✔ 最佳答案
=>
for all positive definite C
A^TB+BA=-C has a positive definite B
i.e.
for all column vector v, where Cv is a well defined product
for all C such that v^TCv>=0
A^TB+BA=-C has a B such that w^TBw>=0 for all w, where Bw is well defined product
let Au = ku, where k is a complex eigen value of A and u is complex eigenvector of A
u^TA^TBu+u^TBAu = - u^TCu <=0 ; since C is positive definite
(Au)^T Bu+ u^TB k u <=0
(ku)^T B u + k u^TBu <=0
(k*+k) u^T B u <=0 ; k* is complex conjugate of k
2 Re[k] u^T B u <=0
Re[k] <=0
therefore, all real part of eigenvalues of A are negative.
<=
if all real part of eigenvalue of A are negative
Re[k] <=0
for every positive defined C
suppose for all negative definite B, such that for all v with Bv is well defined product, we have
A^TB+BA=-C
then
2 Re[k] u^T B u <=0
which make Re[k]>=0
contradicted to Re[k], there exist a positive definite B for every positive definite C