Re: [線代] 矩陣的問題
※ 引述《zako1113 (那個人)》之銘言：
: If A is a skew-symmetric real matrix, prove that (I+A) is invertible.
: (Hint: show that (I+A)X = 0 cannot have non-zero solution X in R^n)
: 用書內的提示
: (I+A)X = 0 for some non-zero X
: => AX = -X
: => -1 is an eigenvalue of A
: 請問之後要怎麼做呢?
設存在非零v
(I+A)v = 0
Av = -v --- (1)
A^T = -A --- (2)
(1) => v^T A^T = -v^T
=> v^T A = v^T
=> v^T A v = v^T v > 0
但是 根據假設 Av = -v
=> - v^T v = v^T v
=> v只能為0 與原假設矛盾
故原命題得證
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 128.220.147.185
推
07/21 19:33, , 1F
07/21 19:33, 1F
討論串 (同標題文章)