[math-fun] Elementary (?) matrix question
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.
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
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
From: Henry Baker <hbaker1@pipeline.com> To: math-fun@mailman.xmission.com Sent: Thu, December 31, 2009 1:40:52 PM Subject: [math-fun] Elementary (?) matrix question 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. _______________________________________________ The only other evident fact is that M.M* is positive indefinite, so its eigenvalues are nonnegative. That there may not be any interesting relation between the eigenvalues of M and those of M.M* is suggested by the following example. Let M = [[1,c],[0,1]], M* = [[1,0],[c*,1]]. M has a double eigenvalue 1. M.M* = [[1 + |c|^2,c],[c*,1]] has eigenvalues b +- sqrt(b^2 - 1), b = 1 + |c|^2/2. These eigenvalues are positive and reciprocals, and can be as large (and small) as you please by choice of c. -- Gene
If you replace "eigenvalues" with "singular values", then there is a very direct relationship. This restatement might make more sense in your context (which we were not told). Veit On Jan 3, 2010, at 12:01 PM, Eugene Salamin wrote:
From: Henry Baker <hbaker1@pipeline.com>
To: math-fun@mailman.xmission.com Sent: Thu, December 31, 2009 1:40:52 PM Subject: [math-fun] Elementary (?) matrix question
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. _______________________________________________ The only other evident fact is that M.M* is positive indefinite, so its eigenvalues are nonnegative. That there may not be any interesting relation between the eigenvalues of M and those of M.M* is suggested by the following example. Let M = [[1,c],[0,1]], M* = [[1,0],[c*,1]]. M has a double eigenvalue 1. M.M* = [[1 + |c|^2,c],[c*,1]] has eigenvalues
b +- sqrt(b^2 - 1), b = 1 + |c|^2/2.
These eigenvalues are positive and reciprocals, and can be as large (and small) as you please by choice of c.
-- Gene
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
participants (4)
-
Eugene Salamin -
Gareth McCaughan -
Henry Baker -
Veit Elser