Bourbaki

Proposition 3

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

Algebra, chapter II, § 10. Matrices

Let E, F be two right (resp. left) A-modules and (ei)i∈I(e_i)_{i \in I}, (fk)k∈K(f_k)_{k \in K} finite bases of E and F respectively. For every linear mapping u of E into F, let M(u)M(u) be the matrix of u with respect to the bases (ei)(e_i) and (fk)(f_k). Then the matrix of tu:F∗→E∗{}^t u : F^* \to E^* with respect to the dual bases (fk∗)(f_k^*) and (ei∗)(e_i^*) is equal to tM(u){}^t M(u).

E is canonically identified with its bidual E∗∗E^{**} and (ei)(e_i) with the dual basis of (ei∗)(e_i^*); then (supposing for example that E and F are right modules)

⟨tu(fk∗),ei⟩=⟨fk∗,u(ei)⟩, \langle {}^t u(f_k^*), e_i \rangle = \langle f_k^*, u(e_i) \rangle,
whence the proposition.

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

Cited by

Cites

Nothing.

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