作者elps ( )
看板Grad-ProbAsk
标题Re: [理工] [线代]-矩阵可逆的问题
时间Tue Sep 22 20:51:59 2009
※ 引述《HP0 (cksh)》之铭言：
: B是任何3乘2的实数矩阵，B转置用B'表示，我想证明以下这个矩阵可逆
: [1 0 0 ]
: [0 1 0 B ]
: [0 0 1 ]
: [ -1 0]
: [ B' 0 -1]
: 请问要如何下手？
[I -B] = [I 0][I -B ]
[B' I] [B' I][0 I+B'B]
=> det[I B] = det[I -B] * (-1)^2
[B' -I] [B' I]
= det[I 0] det[I -B ]
[B' I] [0 I+B'B]
= [det(I)det(I)][det(I)det(I+B'B)]
= det(I+B'B)
claim: I+B'B is positive definite
pf:
For any eigenvalue λ of I+B'B,
let x be an eigenvector of I+B'B w.r.t. λ
=> (I + B'B)x = λx
=> x + B'Bx = λx
=> B'Bx = (λ-1)x
Since B'B is positive semi-definite,
all eigenvalues of B'B are non-negative, i.e., λ-1>=0 => λ>=1>0.
Therefore I+B'B is positive definite, which implies that det(I+B'B)>0.
Hence, [I B] is invertible.
[B' -I]
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1F：推 HP0:谢谢你！！ 09/23 11:30