0
$\begingroup$

I have $2$ vectors with size $8\times1$. How to find linear combination of those two vectors to be positive with positive coefficient?

i.e. $a_{1}.v_{1}+a_{2}.v_{2}\ge 0$ where $v_{1}$ and $v_{1}$ are vectors and $a_{1}$, $a_{2}$ are positive constants.

I have provided with only $v_{1}$, $v_{1}$vectors.

$a_{1}$, $a_{2}$ are any positive unknowns.

  • 0
    How is this related to complex analysis? – 2017-01-23
  • 0
    It was by mistake. It is related with convex analysis. – 2017-01-23
  • 0
    What do you mean by "I have provided with only $v_1,v_1$ vectors" ? – 2017-01-23
  • 0
    It means I have only v1, v2 vectors with me. a1 and a2 are unknowns with positive values. – 2017-01-23

1 Answers 1

1

You want to find an element of the polyhedron $\{ a : a_1 v_1 + a_2 v_2 \geq 0, a \geq 0 \}$. The polyhedron is defined by ten inequalities in $\mathbb{R}^2$. One solution approach is to draw the inequalities. You will see that each inequality can be represented with a line through the origin. The inequalities are of the form $$a_1 \geq -((v_2)_i / (v_1)_i) a_2 \text{ if } (v_1)_i > 0$$ $$a_1 \leq -((v_2)_i / (v_1)_i) a_2 \text{ if } (v_1)_i < 0$$ Therefore: $$\max_i \{ -(v_2)_i / (v_1)_i : (v_1)_i > 0 \}a_2 \leq a_1 \leq \min_i \{ -(v_2)_i / (v_1)_i : (v_1)_i < 0 \} a_2$$ In Matlab you can find the coefficients for $a_2$:

v1=[0.6666 0.4084 -0.1432 0.2781 -0.2427 0.2427 0.2943 -0.2943]';
v2=[0.2970 0.6048 0.3077 -0.1945 0.2893 -0.2893 -0.3508 0.3508]';
[max( -v2(v1>0) ./ v1(v1>0) ) min( -v2(v1<0) ./ v1(v1<0) ) ]

This results in the coefficients 1.192006592501030 and 1.191980971797486, which leads to a contradiction since the first one is larger than the second one. For your other example, you obtain the coefficients 0.3598 and 1.4275. That means that $0.3598 a_2 \leq a_1 \leq 1.4275 a_2$. For any nonnegative $a_2$ you can find a nonnegative $a_1$ that satisfies this inequality.

  • 0
    There is an error because $v_1$ is not 1D. – 2017-01-23
  • 0
    My conclusion is that your question has no solution. Take for example the case where the two first components are negative... – 2017-01-23
  • 0
    Is not possible to solve them by plotting convex polygon of those two vectors in two dimensional plane in MATLAB? – 2017-01-23
  • 0
    v1=[0.6666 0.4084 -0.1432 0.2781 -0.2427 0.2427 0.2943 -0.2943]' v2=[ 0.2970 0.6048 0.3077 -0.1945 0.2893 -0.2893 -0.3508 0.3508]'. These are the two vectors(8x1 size). And my teacher told me that their linear combination does not provide non negative value. I plotted convex polygon with these points onto the MATLAB(2D plane) with v1 vector as x coordinate and v2 vector as y coordinate. I got convex polygon with origin on its edge. – 2017-01-23
  • 0
    Other example which provides positive linear combination is v1=[ 0.5500 0.1068 -0.1758 0.4142 -0.0295 0.5026 0.4097 -0.2498] ' and v2=[0.2603 0.5990 0.3815 -0.0695 0.5128 0.1061 -0.1474 0.3566] ' . And their convex polygon does not contain origin. Indirectly I wanted to say that, is it possible to apply convex polygon (with or without origin lying in the polygon) concept here? – 2017-01-23
  • 0
    @JeanMarie I know $v_1$ is not a scalar, hence the first inequality is a vector inequality. Pratibha: try to draw the 10 inequalities in $a$-space instead of the vectors $v_1$ and $v_2$. For your first case, an inequality is $0.6666 a_1 + 0.2970 a_2 \geq 0$. – 2017-01-23
  • 0
    @LinAlg this remark wasn't intended for you but for the OP. – 2017-01-23
  • 0
    The solution is actually considerably easier. See update. – 2017-01-23
  • 0
    Thank you LinAlg. I tried to solve with your method. And i reached to the solution. But there comes a problem when all the elements of v1 are positive. How can we co relate the inequality for this case? – 2017-01-24
  • 0
    In that case there is no upper bound on $a_1$. – 2017-01-24
  • 0
    You have considered two conditions for v1. What about v2? If elements of v2 are negative then above inequality fails. – 2017-01-24
  • 0
    The Matlab example contains some negative elements in v2, and the inequality does not 'fail'. The edge case that has not been considered yet is when some elements of v1 or v2 are zero. Those either do not put a condition or $a_1$ and $a_2$ (when the corresponding element in the other vector is nonnegative), or they make the polyhedron empty. – 2017-01-24
  • 0
    a1≤−((v2)i/(v1)i)a2 if (v1)i<0. In this case, if v2<0 then a1 will be negative. But it violates the condition i.e. a1>0. – 2017-01-24
  • 0
    Right, so in that case no a1 satisfies the condition and there is no linear combination that answers your question. – 2017-01-24
  • 0
    Thank you very much LinAlg. – 2017-01-24