推 NewFighterZR:第二小題看不太懂..請原諒我高中數學很差Orz 02/11 17:40
※ 編輯: BenLinus 來自: 114.43.195.103 (02/11 18:21)
※ 引述《NewFighterZR (NFT-ZR)》之銘言：
: 2
: 1.Let L:P2->P2 be the linear transformation defined by L(y) = x y''-y'+y
: Compute the matrix M that represents the linear transformation L using the
: 2 2
: ordered basis B={1 ,(x-1),(x-1)} for the domain and B'={1,(x-2),(x-2)}
: for the target space.
L(1) = 0 - 0 + 1 = 1
L(x-1) = 0 - 1 + (x-1) = x-2
2 2 2 2
L((x-1)) = 2x - (2x-2) + (x -2x + 1) = 3x - 4x + 3
2
再運算一下 1(1) + 0(x-2) + 0(x-2) = 1
2
0(1) + 1(x-2) + 0(x-2) = x-2
2 2
7(1) + 8(x-2) + 3(x-2) = 3x - 4x + 3
[1 0 7]
所以 M = [0 1 8]
[0 0 3]
: T 3
: 2.Find the distance of the point X=[4 1 7] of R from the subspace W consisting
: T
: of all vectors of the form [a b b].
[a] [1] [0] [1] [0]
[b] = a[0] + b[1] so W is a plane = span({[0], [1]})
[b] [0] [1] [0] [1]
這題用高中數學好像比較快 XD
(把(1,0,0) 和 (0,1,1) 作外積得 (0,-1,1) 所以平面是 0x -1y +1z = 0)
span 出來的平面方程式 -y + z = 0
再用 點到平面距離公式 算得 3*sqrt(2)
公式是,
點代入平面方程式取絕對值 | 0(4) -1(1) +1(7) | 6
---------------------------- = -------------------- = ------
法向量長度 2 2 2 sqrt(2)
sqrt(0 + (-1) + 1)
有點可恥可是... 蠻 快 的...
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有沒有人要用像樣點的方法幫解... 用高中解法實在可恥 XD
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