please explain your notation
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作者:icky (等级:15 - 最接近神,发帖:7923) 发表:2007-08-31 17:53:16  楼主  关注此帖
请教一个简单的不等式问题由对赌博的思考引起的关于线性不等式组的问题。 对于一个线性不等式组 AX > B                      (1) 其中X = [x1 x2 ... xn]T, A(n*n),B(1*n)为确定的实矩阵。 1。任意给定向量 K = [k1 k2 ... kn],如何确定MX的符号 2。对于向量 L = [l1(t) l2(t) ... ln(t)]T,其中ln(t)=ln1 x + ln2 (ln1, ln2是确定的实数),如何确定满足(1)的t的范围。 3。一般情况,对于 M = [M1(t) M2(t) ... Mn(t)],其中 Mn(t)是关于的确定代数函数,如何确定满足(1)的t的范围。 哪位知道要解决这三个问题应该阅读哪些方面的书籍?谢谢!
please explain your notation
1。任意给定向量 K = [k1 k2 ... kn],如何确定MX的符号

how is K related to MX

2。对于向量 L = [l1(t) l2(t) ... ln(t)]T,其中ln(t)=ln1 x + ln2 (ln1, ln2是确定的实数),如何确定满足(1)的t的范围。

how is L related to equation (1), and what is T here?

3。一般情况,对于 M = [M1(t) M2(t) ... Mn(t)],其中 Mn(t)是关于的确定代数函数,如何确定满足(1)的t的范围。

please explain "Mn(t)是关于的确定代数函数"
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作者:icky (等级:15 - 最接近神,发帖:7923) 发表:2007-09-06 12:40:42  2楼
Sorry... so many typos...1. MX Should be KX 2. Substitute X with L, we get a group of linear inequations for t only. (Should be ln(t) = ln1 t + ln2). t is an real variable. 3. Mn(t) will be some function regarding to t but no longer polynomial or linear, like exp(t) + 2sin(t), or t^5 + ln(t)....
i'm still confused
your number 3 has this function M, but your equation does not contain M, what are you talking about?
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作者:icky (等级:15 - 最接近神,发帖:7923) 发表:2007-09-06 17:22:12  3楼
It simply says if each element of X is a function of variable tWhat will be the range of t which makes AX > B
this is not easy, even for linear function of t
is your matrix A semi-definite positive or semi-definite negative?
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作者:icky (等级:15 - 最接近神,发帖:7923) 发表:2007-09-06 17:22:44  4楼
this is not easy, even for linear function of tis your matrix A semi-definite positive or semi-definite negative?
i mean
does A has that kind of property? if it does, it is still sovable...
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作者:icky (等级:15 - 最接近神,发帖:7923) 发表:2007-09-06 23:49:48  5楼 评分:
I do understand what you mean but what's tricky here is thatAX > B is not equivalent to X > A^-1 * B even if A is PD. Since a > b , c > d => a + c > b + d but not vice versa.
if A is positive definite, then Ax > B eq to x > A^-1 B
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作者:icky (等级:15 - 最接近神,发帖:7923) 发表:2007-09-07 12:38:48  6楼
... Can you prove that?Note what I said, for equation: if a = b then c = d a+b=c+d but for inequation, if a > b then c > d => a+b > c+d but a+b > c+d (not =>) c > d A simple counterexample: A = [ 1 1 ; 1 2] is PD B = [2;3] and AX > B (1) But we can't say X > A^-1 * b = [1;1] (2) Since obviously x1 = 0, x2 = 3 fits (1) but not (2)
oh, ok, you are saying vectors can not be compared directly
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作者:icky (等级:15 - 最接近神,发帖:7923) 发表:2007-09-07 17:23:51  7楼
... Can you prove that?Note what I said, for equation: if a = b then c = d a+b=c+d but for inequation, if a > b then c > d => a+b > c+d but a+b > c+d (not =>) c > d A simple counterexample: A = [ 1 1 ; 1 2] is PD B = [2;3] and AX > B (1) But we can't say X > A^-1 * b = [1;1] (2) Since obviously x1 = 0, x2 = 3 fits (1) but not (2)
how do u define your AX > B ? AX is a vector, B is also a vector
you define AX>B as every element of A is greater than every element of B?

I don't think that's a good definition....
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作者:icky (等级:15 - 最接近神,发帖:7923) 发表:2007-09-08 01:40:40  8楼 评分:
... Can you prove that?Note what I said, for equation: if a = b then c = d a+b=c+d but for inequation, if a > b then c > d => a+b > c+d but a+b > c+d (not =>) c > d A simple counterexample: A = [ 1 1 ; 1 2] is PD B = [2;3] and AX > B (1) But we can't say X > A^-1 * b = [1;1] (2) Since obviously x1 = 0, x2 = 3 fits (1) but not (2)
aiyah....
that's solving simultaneous inequalities, then why do u need to generate to matrix form?

use geometric solutions
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