[線代] 矩陣分析

看板Math作者 (電磁霸主)時間14年前 (2011/11/08 20:08), 編輯推噓2(2推 0噓 5→)
留言7則, 2人參與, 最新討論串1/3 (看更多)
Consider X = AX(t), where A is an n by n matrix. X(t) is an n by 1 vector, n≧2, choose correct ones At -1 (A) e = inverse laplace transform of (sI-A) At At (B) X(t) = e C = Ce for any n by 1 vector C (C) (sI-A) is nonsingular for any scalar s At (D) e is nonsingular for any scalar t d At At At (E) ---e = Ae = e A . dt -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 111.249.167.80
推
11/08 20:12, , 1F
(A)
11/08 20:12, 1F
→
11/08 20:12, , 2F
(B)需要考慮 C 常數向量的樣子，通常是 n*1
11/08 20:12, 2F
→
11/08 20:13, , 3F
(C)一定會有一個 s 會使 (sI-A) singular
11/08 20:13, 3F
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11/08 20:14, , 4F
(D) 不一定
11/08 20:14, 4F
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11/08 20:16, , 5F
(E) 靠泰勒展開 + 微分可以知道 d/dt (e^At) = Ae^At
11/08 20:16, 5F
※ 編輯: dkcheng 來自: 111.249.167.80 (11/08 20:34)
推
11/08 22:19, , 6F
C) depend on char poly of A has a root or not
11/08 22:19, 6F
→
11/08 22:20, , 7F
D) exp(At)exp(-At)=I
11/08 22:20, 7F
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文章代碼(AID): #1EkHlFzm (Math)
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文章代碼(AID): #1EkHlFzm (Math)