Re: [線代] 矩陣的問題
※ 引述《Honor1984 (希望願望成真)》之銘言：
: ※ 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
Another way,
Claim:〈Av,v〉= 0, where A is a skew-symmetric real matrix and v in R^n.
(I+A)X = 0 => AX = -X
And 0 =〈AX,X〉= -〈X,X〉=> X = 0.
∴ I+A is invertible. Q.E.D.
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