Let $A(t)$ be a symmetric $n\times n$ matrix that continuously depend on $t\in [0,1]$. Let $\lambda_1(t)$ stand for the smallest eigenvalue for $A(t)$.

**Question.** Does there exist a Lebesgue measurable vector $v_1(t)$ on $(0,1)$ that is not zero for almost every t and satisfies $A(t)v_1(t)=\lambda_1(t)v_1(t)$ almost everywhere on $(0,1)$ such that
$$ v_1 \in L^{\infty}((0,1);\mathbb R^n)?$$