# Matrix smoothly parametrized by t has eigenvalues (0, $\lambda$), eigenvector $v$. Is $\lambda v$ smooth?

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 $$h$$.
• 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. – user32157 Feb 17 '20 at 21:43
• @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. – Iosif Pinelis Feb 17 '20 at 22:50
• 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. – user32157 Feb 18 '20 at 14:47
• Also consider whether $(a|c)$ and $(b|d)$ are the same. They're clearly parallel but could still give opposite sides, for example $a = 1, b = -1, c= -1, d=1$. – user32157 Feb 18 '20 at 14:49