can be written as some minor of $X$, and vice versa, so you get in addition that every minor of $X$ has magnitude at most $1$.

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Thanks, this is interesting. I was doing something very similar in a specific basis representation, but not recognizing my objects as Plücker coordinates. –  Federico Poloni Mar 18 '11 at 17:23

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