Task:
Is the restriction of a Minkowski-form in $\Bbb R^n$ on a vector subspace $U$ with $\dim(U) = n - 1$ also a Minkowski-form?
Solution:
Since a Minkowski-form has the type $(n - 1, 1)$, there is a vector subspace where the form is positive definite, so it can't be a Minkowski-Form.
I don't quite understand this solution. Does this simply follow from the fact that the Minkowsi-form has the following gramian matrix?
$$ \begin{pmatrix} 1 & 0 & \cdots & 0 & 0 \\ 0 & 1 & \cdots & 0 & 0 \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \cdots & 1 & 0 \\ 0 & 0 & \cdots & 0 & -1 \end{pmatrix} $$
The solution is a bit terse, but basically the point is that there is always at least one counterexample $U$. Namely: If you pick a basis where the form has the matrix representation you show, then $U$ can be the space of all vectors with last element $0$ (which, as required, is $(n-1)$-dimensional), and the form restricted to this $U$ is then just the standard (positive definite) inner product, which is not Minkowski.
Or in physical terms: Pick a nonzero timelike vector and consider the vector subspace $U$ of all the vectors orthogonal to it. These vectors are all spacelike, so the form restricted to $U$ is Euclidean rather than Minkowski.