This exercise is taken from the book Introduction to Lorenztian Geometry: Curves and Surfaces.
Exercise 1.2.13. (Triangles of light). Prove that there are not three vectors $u,v,w\in \mathbb{L}^n$ all of the light type such that $u+v+w=0$ and $\{u,v,w\}$ is linearly independent. Generalize.
Tip. You can use the square matrix of order $k$ with zeros on the diagonal and 1 in the other entries is non-singular if $k\ge 2$.
There is a result that says that: given $u,v \in \mathbb{L}^n$ light-type vectors then $u+v$ and $u-v$ are not light-type and have opposite causal characters. So my question is: How to prove that the sum of a space- or time-type vector with a light-type vector is never the null vector? That is, if $j$ is a space-type vector and $t$ is a light-type vector then $j+t\neq 0$. Also, if $y$ is a time vector then $y+t\neq 0$.
How would the tip's non-singular $k$ order matrix look like?
Thanks for the hel!!!