Bourbaki

Lemma 2

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

Lie Groups and Lie Algebras, chapter III, § 6. Real and complex Lie groups

*Let E be a complex Banach space, u an element of L(E)\mathscr{L}(E), S the spectrum of u in L(E)\mathscr{L}(E) (Spectral Theories, Chapter I, § 1, no. 2) and f a holomorphic complex function on an open neighbourhood Ω\Omega of S. Suppose that f admits in Ω\Omega only a finite number of distinct zeros z1,…,znz_1, \ldots, z_n, of multiplicities h1,…,hnh_1, \ldots, h_n. Then Ker f(u)\mathrm{Ker}\ f(u) is the direct sum of the Ker(u−zi)hi\mathrm{Ker}(u - z_i)^{h_i} for 1⩽i⩽n1 \leqslant i \leqslant n.

(For the definition of f(u)f(u), see Spectral Theories, Chapter I, § 4, no. 8.)

There exists a holomorphic function g on Ω\Omega, everywhere non-zero, such that f(z)=(z−z1)h1…(z−zn)hng(z)f(z) = (z - z_1)^{h_1} \ldots (z - z_n)^{h_n} g(z). Then g(u)g−1(u)=g−1(u)g(u)=1g(u)g^{-1}(u) = g^{-1}(u)g(u) = 1 and hence Ker f(u)=Ker∏i=1n(u−zi)hi\mathrm{Ker}\ f(u) = \mathrm{Ker} \prod_{i=1}^n (u - z_i)^{h_i}. Considering Ker f(u)\mathrm{Ker}\ f(u) as a C[X]\mathbf{C}[X]-module by means of the external law (h,x)↦h(u)x(h, x) \mapsto h(u)x for h∈C[X]h \in \mathbf{C}[X], x∈Ker f(u)x \in \mathrm{Ker}\ f(u), we see that Ker f(u)f(u) is the direct sum of the Ker(u−zi)ni\mathrm{Ker}(u - z_i)^{n_i}, using Algebra, Chapter VII, § 2, no. 1, Proposition 1.

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

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