Let $M$ be a Riemannian manifold and $A_t$ a $(0, 2)$ positive definite tensor defined in $M$ for all $t \in [t_0, t_1)$ and suppose that there exists $p \in M$ and $v \in T_p M$ such that $A(v, v) = 0$ at $(p, t_1)$. Does this imply that there exists a vector $\tilde{v}$ satisfying $A_{ij}\tilde{v}^{j} = 0$?
Context: when one proves the maximum principle for tensors, we define a certain $(0, 2)$ tensor that depends on time like that and it's necessary to prove that $A > 0$ for all $t \in [0, T)$ for a certain $T$. Proceeding by contradiction, we suppose that that's not true, so that there exists a point and instant $(p, t_1)$ such that $A > 0$ for all $t \in [t_0, t_1)$ but $A(v, v) = 0$ at $(p, t_1)$. I get that so far.
But the problem is that $A(v, v) = A_{ij}v^{i} v^{j} \neq A_{ij} v^{j}$. I don't see how $A_{ij}v^{i}v^{j} = 0$ implies that $A_{ij}v^{j} = 0$ (of course, if it were the other way around this would be trivial, but it's not).
From what you have written I'm under the impression that you know that the map $w\mapsto A(w,w) $ has a minimum in $v$ (for fixed $(p, t_1)$).
But then it's derivative in $v$ is zero, and that's just the map $w\mapsto 2 A(v, w)$. So this then implies that $A(v, .) $ is zero.