Bourbaki

Remark

Tag 05WQ, permanent. Cite it as [Bourbaki, Tag 05WQ].

Algebra, chapter II, § 10. Matrices

Let A be a ring and σ\sigma an anti-endomorphism of A (no. 3); consider the two following situations:

(GD) E is a left A-module, F a right A-module and uu a Z\mathbf{Z}-linear mapping of E into F such that u(ax)=u(x)σ(a)u(ax) = u(x)\sigma(a) for a∈A, x∈Ea \in A,\ x \in E; in other words, uu is a semi-linear mapping relative to σ\sigma of the right A0A^0-module E into the right A-module F.

(DG) E is a right A-module, F a left A-module and uu a Z\mathbf{Z}-linear mapping of E into F such that u(xa)=σ(a)u(x)u(xa) = \sigma(a)u(x) for a∈A, x∈Ea \in A,\ x \in E; in other words, uu is a semi-linear mapping relative to σ\sigma of the left A0A^0-module E into the left A-module F.

In the two cases, the matrix M(u)M(u) of uu relative to bases of E and F has its elements in A; if these bases are finite, then, for all x∈Ex \in E, we have the respective formulae

(17 GD)

M(u(x))=M(u).σ(M(x)) M(u(x)) = M(u) . \sigma(M(x))
(17 DG)
tM(u(x))=σ(tM(x)).tM(u), {}^t M(u(x)) = \sigma({}^t M(x)) . {}^t M(u),
the products on the two sides being calculated in A. This follows immediately from (26 D) and (26 G) respectively.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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Cites

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Transcribed from the printed edition, 1998, Springer. Not translated.