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!

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.
