Can We Smoothly Choose an Eigenvector of a Smoothly Parameterized Self-Adjoint Linear Map

Question

$\newcommand{\R}{\mathbf R}$ Let $V$ be a finite dimensional real inner product vector space, and let $f:\R\to L(V)$ be a smooth map, where $L(V)$ denotes the space of all linear maps mapping $V$ into $V$. Assume that for each $t\in \R$, $f(t)$ is self-adjoint. Thus for each $t$ we know that $f(t)$ has a basis consisting of eigenvectors.

Question. Does there necessarily exist a smooth map $g:\R\to V\setminus\{0\}$ such that $g(t)$ is an eigenvector of $f(t)$ for each $t$?

If at $t_0\in \R$ we have $f(t_0)$ has $\dim V$ distinct eigenvalues, then by the proof of the LEMMA in this post we know that there is a smooth $\alpha:(t_0-\epsilon, t_0+\epsilon)\to \R$ such that $\alpha(t)$ is an eigenvalue of $f(t)$ for each $t\in (t_0-\epsilon, t_0+\epsilon)$, where $\epsilon$ is sufficiently small. Thus what we want to know, as a special case of the above question, is whether the map $(t_0-\epsilon, t_0+\epsilon)\to \mathbb P(V)$ given by $t\mapsto \ker(f(t)-\alpha(t) I)$ smooth or not.

Answer 1

No. Consider $$ M(t) = \begin{cases} \begin{pmatrix} e^{1/t} & 0 \\ 0 & 2 \;e^{1/t} \end{pmatrix} & t < 0, \\[6pt] \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} & t = 0, \\[6pt] \begin{pmatrix} 0 & e^{-1/t} \\ e^{-1/t} & 0 \end{pmatrix} & t > 0. \end{cases} $$ Its matrix elements are smooth. The eigenvectors are $ \{ (1,0), (0,1) \} $ for $ t < 0 $ and $ \{ (1,1), (1,-1) \} $ for $ t > 0 $.

I think it has more to do with the regularity of the path at degenerate points (matrices with degenerate eigenspaces.)

Answer 2

In the case that all eigenvalues are different, one can get a smooth family of eigenvectors. The argument here is that for a diagonalizable linear map on a vector space, the projection onto each eigenspace is given by a polynomial in the initial map. So it $M(t)$ is a smooth family of diagonalizable linear maps with eigenvalues $a_1(t),\dots,a_n(t)$, which are all different for all $t$, then you can write the projection onto the $a_i$-eigenspace as $\pi_i(t):=(\prod_{j\neq i}(a_j(t)-a_i(t)))^{-1}\prod_{j\neq i}(M(t)-a_i(t)\operatorname{Id})$. (In the second factor you have a composition of commuting linear maps, which shows that this indeed vanishes on all eigenspaces except the one for $a_i(t)$ and restricts to the identity there.) Hence given $t_0$, you can simply choose an eignvector $v_0$ for the eigenvalue $a_i(t_0)$. Then for $t$ sufficiently close to $t_0$, $v(t):=\pi_i(t)(v_0)$ will be non-zero and thus an eignvector of $M(t)$ with eigenvalue $a_i(t)$. (Indeed the same argument works in the case of higher multiplicities, as long as the multiplicities are constant around $t_0$.