Vector perpendicular to timelike vector must be spacelike? 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!
 A: The answer of user1551 is perfectly fine, but I found a highschool level proof that I want to share here:
Since $v$ is time-like, we follow
$$v_1^2> v_2^2+v_3^2+v_4^2.$$
Assume that $\langle w,w \rangle\leq 0$. Then 
$$w_1^2\geq w_2^2+w_3^2+w_4^2.$$
Now by assumption it is $\langle v,w \rangle=0$ and therefore
$$v_1w_1=v_2w_2+v_3w_3+v_4w_4.$$
Taking the square of this equation gives
$$(v_1w_1)^2=(v_2w_2)^2+(v_3w_3)^2+(v_4w_4)^2+2v_2w_2v_3w_3+2v_2w_2v_4w_4+2v_3w_3v_4w_4.$$
For the mixed terms we can use Cauchy's inequality to conclude
$$(v_1w_1)^2\leq(v_2w_2)^2+(v_3w_3)^2+(v_4w_4)^2+(v_2w_3)^2+(w_2v_3)^2+(v_2w_4)^2+(w_2v_4)^2+(v_3w_4)^2+(w_3v_4)^2.$$
On the other hand, we either have that $w_1=0$, and therefore $w=0$ because of the first equation,
or
$$(v_1w_1)^2>(v_2^2+v_3^2+v_4^2)(w_2^2+w_3^2+w_4^2).$$
Expanding the rhs gives
$$(v_1w_1)^2>(v_2w_2)^2+(v_3w_3)^2+(v_4w_4)^2+(v_2w_3)^2+(w_2v_3)^2+(v_2w_4)^2+(w_2v_4)^2+(v_3w_4)^2+(w_3v_4)^2,$$
which is a contradiction to the statement above.
A: 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.
A: 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.
