I am trying to prove a problem however I am stuck in the middle. So the task is that for a given matrix $A$ with coefficients $a_{ij}$ element of Q, that the rank of this matrix over Q is the same its the rank over R.
My attempt: As there are more solutions in R, the kernel of A over R is larger, so $ker \, A_Q \leq ker \, A_R$ which yields $rank \, A_R \leq rank \, A_Q$. However I can't see how to prove the other way $rank \, A_R \geq rank \, A_Q$.
I have seen seen solutions with tensor products, however I am just asking for a hint with more elementary algebra.
A matrix $A$ having rank $n$ means that there is a non-zero minor determinant of order $n$, but none of a larger order (including the case where no larger order fits inside $A$).
Minors are just determinants of submatrices, and the result of calculating determinants (which are just sums and products of the entries of a matrix) is entirely independent on whether the rational entries in the matrix are interpreted as rational numbers or real numbers.