Every Lorentz form admits an ordered basis with matrix diag(-1,1,1,1)

By [L1] the symmetric bilinear form $\eta$ on the finite-dimensional real space $V$ is congruent to exactly one normal form $\mathrm{diag}(I_p,-I_q,0_r)$ with $p+q+r=\dim V=4$, and by [L2] its inertia is $(3,1,0)$; hence $(p,q,r)=(3,1,0)$ and there is a basis $v_0,v_1,v_2,v_3$ of $V$ with $\eta (v_i,v_j)=\mathrm{diag}(1,1,1,-1{)}_{ij}$ for all $i,j$.

Define the reordering $e_0:=v_3$ and $e_1:=v_0$, $e_2:=v_1$, $e_3:=v_2$. Then $(e_0,e_1,e_2,e_3)$ is an ordered basis of $V$, and evaluating the displayed matrix gives $\eta (e_0,e_0)=-1$, $\eta (e_1,e_1)=\eta (e_2,e_2)=\eta (e_3,e_3)=+1$ and $\eta (e_a,e_b)=0$ for $a\mathrm{\ne }b$, that is, $\eta (e_a,e_b)=\mathrm{diag}(-1,1,1,1{)}_{ab}$: the required ordered basis exists.

Now let $(f_0,f_1,f_2,f_3)$ be any ordered basis with $\eta (f_a,f_b)=\mathrm{diag}(-1,1,1,1{)}_{ab}$. Its matrix is diagonal with diagonal entries $-1,1,1,1$, so in this diagonalizing basis the numbers of positive, negative and zero diagonal entries are $(3,1,0)$; by the uniqueness clause of [L1] the triple $(3,1,0)$ is intrinsic and equals the inertia of $\eta$ given in [L2]. In such a basis exactly one vector, $f_0$, is timelike, because $\eta (f_0,f_0)=-1<0$, and the remaining three are spacelike, because $\eta (f_i,f_i)=+1>0$ for $i=1,2,3$; and this count is exactly the inertia $(3,1,0)$.