A non-degenerate subspace of Minkowski space I am trying to show the following equivalence:
Proposition 1. A subspace $V$ of the Minkowski space $\mathbb{R}^{n+1}$ of signature $(n, 1)$ is non-degenerate if  it contains a vector $v$ with $\langle v, v \rangle <0$.
I guess that this proposition is true but couldn't prove it.
 A: On the orthogonal complement $v^\perp$ of $v$ the Minkowski inner product is positive definite (to get the total signature $(n,1)$). The orthogonal complement $V^\perp$ of $V$ is a subspace of $v^\perp$, therefore it's positive definite, therefore $V^\perp\cap V=0$, therefore $V$ is nondegenerate.
A: $G\in \mathbb R^{n+1 \times n+1}$ with $g_{i,j}= \langle \mathbf e_i, \mathbf e_j\rangle$   for standard basis vectors $\mathbf e_r$
Suppose $\dim V = k$ and select basis vectors $\mathbf v_i$ for $V$. Now let $B$ be an $n+1 \times k$ matrix where the jth column is $\mathbf v_j$.  Running 'thin' QR factorization we have $B = QR$ where $R$ is invertible and $Q$ is tall and skinny.
Your form on the subspace is specified by
$B^T G B = R^T \big(Q^T B Q\big)R$ which is congruent to $G':=Q^T B Q$ (hence they have the same signature).  So if $\mathbf x^T B^T G B\mathbf x \lt 0$ for some $\mathbf x$ then $G'$ has a minimum eigenvalue $\lambda_k' \lt 0$.  The eigenvalues of $G'$ (Cauchy) Interlace with those of $G$ hence the second smallest eigenvalue of $G'$ is $\lambda_{k-1}'\geq \lambda_{n}\gt 0$ and thus $G'$ is invertible, so congruent matrix $B^TGB$ is invertible as well and the form is non-degenerate on $V$.
