I seem to recall that if there aren't any repeated roots, the Vandermonde matrix diagonalizes a companion matrix. Of course these Vandermonde matrices aren't unitary in general. I don't know if companion matrices are normal or not. The 2x2 & 3x3 examples I tried would seem to indicate that they aren't -- at least in general. At 06:56 PM 12/31/2009, Gareth McCaughan wrote:
On Thursday 31 December 2009 21:40:52 Henry Baker wrote:
I hate to be so dense about linear algebra, but here's another matrix question:
Given an _arbitrary_ square matrix M with complex entries, we can compute the matrix product M.M', where M' is the conjugate transpose of M. Clearly, M.M' is Hermitian, so all of its eigenvalues are real.
Question: Is there any interesting relationship between the eigenvalues of M.M' and those of M ?
There's an obvious one from the fact det(M.M')=det(M)det(M').
Also, trace(M.M') is the sum of the squares of the absolute values of the entries of M.
Well, if M is normal -- i.e., commutes with its conjugate transpose -- then it's diagonalizable by a unitary transformation, which readily gives us that the eigenvalues of M.M' are the squared absolute values of the eigenvalues of M.
I don't know whether that holds whenever M is diagonalizable at all; probably there's either a simple proof or a simple counterexample, but I too am dense about linear algebra.
-- g