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 $v(t)$.


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

  • $\begingroup$ 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
  • 1
    $\begingroup$ @user32157 : Now your question has been fully answered. $\endgroup$ – Iosif Pinelis Feb 20 '20 at 1:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.