Proposition 5

Let u be a hermitian endomorphism of E. The eigenvalues of u belong to K, and there exists an orthonormal basis of E formed of eigenvectors of u.

When A is equal to K(i) or to the quaternion field over K, it suffices, by virtue of prop. 4, to prove the first assertion; now, if x is an eigenvector $\neq 0$ of u and a the corresponding eigenvalue, one has, by virtue of the assumption $u = u^*$,

as $\Phi(x, x)$ is a nonzero element of the center of A, it follows that $a = \overline{a}$, hence $a \in K$. Consequently a hermitian matrix has all its eigenvalues in K. Suppose now that A is equal to K. The matrix M of u with respect to an orthonormal basis of E is then symmetric; it is therefore a hermitian matrix if one considers it as a matrix over K(i). The first part of the proof then shows that this matrix has all its eigenvalues in K, and admits therefore, if $E \neq \{0\}$, nonzero eigenvectors in E. It follows that every stable subspace for u and minimal is necessarily of dimension 1, and the conclusion follows immediately, as in prop. 4.