Re: [線代] 特徵值問題

[JohnMash (Paul)]

※ 引述《JohnMash (Paul)》之銘言：
: ※ 引述《SS327 ()》之銘言：
: : A為3階方陣，特徵值為1，2，3
: :        { A  3A }
: :   B =  {2A  2A }，B為6階方陣   ，求B的特徵值
: : 請問這題大概要怎麼下手阿
:     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)



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