# There are no $3$ linearly independent lightlike vectors such that $u+v+w = 0$.

Consider the Lorentz-Minkowski space $$E^n_1$$, also known as $$\mathbb{L}^n$$. I want to prove that there are not lightlike linearly independent vectors $$u, v, w \in E^n_1$$ such that $$u + v + w = 0$$. How to do it? I'm still unfamiliar with the intuition behind such space.

• Does this space have 2 spacelike and one timelike variables? or what if not? – coffeemath Dec 8 '18 at 1:38
• @coffeemath Those are all hypotheses I have. – user71487 Dec 8 '18 at 1:41
• I don't see how this can be true... can't we take one of the vectors from the future light cone, say $\vec{u}$, and the other two to be $-\frac{1}{2}\vec{u}$? Don't we need to say that they are non-coplanar? – RandomMathGuy Dec 8 '18 at 1:41
• @RandomMathGuy The title (but not the body) specifies that the three vectors must be linearly independent. – Travis Willse Dec 8 '18 at 1:44
• @Travis I should learn to read the titles more carefully! – RandomMathGuy Dec 8 '18 at 1:44

Hint Suppose there were. Expand $$0 = [{\bf u} + {\bf v} + {\bf w}] \cdot [{\bf u} - ({\bf v} + {\bf w})]$$ to conclude that $${\bf v} \cdot {\bf w} = 0$$.
Additional hint What is the matrix representation of the bilinear form $$\cdot$$ with respect to the basis $$({\bf u}, {\bf v}, {\bf w})$$?
• Since I wrote this answer, OP edited the question to address the case of $n$-dimensional Minkowski space $\mathbb L^n$ rather than just the special case $n = 3$. The initial hint still leads to a solution for the general case, but of course $({\bf u}, {\bf v}, {\bf w})$ is not a basis of $\mathbb L^n$ for $n \neq 3$. – Travis Willse Dec 8 '18 at 1:48
• Shouldn't I conclude ${\bf u} \cdot {\bf w} = 0$ instead? – user71487 Dec 8 '18 at 1:58
• I don't see how---after all, the expression is symmetric in ${\bf v}$ and ${\bf w}$. – Travis Willse Dec 8 '18 at 2:02
• Expanding gives $0 = {\bf u} \cdot {\bf u} - ({\bf v} + {\bf w}) \cdot ({\bf v} + {\bf w}) = -({\bf v} \cdot {\bf v} + 2 {\bf v} \cdot {\bf w} + {\bf w} \cdot {\bf w}) = -2 {\bf v} \cdot {\bf w}$. – Travis Willse Dec 8 '18 at 2:58
• Not directly anyway. By symmetry of notation, we also have ${\bf u} \cdot {\bf v} = {\bf w} \cdot {\bf u} = 0$, but for $n = 3$ that means all pairs of basis vectors are orthogonal, which implies that $\cdot$ is the zero bilinear form, a contradiction. – Travis Willse Dec 8 '18 at 3:00