On rational rank of matrices If we have a matrix $M\in\Bbb Z^{n\times k}$ of real rank $r\leq\min(n,k)$, does it mean we could find $r$ rows/columns such that every row/column is a $\Bbb Q$ (not just $\Bbb R$ linear) linear combination of these rows/columns?
 A: It is a fundamental property of systems of linear equations with coefficients in a field$~F$ that if they have any solution over an extension field$~K$ of$~F$, then they already have a solution over$~F$. Indeed the method of Gaussian elimination performed over$~K$ will only involve entries in$~F$. So either it will find an inconsistency in the equations (which will exclude any solutions), or it will find a solution that can be described as a particular solution over$~F$, to which may possibly added any solution of the associated homogeneous system. The general homogeneous solutions can in turn be given as linear combinations of certain homogeneous solutions that again can be taken to be defined over$~F$. The only thing that changes when solving over$~K$ is that those linear combinations may be taken with coefficients in$~K$ rather than$~F$.
So concretely for your problem the question of rank is how many linearly dependent columns can be found. Linear dependence of a certain set of columns means the existence of a non-trivial solution of a certain linear system (expressing that a linear combination of those columns is zero). Since the columns have rational entries, it follows from the above that if a set of columns is linearly dependent over$~\Bbb R$, it is already linearly dependent over$~\Bbb Q$ (and the converse is obvious). Once a maximal linearly independent set of columns is found, every remaining column can be expressed as linear combination of them with rational coefficients, again because doing so can be done by solving a system of linear equations.
Therefore it is indeed true that the rank of the matrix over$~\Bbb Q$ is the same as its rank over$~\Bbb R$.
