Proposition 4

Let u be a normal endomorphism of E. When A is equal to K(i) or to the quaternion field over K, there exists an orthonormal basis (for Φ) of E formed by eigenvectors of u. When A = K, u is semisimple and E is direct sum of stable subspaces for u, pairwise orthogonal, and of dimension 1 or 2.

First examine the case where A is commutative (A = K or A = K(i)). Then the subalgebra B = A[u, u*] of $\mathscr{L}_A(E)$ is commutative since u is normal; it is stable under the mapping $\varphi \to \varphi^*$ by virtue of formulas (32) and (33) of § 1, No. 8. The assertion concerning the case A = K then follows immediately from cor. 2 of th. 2. When A = K(i), this corollary also shows that E is direct sum of vector subspaces $Ax_i$ ($i = 1, \ldots, n$) of dimension 1, pairwise orthogonal and stable for u; if one puts $e_i = (\Phi(x_i, x_i))^{-1/2} x_i$, $(e_i)$ is the orthonormal basis sought.

When A is the quaternion field over K, it will suffice similarly to prove, by virtue of th. 2, that every minimal element of the set of subspaces $\neq \{0\}$ of E stable for u and $u^*$ is of dimension 1. Now such a subspace V necessarily contains an eigenvector $x \neq 0$ of u (*), as is seen by observing that the quaternion field A contains K(i) as an algebraically closed subfield, and by restricting to K(i) the field of

(*) If E is a left vector space over a noncommutative field A, and u an endomorphism of E, one still says that a vector $x \neq 0$ of E is an eigenvector for u if there exists $a \in A$ such that $u(x) = ax$; the scalar a is then called an eigenvalue of u. One will note that, for every $b \neq 0$ in A, the vector $bx$ is an eigenvector for u, and that the corresponding eigenvalue is $bab^{-1}$.

scalars of V. It remains therefore only to show that the eigenvector x of u is also an eigenvector of u*. Put $u(x) = ax$ ($a \in A$); one then has $\Phi(u(x), x) = a\Phi(x, x) = \Phi(x, x)a = \Phi(x, \overline{a}x)$ since $\Phi(x, x)$ belongs to the center of A, and on the other hand $\Phi(u(x), x) = \Phi(x, u^*(x))$; it follows that one has $\Phi(x, u^*(x) - \overline{a}x) = 0$, and one can therefore write $u^*(x) = \overline{a}x + z$, where z is a vector orthogonal to x. One therefore has

Now, since u is normal, one has

Consequently one has $\Phi(z, z) = 0$, which, by assumption, entails $z = 0$ and $u^*(x) = \overline{a}x$. QED.