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！[香陵居士 (8-31 16:06, Long long ago)] [ 传统版 | sForum ][登录后回复]1楼

please explain your notation1。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"[icky (8-31 17:53, Long long ago)] [ 传统版 | sForum ][登录后回复]2楼

(引用 icky:please explain your notation1。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)...)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)....[香陵居士 (8-31 18:18, Long long ago)] [ 传统版 | sForum ][登录后回复]3楼

(引用 香陵居士: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 b...)i'm still confusedyour number 3 has this function M, but your equation does not contain M, what are you talking about?[icky (9-6 12:40, Long long ago)] [ 传统版 | sForum ][登录后回复]4楼

(引用 icky:i'm still confusedyour number 3 has this function M, but your equation does not contain M, what are you talking about?)It simply says if each element of X is a function of variable tWhat will be the range of t which makes AX > B[香陵居士 (9-6 15:05, Long long ago)] [ 传统版 | sForum ][登录后回复]5楼

(引用 香陵居士: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 tis your matrix A semi-definite positive or semi-definite negative?[icky (9-6 17:22, Long long ago)] [ 传统版 | sForum ][登录后回复]6楼

(引用 icky:this is not easy, even for linear function of tis your matrix A semi-definite positive or semi-definite negative?)i meandoes A has that kind of property? if it does, it is still sovable...[icky (9-6 17:22, Long long ago)] [ 传统版 | sForum ][登录后回复]7楼

(引用 icky:i meandoes A has that kind of property? if it does, it is still sovable...)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.[香陵居士 (9-6 22:34, Long long ago)] [ 传统版 | sForum ][登录后回复]8楼

(引用 香陵居士: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. Sinc...)if A is positive definite, then Ax > B eq to x > A^-1 B[icky (9-6 23:49, Long long ago)] [ 传统版 | sForum ][登录后回复]9楼

(引用 icky:if A is positive definite, then Ax > B eq to x > A^-1 B)... 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)

[香陵居士 (9-7 2:00, Long long ago)] [ 传统版 | sForum ][登录后回复]10楼

(引用 香陵居士:... 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 => ...)oh, ok, you are saying vectors can not be compared directly[icky (9-7 12:38, Long long ago)] [ 传统版 | sForum ][登录后回复]11楼

(引用 香陵居士:... 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 => ...)how do u define your AX > B ? AX is a vector, B is also a vectoryou define AX>B as every element of A is greater than every element of B?

I don't think that's a good definition....[icky (9-7 17:23, Long long ago)] [ 传统版 | sForum ][登录后回复]12楼

(引用 香陵居士:... 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 => ...)As I describedx1 + x2 > 2
x1 + 2x2 > 3

is expressed as AX > B, where A = [ 1 1 ; 1 2 ]; B = [2 3]T

Maybe it is a formal notation but I just would like to solve the problem.
[香陵居士 (9-8 0:19, Long long ago)] [ 传统版 | sForum ][登录后回复]13楼

(引用 香陵居士:... 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 => ...)aiyah....that's solving simultaneous inequalities, then why do u need to generate to matrix form?

use geometric solutions
[icky (9-8 1:40, Long long ago)] [ 传统版 | sForum ][登录后回复]14楼

(引用 香陵居士:... 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 => ...)Thanks, but the most important thing isGeometrical method is useful but limited to 3 variables (and for 3 variables it is not so obvious to see it). Thus some analytical or at least numerical solution is needed.[香陵居士 (9-8 9:29, Long long ago)] [ 传统版 | sForum ][登录后回复]15楼

1.

[bugzzj (9-24 18:09, Long long ago)] [ 传统版 | sForum ][登录后回复]16楼

2.

[bugzzj (9-24 18:30, Long long ago)] [ 传统版 | sForum ][登录后回复]17楼

(引用 香陵居士: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 b...)I think the examples in 3 can still be treated as linear.

As long as Mn(t) is in the form as show above, it can be treated as linear. The last equation looks hard to solve, but, i guess it is already another problem. [bugzzj (9-25 10:26, Long long ago)] [ 传统版 | sForum ][登录后回复]18楼