Let us consider Minkowski inner product on $\mathbb R^{1+n}$, defined by
$$ \langle v,w \rangle = -v\_0w\_0+\sum\_{j=1}^n v\_j w\_j\quad \,\forall\, v,w \in \mathbb R^{1+n}.$$
We say that a vector $v$ is null if $\langle v,v\rangle =0$. Moreover, given a symmetric matrix $A$, we say that $A$ is null positive definite if $\langle Av,v\rangle >0$ for all non-zero null vectors $v$, and null semi-positive definite if $\langle Av,v\rangle \geq 0$ for all null vectors $v$.


Suppose $A(t)$ is a family of symmetric matrices smoothly depending on $t \in [0,1]$ and suppose that $A(t)$ is null positive definite for all $t \in (0,t\_0)$. Suppose also that $A(t\_0)$ is null semi-positive definite but it is not null positive definite. Does it follow that there exists a null vector $v$ such that
$$ A(t\_0) v =0?$$



A linear algebra question for semi-Euclidean norm
Consider the case $n = 1$.


Let $A = \begin{pmatrix} -1 \\ & -2t\end{pmatrix}$


If $v = (1,1)$, then $Av = (-1,-2t) \neq 0$.
If $w = (1,-1)$, then $Aw = (-1, 2t) \neq 0$.


But $\langle Av,v\rangle = \langle Aw,w\rangle = 1-2t$, and so $A$ is null positive definition for $t \in (0,\frac12)$, null semi-positive definition at $t = 1/2$, but there are no non-trivial null vectors in the kernel of $A$ at $t = 1/2$.


