Remark 22

Suppose that the conditions of No. 3 are fulfilled, and in addition that $A$ is equal to $K(i)$ or to the field of quaternions over $K$. Let $u, v$ be two normal endomorphisms of $E$; let $(E_i)_{1 \leqslant i \leqslant r}$ (resp. $(F_j)_{1 \leqslant j \leqslant s}$) be the decomposition of $E$ into a direct sum of pairwise orthogonal subspaces, such that in $E_i$ (resp. $F_j$) there is an orthonormal basis formed of eigenvectors of $u$ (resp. $v$) for the same eigenvalue $\lambda_i$ (resp. $\mu_j$), and that $\lambda_h$ and $\lambda_i$ (resp. $\mu_j$ and $\mu_k$) are not transformed into one another by an inner automorphism of $A$ if $h \neq i$ (resp. $j \neq k$) (cf. prop. 4 and exerc. 9). In order that an endomorphism $w$ of $E$ be such that $uw = vw$, it is necessary and sufficient that for every $j$ ($1 \leqslant j \leqslant s$), $w(F_j)$ be contained in one of the $E_i$ and that the image by $w$ of every eigenvector of $v$ corresponding to the eigenvalue $\mu_j$ be an eigenvector of $u$ corresponding to the eigenvalue $\mu_j$ (which in particular implies that $\mu_j$ and $\lambda_i$ are transformed into one another by an inner automorphism of $A$). Deduce that if this is so, then $u^*w = wv^*$.