| [ KAIST ] in KIDS 글 쓴 이(By): mkjung (LOVER) 날 짜 (Date): 2000년 8월 11일 금요일 오전 03시 45분 04초 제 목(Title): Re: [질문] 수학문제 기호를 간단하게 하기 위해 A^(-1) 대신 A^ 라고 줄여 쓰겠읍니다. det(xI - AB) = det[A(xA^ - B)] = det[A(xA^B^ - I)B] = det[(xA^B^ - I)BA] = det(xI - BA) char poly이 같으므로 AB와 BA의 eigenvalue는 같습니다. ============== this proof assumes the the existance of inverse.there is another proof without assuming the invertable matrix. |xI + AB| = |xI +BA| is a well known fact. proof follows in the following way. |I + AB| = 1+ tr(AB) + detr_2(AB) + ... detr_n(AB) , where der_n(AB) is the sum of of n by n principal minors. you can express detr_i(AB) by i-th symmetric function of eigenvalues of (AB) as well. note that detr_i(AB)=detr_i(BA). detr is called 'det-trace'. i believe such expansion of the determinant of the sum of two matrix is due to the famous mathematician Cauchy. (I neglected 'x' in the proof but it is a trivial matter.) |