Re: [理工] 線代

看板Grad-ProbAsk作者 (八字-風水-姓名學)時間13年前 (2013/02/03 23:59), 編輯推噓1(100)
留言1則, 1人參與, 最新討論串86/120 (看更多)
※ 引述《KAINTS (RUKAWA)》之銘言: : T/F : 假設A,B兩個矩陣分別是 n*k , k*m,如果A,B皆為行獨立, : 則AB也為行獨立 Suppose that A, B are n*k, k*m matrices with linearly independent columns, respectively. Let ABx = 0 for x in F^m (F, a field). Because of linearly independence of columns of A, that is, ker(A) = {0}, A(Bx) = 0 implies Bx = 0. Similarly, Bx = 0 implies x = 0 since B has linearly independent columns. By the above, ABx = 0 implies x = 0, in other words, AB has linearly independent columns. -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 114.37.135.131

02/04 08:36, , 1F
感謝!!
02/04 08:36, 1F
文章代碼(AID): #1H3ebTcs (Grad-ProbAsk)
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文章代碼(AID): #1H3ebTcs (Grad-ProbAsk)