MATRIX!!!!!!

2008-01-12 5:10 am
Question:

回答 (1)

2008-01-12 12:49 pm
✔ 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


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