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Is there a continuous function $f:\text{Sym}_n \to \mathbb{R}^n$ ($\text{Sym}_n$ is the space of real $n \times n$ symmetric matrices),

such that for every $A \in \text{sym}_n$ : $f(A)$ is an eigenvector of $A$ corresponding to the maximal eigenvalue of $A$. (They are all real since $A$ is symmetric).

If such a function cannot exist, can we do this if restrict ourselves to the (open) submanifold of symmetric positive definite matrices?

Asaf Shachar
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1 Answers1

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No, there isn't (at least assuming $f(A)$ is supposed to always be nonzero and $n>1$). For instance, let $n=2$ and consider the matrices $A_t=\begin{pmatrix} 1+t & 0 \\ 0 & 1\end{pmatrix}$. For $t>0$, $f(A_t)$ must be a multiple of $(1,0)$, and for $t<0$, $f(A_t)$ must be a multiple of $(0,1)$. By continuity of $f$, we get that $f(A_0)$ must be a multiple of both $(1,0)$ and $(0,1)$. But the only way this can happen is if $f(A_0)=(0,0)$.

In fact, even if you restrict to symmetric matrices without repeated eigenvalues, there is still no such $f$. Let us again take $n=2$, identify $\mathbb{R}^2$ with $\mathbb{C}$, and let $A_\theta$ be the matrix which has $e^{i\theta}$ and $e^{i(\theta+\pi/2)}$ as eigenvectors with eigenvalues $2$ and $1$, respectively. Since $f(A_\theta)$ is continuous as a function of $\theta$ and must always be a nonzero multiple of $e^{i\theta}$, the function $g(\theta)= f(A_\theta)/e^{i\theta}$ is a continuous map $\mathbb{R}\to\mathbb{R}\setminus\{0\}$. In particular, $g(\theta)$ must always have the same sign. But $A_{\theta+\pi}=A_\theta$ so $g(\theta+\pi)=-g(\theta)$, so $g$ must change sign. This is a contradiction.

Eric Wofsey
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  • On the other hand, this is the only place where you have a problem. If you avoid repeated eigenvalues and stick to the symmetric case, everything is great. – Ian Dec 01 '15 at 21:39
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    @Ian: That's not true; there is an orientation obstruction even without repeated eigenvalues. I'll add that to my answer in a minute. – Eric Wofsey Dec 01 '15 at 21:41
  • Even in the symmetric case? That's surprising. Can you have problems when the eigenvalues are held fixed? (For instance, can you have a problem as $A$ ranges over $Q diag((1,2,3)) Q^T$ where $Q$ is orthogonal?) – Ian Dec 01 '15 at 21:45
  • Oh, now I see the difficulty, it's the usual one: the map should really be to eigenspaces, not eigenvectors. In your example you rotate the eigenvectors by $\pi$ which puts them back on the same lines as they started on, but turned around. An appropriate quotient operation would get rid of this, I think. – Ian Dec 01 '15 at 22:12