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Roland Bacher
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Consider the symmetric matrices associated to orthogonal reflections of the Euclidean plane. Since a line comes back to itself after a half-turn, you get a family indexed by $[0,\pi]$ describing a closed path in the set of symmetric matrices. A corresponding eigenvector of eigenvalue $1$ and norm $1$, deformed continuously goes into its opposite. This path canWe get thus not be lifted into a continuous part formed by eigenvectorsclosed path (of symmetric matrices) which does not give rise to closed paths of associated eigenvectors.

Consider the symmetric matrices associated to orthogonal reflections of the Euclidean plane. Since a line comes back to itself after a half-turn, you get a family indexed by $[0,\pi]$ describing a closed path in the set of symmetric matrices. A corresponding eigenvector of eigenvalue $1$ and norm $1$, deformed continuously goes into its opposite. This path can thus not be lifted into a continuous part formed by eigenvectors.

Consider the symmetric matrices associated to orthogonal reflections of the Euclidean plane. Since a line comes back to itself after a half-turn, you get a family indexed by $[0,\pi]$ describing a closed path in the set of symmetric matrices. A corresponding eigenvector of eigenvalue $1$ and norm $1$, deformed continuously goes into its opposite. We get thus a closed path (of symmetric matrices) which does not give rise to closed paths of associated eigenvectors.

Source Link
Roland Bacher
  • 17.9k
  • 3
  • 45
  • 113

Consider the symmetric matrices associated to orthogonal reflections of the Euclidean plane. Since a line comes back to itself after a half-turn, you get a family indexed by $[0,\pi]$ describing a closed path in the set of symmetric matrices. A corresponding eigenvector of eigenvalue $1$ and norm $1$, deformed continuously goes into its opposite. This path can thus not be lifted into a continuous part formed by eigenvectors.