Re: [理工] [線代]交大100

[doom8199 (～口卡口卡 修～)]

※ 引述《death888 (ZZZZ)》之銘言：
: 9.Suppose A:3x3 has three distinct eigenvalues 0,1,3 with corresponding
: eigenvectors u,v,w. 何者敘述正確
: A.The rank of A must be 2
    true
: B.The matrix A^2-I is invertible
    false

      observe  (A^2 - I)v = AAv - v
                          = A(1v) - v
                          = 1(v) - v
                          = 0
      since (A^2 - I) has zero eigenvalue
      (A^2 - I) is not invertible

: C.v and w must span the column space of A
    true

      for any c1、c2 εR
      c1*v + c2*w = c1*Av + c2*A(w/3)
                  = A[c1*v + (c2/3)w]
      so span{v,w} € Im{A}  ____(1)

      on the other hand
      for any yεIm{A} , there exist x in R^3, s.t.  y = Ax
      since span{u,v,w} = R^3
      there exists c3、c4、c5 εR , s.t. x = c3*u + c4*v + c5*w

      so y = A(c3*u + c4*v + c5*w)
           = c4*v + 3*c5*w
      it implies Im{A} € span{v,w}   ____(2)

      by (1)(2) , span{v,w} = Im{A}

: D.The least squares error of Ax=2v+3w+u must be ∥u∥^2
    true
                                            2
      e =  min ∥(2v+3w+u) - (c1*v + c2*w)∥
          c1,c2

      when (c1,c2)=(2,3) , e has min=∥u∥^2



: 10.Suppose A:3x3 = xy^T is rank 1 matrix.何者敘述正確
: A.The eigenvector matrix S must be invertible
    false

      consider x=[ 1, 0, 0]
               y=[ 0, 1, 0]

      then A has zero eigenvalue with algebraic multiplicities = 3
      but corresponding  geometric multiplicities = 2

: B.A must have one non-zero eigenvalue
    false

      as same as part (A)

: C.When A is facotrized by SVD into UΣV^T,the only non-zero singular value
:   is ∥x∥∥y∥
    true

      (A^T)A = (yx^T)(xy^T)
             = <x,x>(yy^T)

      then  (A^T)Ay = <x,x>(yy^T)y
                    = <x,x>*<y,y>y
      →  (A^T)A  has an eigenvalue = <x,x>*<y,y> with respect to
          eigenvector = y

      so  A has a singular value = ∥x∥∥y∥
      ( and only one non-zero singular value since rank(A)=1 )


: D.When A is factorized by SVD into UΣV^T,U must include x/∥x∥ as
:   one of its column vectors
    false

      so as -x/∥x∥



: 請問這兩題要怎麼選 腦袋轉不過來
: 覺得觀念考好多orz
: 順便再問幾個選項
: 1. A,B,A+B,都是n*n可逆矩陣, then A^-1 +B^-1 也可逆
: 我是選FALSE 總覺得舉得出反例 不過想不到

    應該是對的

    set X is an inverse matrix of  A^(-1) + B^(-1)

    then  [A^(-1) + B^(-1)]X = I     ( I € C^(nxn)  )

    so  ┌ [A^(-1)]X + Y = I    ____(1)    by introducing a matrix Y
        └ Y = [B^(-1)]X        ____(2)

    by (2),  X = BY

    then by (1),   [A^(-1)]X + Y = I
                → [A^(-1)]BY + Y = I
                → BY + AY = A
                → Y = [(A+B)^(-1)]A    since (A+B) is invertible

     hence X = BY
             = B[(A+B)^(-1)]A

:          ┌     ┐
: 2.If     │a b c│
:       det│d e f│=7
:          │1 3 5│
:          └     ┘
: 第三列改1 0 1的時候 det為9,   then 第三列改1 6 9的時候=5
: (排版麻煩 我用文字敘述)
: true or false?
: 我對題目給的兩種展開 然後把三種coefficient用變數A B C代換
: 但這樣也只有兩條方程式 三個未知數解不出來 有別的辦法嗎?

    這題不需要展開
    determinant 本身可視 每列 row vector 為 linear function
    所以:

    │ a  b  c │     │ a  b  c │   │ a  b  c │
    │ d  e  f │ = 2*│ d  e  f │ - │ d  e  f │
    │ 1  6  9 │     │ 1  3  5 │   │ 1  0  1 │

                  = 2*7 - 9

                  = 5

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◆ From: 140.113.47.130
推 daniel770624:好詳細...
推 death888:第9題的A是怎麼看的 我只想到eigenvalue 有0 所以不可逆
→ death888:RANK一定小於3
推 SkullMaster:我猜是A=P^-1DP ,A~D,相似保rank
→ cha122977:dim[V(0)]=1=nullity(A)  Rank(A)=3-nullity(A)=2
推 death888:了解了 感謝!
推 bahamut5461:口卡口卡修 神人...
推 ai305428d:推