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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 $v(t)$.

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$\newcommand{\la}{\lambda} \newcommand{\R}{\mathbb{R}}$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. 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 $h$.

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  • $\begingroup$ Thanks for working on this. I don't believe this answers the question I'm asking. My $C$ is symmetric and so there is no counter-example with $\lambda = 1$ a constant since in that case the Schur decomposition is smooth and the eigenvector itself is smooth (eg: Prop 2.4 in the paper I linked in the question). Also, I do not believe that your $c$ is smooth at $t = 0$ - it seems to have a $sin(1/t)/2$ term. Please correct me if I am mistaken. $\endgroup$ – user32157 Feb 17 '20 at 21:43
  • $\begingroup$ @user32157 : Indeed, I missed the fact that $c$ was not continuous at $t=0$. I have now rewritten the answer completely, proving that one can choose a continuous field $(\lambda(t)v(t))$, even without assuming the matrix $C(t)$ to be symmetric. This choice seems unique, up to a global sign change. So, if there is a completely smooth version of your field, it must be this, again up to a global sign. However, to show the complete smoothness, it seems lots of calculations will be needed. $\endgroup$ – Iosif Pinelis Feb 17 '20 at 22:50
  • $\begingroup$ I think this is correct, but continuity of the vector field is not the problem; I'm concerned about smoothness. Even differentiability is doable (though showing that's it's even continuously differentiable has stumped me) by the last remark in my question and considering just the points where $\lambda(t)$ goes to zero to infinite order. A bounded function times a function going to zero to infinite order also goes to zero to infinite order, hence has zero derivative at that point. $\endgroup$ – user32157 Feb 18 '20 at 14:47
  • $\begingroup$ Also consider whether $(a|c)$ and $(b|d)$ are the same. They're clearly parallel but could still give opposite sides, for example $a = 1, b = -1, c= -1, d=1$. $\endgroup$ – user32157 Feb 18 '20 at 14:49
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    $\begingroup$ @user32157 : Now your question has been fully answered. $\endgroup$ – Iosif Pinelis Feb 20 '20 at 1:14

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