Corollary 1

Let $u$ be an endomorphism of finite trace of E.

a) The endomorphism $u$ is compact;

b) Let $B = (e_i)_{i\in I}$ be an orthonormal basis of the initial space Ker($u$)$^{\circ}$ of $u$, $C = (f_i)_{i\in I}$ an orthonormal family in E, and $(\alpha_i)_{i\in I}$ a family in $(\mathbf{R}^*_+)^I$ such that $u(e_i) =\alpha_if_i$ for every $i\in I$ (cor. 2 of IV, p. 150). One has

Tr($u$) $=\sum_{i\in I}\alpha_i\langle e_i|f_i\rangle$.

To prove a), one may suppose that $K =\mathbf{C}$ (EVT, V, p. 50 and remark 4 of III, p. 2) and that $u$ is positive (EVT, V, p. 50, def. 8).

From EVT, V, p. 56, cor. 1, there exists an orthonormal basis $B = (e_i)_{i\in I}$ of E such that $u$ is diagonal in the basis B and moreover the family $\lambda$ of eigenvalues of $u$ belongs to $\mathbf{R}^I_+$ and is summable. The family $\lambda$ therefore belongs to $\mathscr{C}_0(I; K)$ (TG, III, p. 38, prop. 1), and consequently the endomorphism $u$ is compact (prop. 2 of IV, p. 148).

Let us prove b). Let $(e_i)_{i\in J}$ be an orthonormal basis of E extending the family B, so that $u(e_i) = 0$ if $i\in J$ - I. Then

Tr($u$) $=\sum_{i\in J}\langle e_i|u(e_i)\rangle =\sum_{i\in I}\langle e_i|u(e_i)\rangle =\sum_{i\in I}\alpha_i\langle e_i|f_i\rangle$.