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 $I$.