Proposition 13

Suppose that $K =\mathbf{C}$. Let $u$ be a compact normal endomorphism of E. For $t\in$ Sp$_s(u)$, let $n_t\geqslant 1$ be the spectral multiplicity of the eigenvalue $t$ of $u$. In order that $u$ be of finite trace, it is necessary and sufficient that the family $(n_tt)_{t\in Sp_s(u)}$ be summable. The trace of $u$ is then the sum of this family.

Let $B = (e_i)_{i\in I}$ be an orthonormal basis of E such that $u$ is diagonal in the basis B (Theorem 1 of IV, p. 149), and let $\lambda = (\lambda_i)_{i\in I}$ be the family of its eigenvalues. Since, for $t\in$ Sp$_s(u)$, the multiplicity $n_t$ is equal to the number of elements $i\in I$ such that $\lambda_i=t$, and since the non-zero elements of $\lambda$ belong to Sp$_s(u)$, the assertion then results from Prop. 12 and the preceding lemma.