Re: [理工] [工數]-矩陣
※ 引述《tim760323 (胖子)》之銘言：
: 證明:
: For any matrix A , e^A is always nonsingular
: 不知道有沒有人會解這題
: 感激不盡!!
let Ax=kx
k is eigenvalue of A
and e^k is eigenvalue of e^A
because e^k ,it's not always equal to zero
s.t. det(e^A)=/=0 =>e^A is nonsingular
--
為者常成.行者常至
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 123.193.214.165
→
11/02 22:08, , 1F
11/02 22:08, 1F
※ 編輯: iyenn 來自: 123.193.214.165 (11/02 22:11)
→
11/02 22:12, , 2F
11/02 22:12, 2F
討論串 (同標題文章)