| [ KAIST ] in KIDS 글 쓴 이(By): mkjung (LOVER) 날 짜 (Date): 2000년 8월 13일 일요일 오전 01시 09분 18초 제 목(Title): Re: [질문] 수학문제 if you are only proving that the eigenvalues of AB and BA are same, then the following proof is the simplest. proof. since detr_k(AB)=detr_k(BA) for k=1,...n and detr_k(AB) is the k-th symmetric functin of eigenvalues. 1st symmetric function of eigenvalues is tr(AB)=tr(BA)= c_1 + ... + c_n where c_1, ..., c_n are eigenvalues. nth symmetric function of eigenvalues is det(AB)=det(BA)= c_1*... *c_n in general i-th symmetric function of eigenvalues is detr_i(AB)=detr_i(BA) = sum of the product of i eigenvalues out of n eigenvalues. hence there will be n choose i number of terms. So we have n simultaneous equations and the only way you can solve the equation is when the eigenvalues are identical. |