Wonder whether anyone has an idea on showing the following or to point out that it is not true:
Let $A(t) \in \Re^{n \times n}$ be differentiable over an interval $I$, and it has a zero eigenvalue for all $t \in I$. Then, there exists an eigenvector $v(t)$ corresponding to the zero eigenvalue of $A(t)$ for $t \in I$ such that $v(t)$ is continuous a.e. on $I$.
Let $I = [0,1]$. Construct the set $A(t) = {\rm{Diag}}(a_{1,1}(t), a_{2,2}(t)) \in \Re^{2 \times 2}$ as follows:
First, let the set $E \subset I$ be the fat Cantor set.
For each $k \geq 1$ in the construction of the set $E$, let the intervals removed at the $k^{th}$ step be $(c_{k,i}, d_{k,i})$, $i = 1, \dotsc, 2^{k-1}$.
For $c < d$, define continuous $f_{c,d}(t)$ and $g_{c,d}(t)$ for $t \in (c,d)$ by \begin{align*} f_{c,d}(t) & = \begin{cases} \frac{2c+d}{d-c} - \frac{3}{d-c}t & , c < t \leq \frac{2c + d}{3} \\ 0 & , \frac{2c+d}{3} \leq t \leq \frac{2d + c}{3} \\ -\frac{2d+c}{d-c} + \frac{3}{d-c}t & , \frac{2d + c}{3} \leq t < d, \end{cases} \\ & & \\ g_{c,d}(t) & = \begin{cases} 0 & , c < t \leq \frac{2c + d}{3} \\ -\frac{2(2c+d)}{d-c} + \frac{6}{d-c}t & , \frac{2c+d}{3} \leq t \leq \frac{c+d}{2} \\ \frac{2(2d+c)}{3} - \frac{6}{d-c}t & , \frac{c+d}{2} \leq t \leq \frac{2d+c}{3} \\ 0 & , \frac{2d+c}{3} \leq t < d. \end{cases} \end{align*}
Define for $t \in I$, $a_{1,1}(t)$, $a_{2,2}(t)$ by \begin{align*} a_{1,1}(t) & = \begin{cases} 1 & , t \in E \\ f_{c_{k,i},d_{k,i}}(t) & , t \in (c_{k,i},d_{k,i}), i = 1, \dotsc, 2^{k-1}, k \geq 1, \end{cases} \\ \\ & & \\ a_{2,2}(t) & = \begin{cases} 0 & , t \in E \\ g_{c_{k,i},d_{k,i}}(t) & , t \in (c_{k,i},d_{k,i}), i = 1, \dotsc, 2^{k-1}, k \geq 1. \end{cases} \end{align*}
We have $a_{1,1}(t)$ and $a_{2,2}(t)$ are continuous, but can be easily adapted to be differentiable, on $I$. Furthermore, define $F \subset I \setminus E$ to be such that \begin{equation*} F = \bigcup_{k \geq 1} \bigcup_{1 \leq i \leq 2^{k-1}} \left[\frac{2 c_{k,i} + d_{k,i}}{3}, \frac{2d_{k,i}+c_{k,i}}{3} \right]. \end{equation*}
For $t \in E$, $0$-eigenvectors of $A(t)$ are of the form $[0, \ast]^T$; for $t \in F$, $0$-eigenvectors of $A(t)$ are of the form $[\ast, 0]^T$; for $t \in I \backslash (E \cup F)$, $0$-eigenvectors of $A(t)$ are of the form $[0, \ast]^T$.
For any $t \in E$. Given any open neighborhood of $t$ in $I$, there exists $t_1 \in F$ in the neighborhood, we see then that any $0$-eigenvector $v(t)$ of $A(t)$ cannot be continuous at $t \in E$, where $m(E) > 0$.
