# 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!

-
Do you know Sylvester's Law? – Zach L. Dec 13 '12 at 15:39
Unfortunately I do not. Assuming I did, how could I prove the statement? – Jonas Dec 13 '12 at 15:48
First, I don't think your statement is quite right. There are null vectors $w$ that are not zero and not timelike that satisfy $\langle v, w \rangle = 0$ for all vectors. But, if you could use Sylvester's Law, and were instead trying to show $\langle v,w \rangle = 0$ implies null or spacelike, then assume such a timelike $w$ exists. $w$ would then have to be timelike, giving two orthgonal timelike vectors. Extend these to an orthogonal basis of the space. This basis would then give signature $(n-2,2)$, while we know by Sylvester's law the signature is unique and should be $(n-1,1)$. – Zach L. Dec 13 '12 at 15:54
@ZachL.: The Minkowski inner product is non-degenerate, so there is no null vector. – 23rd Dec 13 '12 at 15:59
@ZhenLin: Maybe my terminology is non-standard, but by saying that $v$ is a null vector, I mean $\langle v,w\rangle =0$ for every $w$. – 23rd Dec 13 '12 at 16:51

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.
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.