推 bineapple :感謝! 又學到新的一招了 03/06 12:46
※ 引述《bineapple (Bineapple)》之銘言：
: 設
: A = [ 6 10 ]
: [ 10 17 ]
: B = [ 1 0 ]
: [ 0 2 ]
: M = [ a b ]
: [ c d ]
: N = M^T
: 且 ad-bc=1, a>0, NA=BM^(-1)
: 求a b c d
: 想請問這題有什麼比較簡潔的算法嗎?
: 光是代來代去的好像會變很長一串
: 謝謝!
T -1 -1 T -1
M A = B M => A = (M ) B M
令
B' = [1 0 ]
[0 √2]
-1 T
M' = (M )
則
' 'T
A = M B'B'M
T
令 L = M'B' => A = LL
根據(http://libai.math.ncu.edu.tw/webclass/matrix/ch1_5/):
T
若 A 為正定，則 A 可以被分解為如 A=LL 的形式，其中 L 為下三角矩陣且其主對角線
上的項皆為正數（且分解的方式為唯一的）。
det(A) = 2 > 0,a = 6 > 0
11
故A正定
對 A 做 Cholesky 分解
A = [1 0 ] [6 10] = [1 0] [6 0 ] [1 5/3]
[5/3 1 ] [0 1/3] [5/3 1] [0 1/3] [0 1 ]
= [1 0] [1 0 ] [6 0] [1 0 ] [1 5/3]
[5/3 1] [0 1/6] [0 12] [0 1/6] [0 1 ]
= [1 0] [1 0 ] [√6 0] [1 0] [√6 0] [1 0 ] [1 5/3]
[5/3 1] [0 1/6] [ 0 √6] [0 2] [0 √6] [0 1/6] [0 1 ]
= [1 0] [√6 0 ] [1 0] [√6 0 ] [1 5/3]
[5/3 1] [0 1/√6] [0 2] [0 1/√6] [0 1 ]
-1 T -1
(M ) B M
-1
M = [√6 0 ] [1 5/3] = [√6 5√6/3]
[0 1/√6] [0 1 ] [ 0 1/√6 ]
代公式,
故 M = [1/√6 -5√6/3]
[ 0 √6 ]
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