4 ms·

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.