Remark

When there exists a basis of $E$ such that the matrix of $A$ with respect to this basis has real elements (cf. IV, p. 186, Remark 2), the characteristic equation of $A$ has real coefficients. For every $\mathbf{x} = (\xi_k)_{1 \leq k \leq n}$ of $E$, expressed in the basis under consideration, let $\overline{\mathbf{x}} = (\overline{\xi}_k)_{1 \leq k \leq n}$; the map $\mathbf{x} \mapsto \overline{\mathbf{x}}$ is an antilinear involution of $E$. One knows (Alg., VII) that, if $r_k$ is a non-real root of the characteristic equation, and $E_k$ is the corresponding stable subspace, then $\overline{r}_k$ is a characteristic root having the same multiplicity $n_k$ as $r_k$, and the image $E'_k$ of $E_k$ under the map $\mathbf{x} \mapsto \overline{\mathbf{x}}$ is the stable subspace corresponding to $\overline{r}_k$. One deduces from this that if $\mathbf{u}_j$ ($1 \leq j \leq n_k$) are $n_k$ linearly independent integrals with values in $E_k$, then the $2n_k$ integrals $\mathbf{u}_j + \overline{\mathbf{u}}_j,\ i(\mathbf{u}_j - \overline{\mathbf{u}}_j)$ are linearly independent, and have, with respect to the chosen basis of $E$, components which are real functions of $E$. If $r_k$ is a real characteristic root Remark 2 of IV, p. 186, shows that (with the same notation) there are $n_k$ linearly independent integrals $\mathbf{v}_j$ ($1 \leq j \leq n_k$) with values in $E_k$ whose components are real. That way one obtains a fundamental system of integrals of (19) whose components are all real.