作者JohnMash (Paul)
看板Math
标题Re: [线代] 特徵值问题
时间Sun May 27 13:01:08 2012
※ 引述《JohnMash (Paul)》之铭言：
: ※ 引述《SS327 ()》之铭言：
: : A为3阶方阵，特徵值为1，2，3
: : { A 3A }
: : B = {2A 2A }，B为6阶方阵 ，求B的特徵值
: : 请问这题大概要怎麽下手阿
: B = K ◎ A, where ◎ means tensor product
: and K= [ 1 3]
: [ 2 2]
: then B.(k,a) = (K.k , A.a) = λ_k λ_a (k,a) = λ_b (k,a)
: where λ_a = 1,2,3
: and λ_k = -1,4
: hence, λ_b = -1,-2,-3,4,8,12
This is a general result
Let K = [k11 k12]
[k21 k22]
denote K。A = [k11*A k12*A]
[k21*A k22*A]
denote k=[p] and a is 3*1 matrix
[q]
then
k。a = [p*a]
[q*a]
It is easy to see that
(K。A).(k。a)=[(k11*p+k12*q)A.a]
[(k21*p+k22*q)A.a]
=(K.k)。(A.a)
Let K.k=κk, A.a=αa
hence (K。A).(k。a) = κα(k。a)
--
※ 发信站: 批踢踢实业坊(ptt.cc)
◆ From: 27.147.57.77
※ 编辑: JohnMash 来自: 27.147.57.77 (05/28 00:13)
1F：→ SS327 :谢谢~ 05/28 00:22