Suppose $u\in\Hom(E)$ is a self-adjoint linear operator on a Lorentz space $E$. If $u(L)\sbe Z$, then there is an orthonormal basis $x_j$ and real numbers $\l_j$ such that $u(x_j)=\l_j x_j$. In particular $u$ is diagonalizable.
Let $e$ be a unit vector such that $Z=C(e)\cup C(-e)$. W.l.o.g. we may assume that $u(e)\in C(e)$ otherwise consider $-u$. Put
$$
K\colon=\{x\in Z\cup L:\,\la x,e\ra=-1\}
$$
and define $F:K\rar E$ by
$$
F(x)\colon=\frac{u(x)}{-\la u(x),e\ra}
$$
Then $F(x)\in K$ because $u(x)\in Z$ and thus $\la u(x),e\ra\neq0$ and $\la F(x),e\ra=-1$. We furnish $E$ with the topology defined in exam. Then $K$ is compact and convex and $F:K\rar K$ is continuous. By Brouwers fixed point Theorem we can find a point $x_0\in K$ such that $F(x_0)=x_0$, i.e. $u(x_0)=-\la u(x_0),e\ra x_0$. Since $u(L)\sbe Z$ we conclude that $x_0\in C(e)$ and thus $x_0$ is a time-like eigen vector of $u$ with eigen value $\l_0\colon=-\la u(x_0),e\ra > 0$. Finally $u:x_0^\perp\rar x_0^\perp$ is a self-adjoint linear operator on the euclidean space $x_0^\perp$. Hence there is an orthonormal basis $x_1,\ldots,x_n$ of $x_0^\perp$ and $\l_1,\ldots,\l_n\in\R$ such that $u(x_j)=\l_jx_j$.