※ 引述《slasher (you got me)》之銘言：
: 12.
: A = [aij] 是一個 正定 對稱 的 nxn 矩陣
: 下列何者正確?
: (A) aii > 0, for all i
: (B) A 可逆
: (C) aii + ajj - 2*aij > 0, for all i != j
: (D) aii*ajj - (aij)^2 > 0, for all i != j
: (E) 以上皆對
: (A), (B)是對的
: (C), (D)怎麼導? 我不知道答案是什麼
: 有這份的參考答案嗎?^^"
(C)
∵ A is a positive definite matrix
∴ A = transpose(B)‧B (Notice that B is invertible.)
∴ aii = sum of bki^2 (for k = 1 to n)
ajj = sum of bkj^2
aij = sum of bki*bkj
And for each k, bki^2 - 2*bki*bkj + bkj^2 = (bki - bkj)^2 >= 0.
(Notice that any i, j be choosen will not affect this result.)
So, sum up all from k = 1 to n, the result won't be negative.
But if aii + ajj - 2*aij = 0,
this means for all k = 1 to n, (bki - bkj) = 0.
So the ith column of B is equal to the jth column of B
-> B is not invertible.
-> Contradiction.
=> aii + ajj - 2*aij > 0, for all i!= j
(D)
As we mentioned above,
aii = sum of bki^2 (for k = 1 to n)
ajj = sum of bkj^2
aij = sum of bki*bkj
∴ aii*ajj = (b1i^2 + b2i^2 + ... + bki^2)*(b1j^2 + b2j^2 + ... + bkj^2)
aij^2 = (b1i*b1j + b2i*b2j + ... + bki*bkj)^2
And the process of comparison of these two numbers is belong to the area of
discrete math.
So you can try it. :)
(Hint: Try aii*ajj - aij^2, you will get the same inequality in question (C).)
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◆ From: 61.56.159.17
※ 編輯: justmaker 來自: 61.56.159.17 (02/02 13:47)