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](https://epubs.siam.org/doi/pdf/10.1137/S0895479897330182) 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 $v(t)$.



Matrix smoothly parametrized by t has eigenvalues (0, $\lambda$), eigenvector $v$. Is $\lambda v$ smooth?
$\newcommand{\la}{\lambda}
\newcommand{\R}{\mathbb{R}}$Such a smooth field $(\la(t)v(t))$ does not exist in general. 


Indeed, let 
$$C=\begin{bmatrix}f&fg\\fg&fg^2\end{bmatrix},$$
where $f$ and $g$ are the (nonnegative) functions defined in [this answer](https://mathoverflow.net/a/353108/36721). 
The eigenvalues of $C$ are $\la:=f+fg^2$ and $0$.


The eigenvectors of $C$ belonging to the eigenvalue $\la$ that are of length $|\la|[=\la]$ are of the form $w:=hf\sqrt{1+g^2}\,(1,-g)$ for some function $h\colon\R\to\{-1,1\}$. As detailed in the mentioned answer, the first coordinate $hf\sqrt{1+g^2}$ of this vector field $w$ cannot be smooth for any choice of a $\pm$-function $h$. Therefore, the vector field $w$ cannot be smooth for any choice of $h$. 


