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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 $$ A(t_0) v =0?$$

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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 $t = 1/2$.

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