| [ QuizWit ] in KIDS 글 쓴 이(By): iLUSiON (ivenomouth) 날 짜 (Date): 1998년 5월 4일 월요일 오전 12시 34분 31초 제 목(Title): [cap] lindear algebra 2 [ QuizWit ] in KIDS 글 쓴 이(By): guest (wiking) 날 짜 (Date): 1998년 5월 3일 일요일 오후 04시 38분 56초 제 목(Title): Re: linear algebra 2. Here is my version. Given condition. [AB, C] = 0 & [A, C] = [B, C] (= X). Using [AB, C] = A[B, C] + [A, C]B = 0. We have AX = -XB. Now if Kernel(X) does not span the space, Then X'X should have non-zero eigen values while all eigen vectors can be orthonomalized. Let e be one of non-zero eigen-valued eigen vectors (eigenvalue = a). Then <e|X'AX|e> = -a<e|B|e>, which is contradictory to positive definiteness of A & B. Therefore Kernel(X) spans the whole space. PS> The above solution by another guest wronly assumed that X'AX is symmetric. [환상] 하하..이구... 제가 문제를 약간 잘못올렸군요. 원래 A,B are symmetric but C is not necessarily symmetric. But I think the condition can be relaxed. Nice! Wink, can you relax the condition on A,B even more ? ^^ In general AX+BX=0 has a unique solution X=0 iff. A,B has common eigen- values. This equation is called a special case of Sylvester's matrix equation. 1010101010101010101010101000000010101010010101010101000000010001111110101001010 0101010011111010101000100010110101010101010101001010101010101010101111111101010 1011111100010101010101010101101010101011111101010111101010001110101010001010110 http://www.math.mcgill.ca/~chung |