Bourbaki

Proposition 2

Tag 0644, permanent. Cite it as [Bourbaki, Tag 0644].

Algebra, chapter III, § 8. Determinants

Let MM be a free AA-module with a finite basis (ei)1≤i≤n(e_i)_{1 \leq i \leq n} and vv an endomorphism of MM. For every sequence (xi)1≤i≤n(x_i)_{1 \leq i \leq n} of nn elements of MM,

det⁡(v(x1),…,v(xn))=(det⁡v)det⁡(x1,…,xn). \det(v(x_1), \ldots, v(x_n)) = (\det v) \det(x_1, \ldots, x_n).

If uu is the endomorphism of MM such that u(ei)=xiu(e_i) = x_i for all ii, then

v(xi)=(v∘u)(ei) v(x_i) = (v \circ u)(e_i)

and (5) therefore follows from (2).

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

Cited by

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Cites

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