[線代] 矩陣分析

[dkcheng (電磁霸主)]

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

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◆ From: 111.249.167.80
推 ntust661    :(A)
→ ntust661    :(B)需要考慮 C 常數向量的樣子，通常是 n*1
→ ntust661    :(C)一定會有一個 s 會使 (sI-A) singular
→ ntust661    :(D) 不一定
→ ntust661    :(E) 靠泰勒展開 + 微分可以知道 d/dt (e^At) = Ae^At
※ 編輯: dkcheng         來自: 111.249.167.80       (11/08 20:34)
推 zombiea     :C) depend on char poly of A has a root or not
→ zombiea     :D) exp(At)exp(-At)=I