Proposition 10

Let A be a commutative field endowed with an automorphism J, E and F two finite-dimensional vector spaces over A, and $\Phi$ a sesquilinear form for J on $E \times F$. If $\Phi$ is nondegenerate, then its extension $\Phi_{(m)}$ to the m-th exterior powers is nondegenerate, and the inverse form of $\Phi_{(m)}$ is the extension to the m-th exterior powers of the inverse form $\widehat{\Phi}$ of $\Phi$.

Indeed, since $s_\Phi$ and $d_\Phi$ are bijective by assumption (Prop. 6, No. 6), the same is true of their exterior powers (Chap. III, § 5, No. 7). On the other hand the canonical mappings $k_s$ and $k_d$ are bijective (Chap. III, § 8, No. 2, Th. 1). Hence, by virtue of (38), $s_{\Phi(m)}$ and $d_{\Phi(m)}$ are bijective, which proves that $\Phi_{(m)}$ is nondegenerate (Prop. 6, No. 6). In the second assertion we have implicitly identified $\wedge^m F^*$ with $(\wedge^m F)^*$ and $\wedge^m E^*$ with $(\wedge^m E)^*$ by means of the mappings $k_s$ and $k_d$, which are here isomorphisms (loc. cit.). The inverse forms considered in the statement exist since $s_\Phi, d_\Phi, s_{\Phi(m)}, d_{\Phi(m)}$ are bijective (No. 7). Let then $x' = x'_1 \wedge \ldots \wedge x'_m$ and $y' = y'_1 \wedge \ldots \wedge y'_m$ ($x'_i \in E^*, y'_i \in F^*, i = 1, \ldots, m$). By definition of the inverse forms (No. 7) and in view of (38), we have

whence our second assertion.