Proposition 9

If the elements of a matrix $X$ with $m$ rows and $n$ columns belong to a subfield $K_0$ of a field $K$, the rank of $X$ with respect to $K_0$ is equal to the rank of $X$ with respect to $K$.

Let $F_0$ be the right vector $K_0$-space generated by the canonical basis of the right vector $K$-space $E = K_d^m$; by hypothesis the columns of $X$ belong to $E_0$. Let $V_0$ (resp. $V$) be the vector sub-$K_0$-space of $F_0$ (resp. the vector sub-$K$-space of $E$) generated by these columns. Then $V = V_0 \otimes_{K_0} K$ (§ 8, no. 2, Proposition 2) and hence $\dim_K V = \dim_{K_0} V_0$.