Remark 18

The conditions of No. 3 are assumed to be satisfied.

a) For every endomorphism $u$ of $E$, let $(e_i)$ be an orthonormal basis of $E$ formed by eigenvectors of $u^*u$, and let $\rho_i$ be the eigenvalue of $u^*u$ corresponding to the vector $e_i$; put $s(u) = (\sum_{i=1}^n \rho_i)^{1/2}$ (square root of $\operatorname{Tr}(u^*u)$ when $A$ is commutative); one has $s(u^*) = s(u)$. If $u, v$ are two endomorphisms of $E$, $U, V$ their matrices with respect to the same orthonormal basis of $E$, show that for the matrix $UV^* + VU^*$, with elements in $K$, one has $(\operatorname{Tr}(UV^* + VU^*))^2 \leq 4s(u)s(v)$ (observe that $(u^* + \lambda v^*)(u + \lambda v)$ is positive hermitian for every $\lambda \in K$); deduce that $s(u + v) \leq s(u) + s(v)$. If in addition $A$ is commutative, one has

(write $u$ as a product by using Ex. 14 b)).

b) Suppose in addition that $A$ is commutative (thus equal to $K$ or $K(i)$). If $H_1, H_2$ are two square matrices hermitian positive, now prove that one has $\operatorname{Tr}(H_1H_2) \geq 0$ (reduce to the case where $H_2$ is a diagonal matrix). Deduce that one has (notations of Exercise 17)

Conclude from these inequalities that for two arbitrary endomorphisms $u, v$ of $E$, one has then $s(uv) \leq s(u)s(v)$.

c) Assuming still that $A$ is commutative, let $\prod_{i=1}^n (x - \lambda_i)$ be the decomposition into linear factors of the characteristic polynomial of an arbitrary endomorphism $u$ of $E$; now prove that one has $\sum_{i=1}^n |\lambda_i|^2 \leq (s(u))^2$ and that, for the two members of this inequality to be equal, it is necessary and sufficient that $u$ be normal (use Exercise 12).