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From: J on 3 Jun 2010 19:16
I'm well aware of eigs() for standard eigenvalue problems, but is there a function (or a way to use eigs() maybe?) to solve the following eigenvalue problem:

A*x = c*conj(x)

?

Here, A is a 2x2 matrix with complex entires, x is a 2x1 complex "pseudo eigen vector", and c is a constant eigenvalue. Note the only difference from the standard eigenvalue problem is that conj(x) appears on the RHS instead of x.
From: Roger Stafford on 3 Jun 2010 21:27
"J " <philNOSPAMsonj(a)spammenot.hotmailDcom> wrote in message <hu9d3m$fqq$1(a)fred.mathworks.com>...
> I'm well aware of eigs() for standard eigenvalue problems, but is there a function (or a way to use eigs() maybe?) to solve the following eigenvalue problem:
>
> A*x = c*conj(x)
>
> ?
>
> Here, A is a 2x2 matrix with complex entires, x is a 2x1 complex "pseudo eigen vector", and c is a constant eigenvalue. Note the only difference from the standard eigenvalue problem is that conj(x) appears on the RHS instead of x.
- - - - - -
Since you postulated that your pseudo eigenvalue problem might be obtained from the eig or eigs functions, I would guess you are expecting it to possess similar properties. However I don't think that is true.

Suppose we consider the simplest case of all where A is 1 x 1. Let A = 5, for example. In the ordinary eigenvector case the unique eigenvalue would be the same value, 5. However in your case the pseudo eigenvalue can be any complex number with magnitude 5 and for each one there is a corresponding pseudo eigenvector. This is a rather different property from the other. An n x n matrix will always have just n eigenvalues which are solutions to the n-th degree characteristic polynomial. However it appears that your pseudo eigenvalues have no such limitation.

I am sure that it should be possible to completerly analyze the 2 x 2 case and determine just what kinds of freedom there are to its eigenvalues and eigenvectors, but I haven't had time to do so yet.

Roger Stafford
From: Bruno Luong on 4 Jun 2010 02:18
"J " <philNOSPAMsonj(a)spammenot.hotmailDcom> wrote in message <hu9d3m$fqq$1(a)fred.mathworks.com>...
> I'm well aware of eigs() for standard eigenvalue problems, but is there a function (or a way to use eigs() maybe?) to solve the following eigenvalue problem:
>
> A*x = c*conj(x)
>
> ?
>
> Here, A is a 2x2 matrix with complex entires, x is a 2x1 complex "pseudo eigen vector", and c is a constant eigenvalue. Note the only difference from the standard eigenvalue problem is that conj(x) appears on the RHS instead of x.

% There is 2xn pseudo-eigen value:

% Data
A=rand(2)+1i*rand(2);

% Engine
AA =[real(A) -imag(A);
imag(A) real(A)];
C = blkdiag(eye(2),-eye(2));

[xx D] = eig(AA,C);

for n=1:4
% pseudo eigen-vector
x = xx(1:2,n)+1i*xx(3:4,n);
lambda = D(n,n);
% check, this is zero about numerical accuracy
norm(A*x - lambda*conj(x))
end

% Bruno
From: Bruno Luong on 4 Jun 2010 02:40
I would like to note you have to pick only the REAL eigen value of AA/C

The opposite eigen value can be combined, providing the continuum set of complex eigen-vector as Roger has pointed out.

Bruno
From: Roger Stafford on 5 Jun 2010 00:42
"Bruno Luong" <b.luong(a)fogale.findmycountry> wrote in message <hua747$s82$1(a)fred.mathworks.com>...
> I would like to note you have to pick only the REAL eigen value of AA/C
>
> The opposite eigen value can be combined, providing the continuum set of complex eigen-vector as Roger has pointed out.
>
> Bruno
- - - - - - -
In fact these pseudo-eigenvalues have the same characteristics as in the one dimensional case I described earlier. Rather than there being a unique set of n eigenvalues in n dimensions, each possible eigenvalue is arbitrary to a multiplication by a complex number of magnitude one. In other words, only the complex magnitude (abs) of each eigenvalue is uniquely determined for a given matrix. They form an n-dimensional continuum.

I congratulate you Bruno for finding a way to analyze the general case. I was very slowly, painfully arriving at the answer for just two dimensions when your very simple analysis appeared. I still don't know how you discovered that method. It saved me a lot of time.

Roger Stafford
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