# spectral theorem - why does it only apply to a symmetric matrix?

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.

• What happened to your previous question? – Lord Shark the Unknown Aug 24 '18 at 9:03
• "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. – Lord Shark the Unknown Aug 24 '18 at 9:05
• I think you need to remove some of your "non-" to make this question correct. – Arthur Aug 24 '18 at 9:05
• 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...) – Rahul Aug 24 '18 at 9:09
• 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 9:21