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 function $v_1(t)$ on $(0,1)$ satisfying $A(t)v_1(t)=\lambda_1(t)v_1(t)$ almost everywhere on $(0,1)$ such that $$ v_1 \in L^{\infty}((0,1);\mathbb R^n)?$$

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 $$ v_1 \in L^{\infty}((0,1);\mathbb R^n)?$$

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 $A(t)$ and let $P_1(t)$ be the projection onto its associated eigenspace.

It is classical that $P_1(t)$ is not in general continuous but I would like to know whether we can show that it must be Lebesgue measurable and more precisely if we have $$ P_1 \in L^{\infty}((0,1);\mathbb R^n)?$$

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 function $v_1(t)$ on $(0,1)$ satisfying $A(t)v_1(t)=\lambda_1(t)v_1(t)$ almost everywhere on $(0,1)$ such that $$ v_1 \in L^{\infty}((0,1);\mathbb R^n)?$$