Re: Matrix equation XA+AX=B

On 11.04.2008 11:01, sanyi wrote:
Thanks for answers.
Since any special matrix class is probably good enough (I'm looking
for a reasonable theorem with "easily" verifiable assumptions),so
from now assume that

A,X,T are self-adjoint matrices, A is invertable and positive
definite.

Does exist such X, if yes, is it unique?

I will only assume A to be positive definite.

Uniqueness:

It suffices to show that if XA + AX = 0, then X = 0. For this let x an eigenvector of A w.r.t. to the eigenvalue mu>0: Ax = mu.x . Then Xx is an eigenvector of A w.r.t. to the eigenvalue -mu unless Xx is null. Since A is assumed to be positive definite, A has only strictly positive eigenvalues, such that Xx is null. Since A is diagonalizable, every vector x is a sum of eigenvectors of A. Hence X = 0.

I think that this argument can be refined such that the assumption on A can weakened to that A has positive eigenvalues only (use Jordan decomposition).

[ Moderator's note:
Yes, in fact a necessary and sufficient condition (over a field in which A has all its eigenvalues) is that the sets of eigenvalues of A and -A are disjoint.
This uses the following facts:
(1) (A-rI)^k X = X (-A-rI)^k
(2) Every vector is a linear combination of generalized eigenvectors
of A, i.e. vectors x such that (A-rI)^k x = 0 for some r and k.
(3) If Au = ru and transpose(v)A = -r transpose(v), then X = u transpose(v) satisfies AX + XA = 0.

- RI ]

Existence:

Consider the linear map f(X):= AX + XA on the space of square matrices MM or symmetric square matrices MM_sym. Then f induces an endomorphism of MM and MM_sym, respectively. Since f is injective (by uniqueness), f is surjective and thus bijective in both cases.

HTH.

Best wishes,
J.


.

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