It is a well-known fact that the orthogonal complement of the span of a timelike vector is a spacelike subspace (i.e all the non-zero vectors are spacelike). This is precisely what makes SR and GR go around, because it says that every observer (whose tangent vector is a (future-directed normalized) timelike vector) can instantaneously set up a reference frame in which they are not moving, i.e instantaneously, they have a notion of time vs space whereby at that instant, they’re moving purely along the time axis, and their notion of space is ‘orthogonal’ to that of time. As such, no null vectors can reside in this orthogonal complement.
Now, let’s prove the following claim (btw your question is not really about differential geometry, it is linear algebra which is why I formulate and prove it as such)
Theorem.
Let $(V,g)$ be a real $(1+n)$-dimensional Minkowski inner product space, with signature $(-,+,\dots, +)$. Suppose $t\in V$ is timelike, and set $T=\text{span}\{t\}$. Then the orthogonal complement $T^{\perp}$ is a spacelike subspace of $V$ (and furthermore we have an orthogonal direct sum decomposition $V=T\oplus T^{\perp}$).
Proof #1: (The ‘Gram-Schmidt’ approach).
Without loss of generality, we may rescale $t$ and assume that $g(t,t)=-1$. One way to argue is to invoke another theorem, which is that you can always “extend to a $g$-orthonormal basis”, which is essentially a Minkowski-version of the usual Gram-Schmidt argument. This means there exist vectors $\{\xi_1,\dots, \xi_n\}$ such that $\{t,\xi_1,\dots, \xi_n\}$ are $g$-orthonormal, in the sense that $g(t,t)=-1$, $g(t,\xi_i)=0$ and $g(\xi_i,\xi_j)=\delta_{ij}$. From this it immediately follows that for any $v\in V$, if $g(t,v)=0$ then $v$ must be in the span of $\{\xi_1,\dots,\xi_n\}$ and hence is either zero or spacelike. Actually, from here is also easy to show that we do indeed have the claimed orthogonal direct sum decomposition $V=T\oplus T^{\perp}$.
Proof #2: (The ‘follow your nose’ approach).
Now, let’s say for whatever reason you don’t like Gram-Schmidt, and suppose the only thing you know about Minkowski vector spaces is that there exists an orthonormal basis $\{e_0,e_1\dots, e_n\}$ with $g(e_{\mu},e_{\nu})=\eta_{\mu\nu}$. Notice I make no assumption about the relationship between $e_0$ and $t$ (unlike the above proof). So, now suppose $t\in V$ is timelike and $v\in V$ is orthogonal to it, $g(t,v)=0$. Relative to the above basis, let us introduce components $t=t^0e_0+t^ie_i$ and $v=v^0e_0+v^je_j$. Now, let us take care of some obvious remarks using casework
- if $v=0$, then ok there’s nothing interesting to say (it certainly belongs to the subspace $T^{\perp}$)
- if $v\neq 0$, note that there must be some $i\in\{1,\dots, n\}$ such that $v^i\neq 0$, because otherwise, $v=v^0e_0$, and by non-zeroness, we must have $v^0\neq 0$, and so $g(t,v)=-t^0v^0\neq 0$ (note that $t^0\neq 0$ because otherwise $t$ cannot be timelike) which contradicts our assumption.
Then, writing out $g(t,v)=0$, we get $-(t^0v^0)=\sum_{i=1}^nt^iv^i.$
So, squaring this and using the Cauchy-Schwarz inequality, we get
\begin{align}
(t^0v^0)^2&=\left[\sum_{i=1}^nt^iv^i\right]^2
\leq\left[\sum_{i=1}^n(t^i)^2\right]\cdot\left[\sum_{i=1}^n(v^i)^2\right]
< (t^0)^2\cdot \sum_{i=1}^n(v^i)^2,
\end{align}
where we get the strict inequality because $t$ being timelike implies $-(t^0)^2+\sum_{i=1}^n(t^i)^2<0$, i.e $\sum_{i=1}^n(t^i)^2<(t^0)^2$; also we showed above that none of the $v^i$ are zero so the sum of the squares is strictly positive. Now, we can divide by $(t^0)^2$ on both sides (again, recall it is non-zero so this is valid) to get
\begin{align}
(v^0)^2&<\sum_{i=1}^n(v^i)^2.
\end{align}
By subtracting $(v^0)^2$ from both sides, this says exactly that $0<g(v,v)$, i.e that $v$ is spacelike. So, this completes our second proof.
All that remains is the proof of the direct sum decomposition; see below for the proof.
Proof #3: (The ‘by definition’ approach).
The third “proof” I offer you is actually a definition/ equivalent characterization of a Minkowski inner product. A Minkowski-inner product on a finite-dimensional real vector space $V$ can be equivalently defined as a bilinear map $g:V\times V\to\Bbb{R}$ such that
- $g$ is symmetric
- $g$ is non-degenerate
- there exists a vector $t\in V$ such that $g(t,t)<0$ (i.e a timelike vector exists) and for every non-zero $v\in V$ such that $g(t,v)=0$, we necessarily have $g(v,v)>0$.
The third condition encapsulates the Lorentzian signature of a pseudo-inner product, so if you had taken this as your definition, then the orthogonality claim would be trivial so, the only thing you’d have to show is the claim about direct sums.
For the direct sum claim, once again suppose for simplicity that $g(t,t)=-1$. Now, for any $v\in V$, let us introduce $\tau=-g(t,v)t\in T$ and $\sigma=v-\tau=v+g(t,v)t$. Observe that
\begin{align}
g(t,\sigma)&=g\bigg(t,v+g(t,v)t\bigg)=g(t,v)+g(t,v)g(t,t)=g(t,v)-g(t,v)=0.
\end{align}
Hence, we have decomposed $v$ into $\tau+\sigma$ whereby $\tau\in T$ and $\sigma\in T^{\perp}$.
Lastly, to show directness, we only have to prove that if $v\in T\cap T^{\perp}$ then $v=0$. But, this is also easy. Since $v\in T$, there must exist some constant $c\in\Bbb{R}$ such that $v=ct$. Next, $v\in T^{\perp}$ implies $g(v,t)=0$, or equivalently, $0=g(ct,t)=cg(t,t)=-c$, and hence $c=0$, which implies $v=0\cdot t=0$. This completes the proof of directness $V=T\oplus T^{\perp}$.
