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推 lai90043:感謝解答 還想請問(A^2)x中的A^2怎麼來的? 08/19 23:55
→ mqazz1:因為projection matrix的定義 A = A^2 08/20 20:05
推 a81288653:在平面上的向量,正交投影兩次與正交投影一次的結果相同 08/21 20:29
※ 引述《lai90043 (賴伯)》之銘言：
: Let W = span{(1,2,3),(4,5,6)} be a subspace in R^3
: find a basis for W的正交補集 ?
+
令(a,b,c)為orthogonal basis for W
< (1,2,3),(a,b,c) > = 0 => a+2b+3c=0
< (4,5,6),(a,b,c) > = 0 => 4a+5b+6c=0
解聯立
b = -2c
a = c
orthogonal basis為(a,b,c) = (c, -2c, c)
取 c=1 => (1, -2, 1)
: Let T:R^3 → R^3 be the projection of the space R^3
: on the plane x+2y+3z = 0
: find the eigenvalue of T
projection matrix的eigenvalue只有0和1
存在x不等於0 使得 Ax = λx
λx = Ax = (A^2)x = A(Ax) = A(λx) = λ(Ax) = λ(λx) = (λ^2)x
2
(λ - λ)x = 0
2
λ - λ = λ(λ-1) = 0
λ為0或1
: ========================分隔線=====================
: ANS:
: 1. (1,-2,1)
: 2. 1,0
: 請這問兩題該怎麼解呢?
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※ 編輯: mqazz1 來自: 140.118.110.186 (08/19 09:11)