Let $T=\{\alpha\_1,\dots,\alpha\_t\}$ be a set of distinct integers.


If $A\in T^{n\times m}$ is an $n\times m$ matrix with entries from $T$ then 



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> does $rank\_{\Bbb R\_+}A=rank\_{\Bbb Q\_+}A$ hold true?
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> 



On ranks of finite element matrices
Yes it's true. The inequality $\le$ is trivial. For the nontrivial inequality, we have some family of nonnegative rational vectors $v\_1,\dots,v\_n$ (by nonnegative vectors I mean with nonnegative coefficients, I apology to use this poor and absurd English terminology), we assume that there exist nonnegative real vectors $w\_1,\dots,w\_d$ such that for every $i$, we can write $v\_i=\sum\_j\lambda\_{ij}w\_j$ with $\lambda\_{ij}$.


We consider this as a system of unknowns (all $\lambda\_{ij}$ and all coefficients of the $w\_j$ being unknowns), with the linear conditions $v\_i=\sum\_j\lambda\_{ij}w\_j$ and the inequalities saying that all these unknowns are nonnegative. This is a system with rational coefficients. Hence it defines a rational polyhedron. Since it has a real point, it has a rational point. Hence we can find $d$ nonnegative rational vectors such that every $v\_i$ is a nonnegative rational combination of those. 


