- #1
ForMyThunder
- 149
- 0
So there's a circular helix parametrized by [tex]\vec x(t)=(a\cos(\alpha t), a\sin(\alpha t), bt)[/tex] and you have the matrix [tex]K[/tex] given in the Frenet-Serret formulas. In the book I'm reading it says that [tex]-\alpha^2[/tex] is the nonzero eigenvalue of [tex]K^2[/tex]. Can someone explain how they know this is? I understand that you can compute the eigenvalues of the matrix to verify this but how can you say this without computation?