Vector perpendicular to timelike vector must be spacelike?

Question

Given $\mathbb{R}^4$, we define the Minkowski inner product on it by $$ \langle v,w \rangle = -v_1w_1 + v_2w_2 + v_3w_3 + v_4w_4$$ We say a vector is spacelike if $ \langle v,v\rangle >0 $, and it is timelike if $ \langle v,v \rangle < 0 $.

How can I show that if $v$ is timelike and $ \langle v,w \rangle = 0$ , then $w$ is either the zero vector or spacelike? I've tried to use the polarization identity, but don't have any information regarding the $\langle v+w,v+w \rangle$ term in the identity.

Context: I'm reading a book on Riemannian geometry, and the book gives a proof of a more general result: if $z$ is timelike, then its perpendicular subspace $z^\perp$ is spacelike. It does so using arguments regarding the degeneracy index of the subspace, which I don't fully understand. Since the statement above seems fairly elementary, I was wondering whether it would be possible to give an elementary proof of it as well.

Any help is appreciated!

Answer 1

Let $\langle v,v\rangle=-\lambda^2$. Normalize it by $\frac1\lambda$, we get $\langle v,v\rangle=-1$. Hence we can extend $\{v\}$ to a "orthonormal" basis $\{v,\,u_1,u_2,u_3\}$ of $\mathbb{R}^4$ such that $\langle u_i, u_i\rangle=1$ and $\langle v, u_i\rangle=\langle u_i, u_j\rangle=0$ for every $i\not=j$ (see here for the reason.) Now the rest is trivial.

Answer 2

The accepted answer by @user1551 is certainly good, but an intuitive physical explanation may be needed, I think.

A timelike vector in special relativity can be thought of as some kind of velocity of some object. And we can find a particular reference frame in which the object is at rest, i.e. with only time component non-zero. With appropriate normalization, the coordinate components of the timelike vector $v$ are $$ v=(1,0,0,0) $$ which means $v$ is actually the first basis vector of this reference frame. And the other three basis vectors were already there when we specified this frame. So, "extending the timelike vector $v$ to an orthonormal basis" physically means a choice of inertial reference frame.

What then follows is trivial. Since $v$'s only non-zero component is the time component and $\langle v,w \rangle=0$, $w$'s time component must be zero. Then it's either the zero vector or spacelike.