Remark 1

When A is commutative, (45 G) reduces to (45 D), and (45 DG) to (45 GD), by means of the formula ${}^t(M'M'') = {}^tM'' \cdot {}^tM'$ (cf. (42)), where the two members are here calculated in A. 2) Let E, F, G be three left modules having finite bases, and $u : E \to F,\ v : F \to G$ A-linear mappings. It follows from (45 G) that one has

Indeed, for whatever $x \in E$,

whence (46).

Recall that, in the case of right modules, one has

III. — We shall henceforth denote by A a ring, by B a ring (resp. a ring endowed with an antiautomorphism J, for which one puts $J' = J^{-1}$), by E a left A-module having a finite basis $(e_i)_{i \in I}$, and by F a right (resp. left) B-module having a finite basis $(f_k)_{k \in K}$. We denote by $(e_i^*)$ and $(f_k^*)$ the dual bases of E* and F*. Unless express mention to the contrary the matrices considered are taken with respect to these bases.

Let G be an (A, B)-bimodule (no 1), Φ a bilinear mapping (resp. a right sesquilinear mapping for J) from $E \times F$ into G, and $R = (\Phi(e_i, f_k))$ the matrix of Φ. Then, for $x \in E$ and $y \in F$, formula (6) of no 1 (resp. (8) of no 2), is written, with the above conventions,

(47) $\Phi(x, y) = {}^t M(x) \cdot R \cdot M(y)$ (resp. $\Phi(x, y) = {}^t M(x) \cdot R \cdot M(y)^J$),

where the products are calculated according to the mappings which define the structure of the (A, B)-bimodule of G ; in particular, if $A = B = G$ (in which case Φ is a form), the products are calculated in A.

Let $E’$ be a left A-module having a finite basis $(e'_s)_{s \in S}$, $F’$ a right (resp. left) A-module having a finite basis $(f'_t)_{t \in T}$, $u : E \to E'$ and $\varphi : F \to F'$ A-linear mappings, and $\Phi'$ a bilinear (resp. right sesquilinear for $J$) mapping from $E' \times F'$ into $G$. Let $\Phi$ denote the inverse image of $\Phi'$ (with respect to $u$ and $\varphi$), $U, V, R, R'$ the matrices of $u, \varphi, \Phi, \Phi'$ with respect to the bases under consideration. We then have

(48) $R = {}^t U \cdot R' \cdot V$ (resp. $R = {}^t U \cdot R' \cdot V^J$),

the products being calculated as in (47). Indeed, for any $x \in E$ and $y \in F$, we have by definition $\Phi(x, y) = \Phi'(u(x), v(y))$, whence, according to (47),

(resp. $'M(x).R.M(y)^J = 'M(u(x)).R'.M(v(y))^J$);

according to (45 G) and (45 D) (resp. (45 G)) and (43) we deduce

(resp. $'M(x).R.M(y)^J = 'M(x).'U.R'.('M(y).'V)^J$ $= 'M(x).'U.R'.V^J.M(y)^J$);

this proves our assertion.

IV. — We suppose here that the rings $A$ and $B$ are equal, and we denote by $\Phi$ a bilinear (resp. right sesquilinear for $J$) form on $E \times F$, and by $R$ its matrix. Let us calculate the matrices of the mappings $s_\Phi$ and $d_\Phi$ associated with $\Phi$, which we shall denote by $s$ and $d$ for simplicity. Since we have $\Phi(x, y) = \langle y, s(x) \rangle = \langle x, d(y) \rangle$ according to (23), No. 6 (resp. $\Phi(x, y) = \langle x, d(y) \rangle = \langle y, s(x) \rangle^J$ according to (24), No. 6), we have $\Phi(e_i, f_k) = \langle f_k, s(e_i) \rangle = \langle e_i, d(f_k) \rangle$ (resp. $\Phi(e_i, f_k) = \langle e_i, d(f_k) \rangle = \langle f_k, s(e_i) \rangle^J$), whence, according to (44) and since $(e_i)$ is the dual basis of $(e_i^*)$ and $(f_k)$ the dual basis of $(f_k^*)$:

(49) $M(d) = R, \ M(s) = 'R$ (resp. $M(d) = R, \ M(s) = 'R^{J'}$).