Lemma 2

Tag 0ASW, permanent. Cite it as [Bourbaki, Tag 0ASW].

Lie Groups and Lie Algebras, chapter V, § 5. Invariants in the symmetric algebra

Let K be a commutative field, V a finite dimensional vector space over K, $S = \bigoplus_{n \geq 0} S_n$ the symmetric algebra of V, s an endomorphism of V, and $s^{(n)}$ the canonical extension of s to $S_n$ . Then, with T denoting an indeterminate, we have in $K[[T]]$

\sum_{n=0}^{\infty} \mathrm{Tr}(s^{(n)}) T^n = (\det(1 - sT))^{-1}.

Extending the base field if necessary, we can assume that K is algebraically closed. Let $(e_1, \ldots, e_r)$ be a basis of V with respect to which the matrix of s is lower triangular, and let $\lambda_1, \ldots, \lambda_r$ be the diagonal elements of this matrix. With respect to the basis $(e_1^{i(1)} \ldots e_r^{i(r)})_{i(1) + \cdots + i(r) = n}$ of $S_n$ , ordered lexicographically, the matrix of $s^{(n)}$ is lower triangular and its diagonal elements are the products $\lambda_1^{i(1)} \ldots \lambda_r^{i(r)}$ . Thus

\mathrm{Tr}(s^{(n)}) = \sum_{i(1) + \cdots + i(r) = n} \lambda_1^{i(1)} \ldots \lambda_r^{i(r)},

and consequently

\begin{align*} \sum_{n=0}^{\infty} \mathrm{Tr}(s^{(n)}) T^n &= \left( \sum_{n=0}^{\infty} \lambda_1^n T^n \right) \left( \sum_{n=0}^{\infty} \lambda_2^n T^n \right) \ldots \left( \sum_{n=0}^{\infty} \lambda_r^n T^n \right) \\ &= (1 - \lambda_1 T)^{-1} (1 - \lambda_2 T)^{-1} \ldots (1 - \lambda_r T)^{-1} \\ &= (\det(1 - sT))^{-1}. \end{align*}

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

Cited by

Lemma 3, § 5

Cites

Nothing.