Re: Newbie question on Optimization

In article <ef307e3.5@xxxxxxxxxxxxxxxx>, Desmond <denpotXXX@xxxxxxxxx> wrote:

Since my vectors w and h are complex as well as R matrix, would the
implementation and results be any different from the real-valued
assumption you made?

Desmond
-------------------------
In answer to your question yesterday as to whether the Lagrange
multiplier technique would carry over to complex values for R, w, and h,
the answer is yes, provided that R is a Hermitian matrix - that is,
provided R = R'. As I pointed out yesterday, that condition on R is
necessary to ensure that w'*R*w always remain a real-valued quantity, so
that you have a well-defined problem.

However, I have revised the matrix, A, just a little. One can show that
at a minimum (or maximum, or stationary,) value of w'*R*w, we must have

R*w = lambda*h

for some scalar lambda, and of course

h'*w = 1,

where the Lagrange multiplier, lambda, can now be complex-valued.

Hence, define the (n+1) x (n+1) and (n+1) x 1 arrays

A = [R,-h;h',0];
b = [zeros(n,1);1];

where n = length(w). Then w and lambda must satisfy

A*[w;lambda] == b.

Thus the solution to:

x = A\b

gives w = x(1:n) and lambda = x(n+1).

As has been pointed out, this may or may not be a minimum, but if one
exists, it must be this value.

The previous A I gave you will also work, since R+R'= 2*R, and will give
twice the value of lambda, but that doesn't affect the value of w.

(Remove "xyzzy" and ".invalid" to send me email.)
Roger Stafford
.

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