Skip to main content
9 events
when toggle format what by license comment
May 6, 2014 at 13:34 vote accept Ozz
May 5, 2014 at 9:37 history edited Ozz CC BY-SA 3.0
added 6 characters in body; edited title
May 5, 2014 at 8:07 answer added Geoff Robinson timeline score: 1
May 5, 2014 at 0:25 comment added Nate Eldredge @Geoff: Indeed, up to change of orthonormal basis, it suffices to take $\theta=\pi/2$.
May 5, 2014 at 0:16 comment added Geoff Robinson I think it was known to Frobenius in the sense that he knew how to reduce to that form directly with real matrices without using complex matrices. If you use complex matrices, it's elementary linear algebra to see that it suffices to do the $2$-dimensional case, which is fairly easy to do directly.
May 5, 2014 at 0:01 comment added Geoff Robinson Yes, I meant to indicate that whenever you have the complex diagonal form of the matrix, arranged so that complex conjugate eigenvalues succeed each other, you can replace each $2 \times 2$ diagonal sub-block corresponding to those eigenvalues with the $2 \times 2$ corresponding scaled rotation matrix. This works for diagonalizable matrices of any size.
May 4, 2014 at 23:55 comment added Ozz Thanks, Geoff! Would you remember a reference for this? That matrix looks like a Givens rotation. Can I use Givens rotations for higher dimensional parametrizations?
May 4, 2014 at 23:45 comment added Geoff Robinson Yes, there is, and it was known to Frobenius if my memory is correct. Replace $2 \times 2$ diagonal matrix with eigenvalues $re^{i\theta}$ and $r e^{-i \theta}$ by the matrix $\left( \begin{array}{clcr} r\cos \theta & r\sin \theta\\-r\sin\theta & r\cos \theta \end{array} \right). $
May 4, 2014 at 23:33 history asked Ozz CC BY-SA 3.0