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$\newcommand{\R}{\mathbf R}$ 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.

caffeinemachine
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    The eigenvectors vary smoothly so long as the eigenvalues are distinct. This can be proven by applying the implicit function theorem to the eigenvalue equation. When the eigenvalues "cross", it is no longer possible to uniquely define the eigenvectors, as anything in the span of the eigenspace will be an eigenvector – Nick Alger Jun 07 '17 at 19:49
  • @NickAlger I am not able to execute your hint. Let us take $V=\R^n$ for simplicity. And we have a smooth map $f:\R\to L(V)$ such that $f(t)$ is self-adjoint for each $t$. Let $t_0$ be such that $f(t_0)$ has $n$ distinct eigenvalues. Let $\lambda:J\to \R$ be a smooth map such that $\lambda(t)$ is an eigenvalue of $f(t)$ for each $t\in J$, where $J$ is a small enough interval centered at $t_0$. Now we have the map $F:S^{n-1}\times J\to \R^n$ defined as $F(x, t)=(f(t)-\lambda(t)I)x$. We know that for each $t\in J$, there are exactly two points $x, x'\in S^{n-1}$ such that $F(x, t)=F(x', t)=0$. – caffeinemachine Jun 08 '17 at 05:59
  • ... (contd.) What we want to show is that $F^{-1}(0)$ is a submanifold of $S^{n-1}\times \R$ of dimension $1$. This would follows if we can show that at each point of $F^{-1}(0)$ we have the rank of $F$ is $n-1$ and that this is the maximum rank that $F$ takes at any point. But I am unable to see how to do this. Do you have another approach in your mind. If possible, can you please post your solution as an answer here. Thanks. – caffeinemachine Jun 08 '17 at 06:01
  • The paper by Nico van der AA is good: http://alexandria.tue.nl/repository/books/616489.pdf – Nick Alger Jun 08 '17 at 13:13
  • @NickAlger Can you please point out where is the relevant information in the paper. I skimmed through the paper but couldn't see any place where my problem was discussed. Thanks. – caffeinemachine Jun 08 '17 at 17:32

2 Answers2

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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 $t_0$.

Calvin Khor
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Andreas Cap
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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.)

user81327
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