Let $C(t)$ be a symmetric, two-by-two real matrix whose entries are smooth functions of $t \in \mathbb{R}$. Suppose that $C(t)$ point-wise has eigenvalues $\lambda$ and $0$. Then $\lambda(t)$ is a smooth function too (since $\lambda(t)$ is the trace of $C(t)$). However, in general the unit-length eigenvector $v(t)$ corresponding to $\lambda(t)$ is not smooth. The problematic points are wherever $\lambda(t) = 0$ where $v(t)$ is not even well-defined even up to sign.

**Is $\lambda(t) v(t)$ a smooth vector field?**

Since $v(t)$ is only ever defined up to sign, I really mean to ask whether there is a smooth vector field of length $\left|\lambda(t)\right|$ that is point-wise an eigenvector of $C(t)$ of eigenvalue $\lambda(t)$.

Dieci and Eirola, 1999 Theorem 3.3 implies that if $\lambda(t)$ never goes to zero to infinite order, then $v(t)$ is in fact smooth. So we are interested in the case when $\lambda(t)$ goes to zero to infinite order and are hoping that $\lambda(t)$ then goes to zero fast enough to kill off any problems occurring in $v(t)$.