※ [本文轉錄自 Dirichlet 信箱]
作者: pretend.bbs@lalala.twbbs.org ("Weak and only weak")
標題: Hadamard 不等式
時間: Wed Jul 23 07:12:55 2014
作者: pretend (數學所英文組) 站內: pretend
標題: Hadamard 不等式
時間: Wed Aug 17 03:39:33 2005
《Hadamard 不等式》
A 是 n 階實方陣 , 且 A 所有行向量長度皆為 1
則 [det(A)]^2 ≦ 1
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pf : 先說明一下符號 ... 證明有點長 orz
A = (a_ij), 1≦i,j≦n
A_ij : 餘因子
以下是用乘數法則的證法 ...
f(A) = det(A)
Constraint: (a_1j)^2 + (a_2j)^2 + ... + (a_nj)^2 = 1, 1≦j≦n
設 g_j = (a_1j)^2 + (a_2j)^2 + ... + (a_nj)^2 - 1
n
▽f = Σλ_j▽g_j, 將左式寫開來 =>
j=1
(A_11 , ... , A_n1 , A_12 , ... , A_n2 , ... , A_n1 , ... , A_nn)
= 2(λ_1 a_11 , ... , λ_1 a_n1 ,λ_2 a_12 , ... , λ_2 a_n1 , ... ,
λ_n a_n1 , ... , λ_n a_nn)
整理一下 => A_ij = 2λ_j(a_ij), 1≦i,j≦n
Case1. 若某個 λ_j = 0 => A_ij = 0 for each i => det(A) = 0
Case2. 若所有 λ_j ≠ 0 => a_ij = A_ij/2λ_j (代入限制條件 g_j = 0)
g_j = a_1j(A_1j/2λ_j) + a_2j(A_2j/2λ_j) + ... + a_nj(A_nj/2λ_j) - 1
= (det(A)/2λ_j) - 1 = 0
=> det(A) = 2λ_j
注意到線代中的一個性質 : det(adj(A)) = det(A)^(n-1) ... (*)
又 a_ij = A_ij/2λ_j => det(adj(A)) = det([Aij]) = det([2λ_ja_ij])
= (2^n)(λ_1)(λ_2) ... (λ_n) det([aij])
= (2^n)(λ_1)(λ_2) ... (λ_n) det(A) ... 代入 (*) 式
=> (2^n)(λ_1)(λ_2) ... (λ_n) = det(A)^(n-2)
又 det(A) = 2λ_k => det(A)^n = (2^n)(λ_1)(λ_2) ... (λ_n)
=> det(A)^n = det(A)^(n-2)
=> det(A)^2 = 1
故只有 0 與 1 可能是 [f(A)]^2 = det(A)^2 的極值
事實上, det(A)^2 ≦ 1 ... Q.E.D.
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→ pretend 修改本文於 Wed Aug 17 16:40:28 2005
※ 發信站: 批踢踢實業坊(ptt.cc)
※ 轉錄者: Dirichlet (111.255.83.4), 07/23/2014 07:26:52