推 mqazz1 :謝謝 另外我還有一些疑問 PO在原文上..不知能否請問 07/06 23:07
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 112.104.143.85
※ 編輯: JohnMash 來自: 112.104.143.85 (08/29 14:48)
T T
Suppose w = [ 2, -1, 0, 2, 1 ] and the matrix A = I + αww is singular,
then (α, rank(A)) = ?
denote w=w_5
w_i⊥w_5, for i=1,2,3,4
w_5'.w_5=10
if A.(Σ[1,5] b_i w_i)=0
then Σ[1,5] b_i w_i +10αb_5 w_5=0
b_1=b_2=b_3=b_4=0
b_5(1+10α)=0
α=-1/10
Nullity=1, Rank=5-1=4
※ 引述《JohnMash (Paul)》之銘言：
※ 引述《mqazz1 (無法顯示)》之銘言：
: [1] [5]
: [2] [4] [v^T]
: u = [3] v = [3] A = I + [u v][u^T]
: [4] [2]
: [5] [1]
: find all the eigenvalue of A
|u|^2=|v|^2=α=55,(u'.v)=(v'.u)=β=35, where ' means transpose
A=I+u.v'+v.u'
w1,w2,w3,u,v are a basis
such that wi⊥u, wi⊥v for i=1,2,3
A.wi=wi, i=1,2,3
A.(u+v)=u+v+u.v'.(u+v)+v.u'.(u+v)
=u+v+(v'.u)u+αu+αv+(u'.v)v
=(1+α+β)(u+v)=91(u+v)
A.(u-v)=u-v+u.v'.(u-v)+v.u'.(u-v)
=u-v+(v'.u)u-αu+αv-(u'.v)v
=(1-α+β)(u-v)=-19(u-v)
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 112.104.170.71
※ 編輯: JohnMash 來自: 112.104.170.71 (07/06 22:23)