Bourbaki

Definition 2

Tag 0253, permanent. Cite it as [Bourbaki, Tag 0253].

Théories spectrales, chapter I, § 1. Spectres et caractères

Let A be a unital algebra over K and x∈Ax\in A. For every λ∈K\lambda \in K - Sp(xx), one sets

R(x,λ)=(λe−x)−1 R(x, \lambda ) = (\lambda e-x)^{-1}

The mapping of K - Sp(xx) into A given by λ↦R(x,λ)\lambda \mapsto R(x, \lambda ) is called the resolvent of xx.

For fixed xx, the values of R(x,λ)R(x, \lambda ) are pairwise permutable. If λ,μ∈K\lambda , \mu\in K, one has:

(λ−μ)e=(λe−x)−(μe−x) (\lambda -\mu)e= (\lambda e-x)-(\mu e-x)

hence, if λ,μ∈K\lambda , \mu\in K - Sp(xx), one has the relation

(λ−μ)R(x,λ)R(x,μ)=R(x,μ)−R(x,λ)(1) (\lambda -\mu)R(x, \lambda )R(x, \mu) = R(x, \mu)-R(x, \lambda ) \tag{1}

If x,y∈Ax, y\in A and λ∈K\lambda \in K, one has:

y−x=(λe−x)−(λe−y) y-x= (\lambda e-x)-(\lambda e-y)

hence, if λ∈K\lambda \in K - (Sp(xx)∪\cup Sp(yy)), one has the relation

R(y,λ)(y−x)R(x,λ)=R(y,λ)−R(x,λ)(2) R(y, \lambda )(y-x)R(x, \lambda ) = R(y, \lambda )-R(x, \lambda ) \tag{2}

Bourbaki sets the proof directly after the statement, and the corpus does not separate them, so what is above is the statement and its proof together.

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Cites

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Machine translation of the English, by gpt-5.4, gpt-5-6-mini, against glossary version 34. Not checked by a person. That model is a cut down version of the one the rest of this language was written by.