2

The real spectral theorem asserts that any symmetric matrix can be decomposed into a composition of rotations, reflections and scaling. Why can't a non-symmetric matrix be represented as such? Are there other types of operations other than rotations, reflections and scaling that explain why non-symmetric matrices are left out of this theorem? I understand that the singular value decomposition says that you can decompose a matrix into 3 other matrices - but the matrices are complex and I don't know if you can interpret a complex matrix as a rotation etc. I apologise if this is a silly question and please let me know if the question requires further clarification.

Christian
  • 453
  • 1
    What happened to your previous question? – Angina Seng Aug 24 '18 at 09:03
  • I deleted it because I decided it was a silly question. I changed my mind! – Christian Aug 24 '18 at 09:03
  • 1
    "The real spectral theorem asserts that any non-symmetric matrix can be decomposed into a composition of rotations, reflections and scaling." That doesn't appear to be any spectral theorem that I know. – Angina Seng Aug 24 '18 at 09:05
  • 1
    I think you need to remove some of your "non-" to make this question correct. – Arthur Aug 24 '18 at 09:05
  • I amended the question and removed 'non-' from first sentence – Christian Aug 24 '18 at 09:07
  • 2
    You could just have undeleted your previous question. Anyway, all the matrices in the singular value decomposition are still real, not complex. (Now I suppose you'll delete your question again...) –  Aug 24 '18 at 09:09
  • OK thanks. My lecture notes states that the matrices are real or complex and not necessarily complex - thus why i was confused. – Christian Aug 24 '18 at 09:12
  • If you want to generate all the matrices an alternative is to use the elementary transformations: scaling of one coordinate, swapping two coordinates (that's a reflection w.r.t a certain hyperplane over the reals) AND shearings/skews. Taking the risk of blowing my own trumpet, take a look at this to see how you get a simple rotation as a composition of scalings and shearings. Similar decompositions are possible in higher dimensions. – Jyrki Lahtonen Aug 24 '18 at 09:16
  • 2
    And, rotation matrices are not symmetric so I don't quite see what you mean. My educated guess is that you want to apply the fact that a symmetric real matrix has an orthonormal system of eigenvectors. But, then you are really conjugating a diagonal matrix (= a composition of scalings) with a rotation rather than just taking arbitrary compositions of all those transformations. – Jyrki Lahtonen Aug 24 '18 at 09:21
  • I mean, I guess that by spectral theorem you refer to that theorem stating that given a symmetric real matrix there is an orthonormal basis consisting of its eigenvectors. I'm sure there is a well-defined question underneath, but I'm not sure exactly which question it is :-) – Jyrki Lahtonen Aug 24 '18 at 14:32

0 Answers0