Remark 2

When $A$ is a field, the linear mappings $s$ and $d$ have the same rank. We see here that their matrices $M(s)$ and $M(d)$ have the same rank; indeed, a matrix over $A$ and its transpose have the same rank (chap. II, § 6, No. 7, prop. 3) and, when $\Phi$ is sesquilinear, the equality of the ranks of $R$ over $A$ and of $'R$ over $A^0$ (ibid.) and the fact that $J'$ is an isomorphism of $A^0$ onto $A$, entail the equality of the ranks of $R$ and of $'R^J$ over $A$.