3 ms·

That doesn't work. It's complex symmetric, but not real symmetric or Hermitian, so you can't use the spectral theorem. This statement here: It acts on R^N b

That doesn't work. It's complex symmetric, but not real symmetric or Hermitian, so you can't use the spectral theorem. This statement here: It acts on R^N by independently scaling the coordinates of an orthogonal basis is precisely the spectral theorem, which you can't use here. [edit] Also, the corresponding claim for Hermitian (and real symmetric) matrices isn't true either. If an eigenvalue of a Hermitian matrix H is negative, then no square root of it can be Hermitian (or real symmetric).

WoahNoun 4y ago
Ah I missed the complex part. Clearly a complex matrix doesn't act on R^N by scaling the coordinates.
- ogogmad 4y ago
  Your incorrect solution is the motivation for my "Another non-solution".
  - WoahNoun 4y ago
    For real symmetric matrices, the n-dimensional coordinate scaling is still a nice visual picture of what's happening in the real case of the spectral theorem.