# Is a vector space automatically spacelike if it has a basis of spacelike vectors?

I am studying Kerr Spacetime and I am not sure about something used in a proof I am trying to understand.

I am wondering, if you consider a 4-dimensional Lorentzian manifold $$\mathcal{M}$$ and $$X_i \in \mathfrak{X}(\mathcal{M})$$, $$i=1,\dots 3$$ spacelike. Consider the hypersurface $$N$$ where $$t=t_0=const$$ and suppose the vectors $$X_i$$ form a basis of $$T_p(N)$$ at each point. Does this make $$T_p(N)$$ (and therefore $$N$$) spacelike already? Because in the proof they say that the $$X_i$$ are also mutually orthogonal and I am not sure if this is actually needed here.

• (1,0)=(1,2)+(1,-2). May 11, 2019 at 13:59

## 1 Answer

No, e.g. in Minkowski space we can reconstruct timelike vectors as linear combinations of spacelike vectors.

However, orthogonality of a spacelike basis is a sufficient condition to ensure that any linear combination is again spacelike.

• A possibly useful necessary and sufficient condition for three vectors to span a spacelike 3-dimensional subspace of Minkowski space would be that the $3\times3$ matrix of their inner products has all its eigenvalues positive. May 12, 2019 at 0:52