Remark 3

If $M$ and $N$ are right $A$-modules having finite bases $(m_i)$ and $(n_k)$, $\Phi$ a left sesquilinear form for $J$ on $M \times N$ (No. 6, Remark), $s$ and $d$ its associated mappings, and $R = (\Phi(m_i, n_k))$ its matrix, the formulas (26) of No. 6 show that one has

Suppose now that the mappings $s$ and $d$ associated to $\Phi$ are bijective and calculate the matrix $\hat{R}$ of the inverse form of $\Phi$ (No. 7). When $\Phi$ is bilinear, $\Phi$ is the inverse image of $\hat{\Phi}$ relative to the linear mappings $s : E \to F^*$ and $d : F \to E^*$; one therefore has, by virtue of (48) and (49), $R = R . \hat{R} . R$, whence, since $R$ is invertible ($d$ being bijective), $\hat{R} = R^{-1}$. This formula extends to the case where $\Phi$ is sesquilinear, because, if one considers $\Phi$ as a bilinear form on $E \times F^J$, and if one identifies $(F^J)^*$ with $(F^*)^J$ (cf. No. 9), the inverse form of this bilinear form coincides with $\hat{\Phi}$ considered as a bilinear form on $(F^*)^J \times E^*$. In both cases the matrix of the inverse form of $\Phi$ is the inverse of the matrix of $\Phi$.

Let finally $E'$ be a left A-module, $F'$ a right A-module (resp. left A-module), both admitting finite bases $(e'_s)$ and $(f'_t)$; let $\Phi'$ be a bilinear form (resp. sesquilinear form for J) on $E' \times F'$, and let $R'$ be its matrix. Suppose $s_\Phi$ and $d_\Phi$ are bijective. Let $u : E \to E'$ and $\nu : F \to F'$ be linear mappings, $u^* : F' \to F$ and $\nu^* : E' \to E$ their adjoints (No. 8, prop. 7); denote by $U, V, U^*, V^*$ the matrices of $u, \nu, u^*, \nu^*$ with respect to the given bases. One then has

(resp. $U^{*J} = R^{-1} . {}^t U . R', \ {}^t V^* = R'. V^J . R^{-1}$).

Indeed, for all $x \in E$ and $y \in F'$, we have $\Phi'(u(x), y) = \Phi(x, u^*(y))$ (No. 8, def. 10). Whence, when $\Phi$ is bilinear, by virtue of (47), ${}^t M(u(x)) . R' . M(y) = {}^t M(x) . R . M(u^*(y))$; this gives, by virtue of (45 G) and (45 D), ${}^t M(x) . {}^t U . R' . M(y) = {}^t M(x) . R . U^* . M(y)$, whence ${}^t U . R' = R . U^*$ and the first formula announced since, $d$ being bijective, $R$ is invertible. When $\Phi$ is sesquilinear (47) and (45 G) give ${}^t M(x) . {}^t U . R' . M(y)^J = {}^t M(x) . R . {}^t ({}^t M(y) . {}^t U^*)^J$; now, by (43), we have $({}^t M(y) . {}^t U^*)^J = {}^t ({}^t U^{*J} . {}^t M(y)^J)$, whence ${}^t ({}^t M(y) . {}^t U^*)^J = U^{*J} . M(y)^J$; it follows therefore that ${}^t M(x) . {}^t U . R' . M(y)^J = {}^t M(x) . R . U^{*J} . M(y)^J$, whence ${}^t U . R' = R . U^{*J}$, and $U^{*J} = R^{-1} . {}^t U . R'$. The verification of the formulas for $V^*$ is analogous.

Exercises. — 1) Let A be a commutative field, E a vector space over A admitting an infinite countable basis $(e_n)_{n \geqslant 1}$. A bilinear form $\Phi$ is defined on E by setting $\Phi(e_{i+1}, e_i) = 1$ for $i \geqslant 1$, $\Phi(e_k, e_j) = 0$ for $k \neq j + 1$ and $j \geqslant 1$. Prove that the linear mapping $d_\Phi$ associated on the right to $\Phi$ is injective, but that the linear mapping $s_\Phi$ associated on the left to $\Phi$ is not injective.

2) Let E be the $\mathbf{Z}$-module direct sum of $\mathbf{Z}$ and of $\mathbf{Z}/(2)$, and let $E^*$ be its dual (isomorphic to $\mathbf{Z}$). Prove that the bilinear form $(x, x') \to \langle x, x' \rangle$ on $E \times E^*$ is such that the linear mapping associated on the right is injective, but the linear mapping associated on the left is not.