Proposition 12

Let $B = (e_i)_{i\in I}$ be an orthonormal basis of E. Let $u\in \mathscr{D}_B(E)$ and let us denote by $\lambda = (\lambda_i)_{i\in I}$ the family of its eigenvalues. The endomorphism $u$ is of finite trace if and only if the family $\lambda$ is summable in K. One then has Tr($u$) $=\sum\lambda_i$.

Replacing $u$ by $u_{(\mathbf{C})}$ if necessary, one may suppose that $K =\mathbf{C}$.

Suppose first that $u$ is of finite trace. According to EVT, V, p. 50 and p. 49, formula (25), the family $(\langle e_i|u(e_i)\rangle )_{i\in I}= (\lambda_i)_{i\in I}$ is summable.

Conversely, suppose that the family $\lambda$ is summable.

Each of the families $(\mathscr{R}(\lambda_i)^+), (\mathscr{R}(\lambda_i)^-), (\mathscr{I}(\lambda_i)^+), (\mathscr{I}(\lambda_i)^-)$ is then summable. The endomorphism $u$ is a linear combination of the elements of $\mathscr{D}_B(E)$ having these families as eigenvalues. These elements of $\mathscr{D}_B(E)$ are positive. Since, by definition, the space of endomorphisms of finite trace is generated by the positive endomorphisms of finite trace, one may therefore suppose that $\lambda_i\geqslant 0$ for every $i$. Since $\langle e_i|u(e_i)\rangle =\lambda_i$, the family $(\langle e_i|u(e_i)\rangle )_{i\in I}$ is summable, therefore $u$ is of finite trace (EVT, V, p. 48, lemma 3).

Finally, if $u$ is of finite trace, then according to EVT, V, p. 50, one has

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