推 tokyo291:感謝您! 02/18 22:26
※ 引述《tokyo291 (工口工口)》之銘言：
: 1. Suppose q1,q2,q3,q4 are orthonormal vectors in R^4.
: Let A=[q1 q2 q3 q4], B=[q1+q2 q2+q3 q3+q4 q4+q1]
: and C=[q2 q3 q4 q1]. Find all possible values for the 4 by 4 determinats
: detA, detB, detA*detC.
: 我的算法是利用q1 q2 q3 q4彼此獨立所以經由列運算可以轉成單位矩陣
: 故detA=1,不曉得這樣子想對不對?
2 T T
(detA) = det(A)*det(A) = det(A)*det(A ) = det(AA ) = det I = 1
=> detA = ±1
detB = det[q1+q2 q2+q3 q3+q4 q4+q1] = det[q1+q2 q2+q3 q3-q1 q4+q1]
^^^^^行運算
= det[q1+q2 q2+q1 q3-q1 q4+q1] = 0
^^^^^行運算again
detC = det[q2 q3 q4 q1] = -det[q3 q2 q4 q1] = det[q4 q2 q3 q1]
= -det[q1 q2 q3 q4] = -detA
2
故 detA*detC = -(detA) = -1
: 2. If A is 3 by 3 symmetric positive definite, then Aqi=λiqi
: with eigenvalues λi and orthonormal eigenvectors qi. Suppose
: x=c1*q1+c2*q2+c3*q3. Assume λ1<=λ2<=λ3. What c's will make the ratio
: x^TAx/(x^Tx) as large as possible? What is the maximum of the ratio
: x^TAx/(x^Tx)?
印象中最大值是最大的eigenvalue，所以答案是(0,0,1)
Ax = A(c1q1+c2q2+c3q3) = c1λ1q1 + c2λ2q2 + c3λ3q3
T 2 2 2
x Ax = c1 λ1 + c2 λ2 + c3 λ3
T 2 2 2
x x = c1 + c2 + c3
2 2 2
T T c1 λ1 + c2 λ2 + c3 λ3
x Ax/(x x) = ──────────── ≦ λ3
2 2 2
c1 + c2 + c3
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