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EASY QUESTION solving system of linear inequalities

  1. Aug 28, 2012 #1
    Hello, I am stuck on something so simple. The problem is i have a great difficulty with the geometric interpretation of things.

    So if I have a system of linear equations in three unknowns, like for example this:

    -x - 2y + z = 0
    x - 3y - 2z = 0

    this is just a simple system of homogeneous equations. I can use simple Gaussian elimination to solve it. There's an infinite number of solutions to this and I can see the geometric reason: both of these equations represent a plane in space through the origin. The intersection of these two planes is a line, and there are an infinite number of points on that line.

    I am starting to have difficulty when I think of solving this:

    -x - 2y + z > 0
    x - 3y - 2z > 0

    Here it's the same equations only inequalities. I don't know how to envision this. When I think of the first inequality for example, I think this will be all the points (x,y,z) in the plane (-x -2y + z=0) that are above the line (2y + x). But I'm not sure. How do I represent a solution that satisfies these inequalities? I imagine I first solve the associated homogeneous equations that are above. Then where do I go from there?
    Last edited: Aug 28, 2012
  2. jcsd
  3. Aug 28, 2012 #2


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    Science Advisor

    Hey fleazo.

    The best to think about this is to get an idea of the half-space.

    The half-space is basically a way to say that you have a linear classification object (i.e. a plane in some number of positive dimensions that your matrix represents) and then you figure out what lies below it and in front of it.

    Here is the basic idea of how you do with geometry:

    The general plane equation is given by n . (r - r0) = 0 where n is the normal vector to the plane, r is a general variable point and r0 is one point on the plane.

    If something is positive (i.e. your whole thing is > 0) then it means that n . (r - r0) > 0. If it's < 0, then it means n . (r - r0) < 0).

    As long as you have your normal defined to be the way you intend it, then you can simply find out when r > 0, r < 0, and equal to 0 (i.e. lies on the plane).

    So in the context of your question, the first thing you have to do is find the object and this is done by treating your system to be equal to 0 rather than less than or greater than zero.

    Once you do this, you can then resort to using the above classification to say whether a point will be greater or less than zero if it's not on the plane.

    If you want to do this visually, you draw or get a graph of the linear object and then see how it divides the space (or the subspace of your space: especially if you have linear dependence or something similar).

    Once you do that, obtain an equation of your linear object in the form of n . (r - r0) and test whether it's positive or negative.

    As long as you get an expression for ax + by + cz + d = 0, you will always be able to do the above.
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