Bourbaki

Proposition 10

Tag 0764, permanent. Cite it as [Bourbaki, Tag 0764].

Algebra, chapter VII, § 5. Endomorphisms of vector spaces

— Let EE be a vector space of finite dimension nn over a commutative field KK, let uu be an endomorphism of EE, and let χu(X)=∏i=1n(X−αi)\chi_u(X) = \prod_{i=1}^n (X - \alpha_i) be a decomposition into linear factors of its characteristic polynomial (in a suitable extension of KK, cf. V, p. 21). If qq is a polynomial with coefficients in KK, then the characteristic polynomial of q(u)q(u) is given by

(6)

χq(u)(X)=∏i=1n(X−q(αi)) , \chi_{q(u)}(X) = \prod_{i=1}^n (X - q(\alpha_i))\ ,

and its trace and determinant are given by

(7)

Tr⁡(q(u))=∑i=1nq(αi), \operatorname{Tr}(q(u)) = \sum_{i=1}^n q(\alpha_i),
(8)
det⁡(q(u))=∏i=1nq(αi). \det(q(u)) = \prod_{i=1}^n q(\alpha_i).

It is clear that (7) and (8) follow from (6) by virtue of (5). To prove the formula (6), we may assume that K is algebraically closed. We then take a basis for E with respect to which the matrix UU of uu is lower triangular (VIIVII, p. 35, Cor. to Prop. 8); we will make use of the following easy lemma:

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

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Transcribed from the printed edition, 2003, Springer. Not translated.