Lemma 2

For an endomorphism $u$ of $E$ to be such that $u^2 = 1$, it is necessary and sufficient that $\frac{1}{2}(1 - u)$ be a projector in $E$; then $u$ is the difference of the two projectors $\frac{1}{2}(1 + u)$ and $\frac{1}{2}(1 - u)$.

Indeed, in the ring $\mathscr{L}(E)$, the relation $\left( \frac{1}{2}(1 - u) \right)^2 = \frac{1}{2}(1 - u)$ is equivalent to $u^2 = 1$. The rest is trivial.

An endomorphism $u$ of $E$ such that $u^2 = 1$ (which is then necessarily an automorphism of $E$ equal to its inverse) is called an involution. Put $\varphi = \frac{1}{2}(1 - u)$, $U^- = \varphi(E)$, $U^+ = \varphi^{-1}(0)$ ($= \omega(E)$ on putting $\omega = \frac{1}{2}(1 + u)$); it is known that $E$ is the direct sum of $U^+$ and $U^-$ (Chap. VIII, § 1, No. 1), and we have $u(x) = x$ in $U^+$, $u(x) = -x$ in $U^-$. When $A$ is a field and $E$ is finite-dimensional, it follows, since $A$ is of characteristic $\neq 2$, that the only eigenvectors $\neq 0$ of $u$ are the elements ≠ 0 in U^+ or in U^-; they correspond respectively to the eigenvalues + 1 and -1.