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Do Equations in More Than Three Variables Represent Graphs in Higher Dimensions?

  1. Dec 12, 2007 #1
    Hey, first of all, I'd like to apologize if I'm posting this in the wrong forum. I wasn't sure whether I should post it here or in the mathematics forum. Recently I was going through an Algebra book, and I saw a chapter on solving linear equations in three variables. The book explained how these sets of equations can be solved using elimination, matrices, Cramer's rule, etc. Anyway, I worked several of these problems. When I was finished, just out of curiosity, I wondered what it would be like if I set up five sets of equations with five variables, and solved them. I created one, and using my preferred technique, elimination, I went about solving it and it went MUCH more smoothly than I expected it would. When I was finished, I had the values that I had started out with, and it had all worked out fine. I used five variables X, Y, Z, K, and J. When I had finished, I went back to the book I was using and saw that linear sentences in two variables represented graphs in two dimensions, and linear equations in three variables represent graphs in three dimensions. So here's my question: Do sets of equations with more than three variables represent graphs in higher dimensions?


    P.S: Here are the equations I worked. Feel free to point out anything that I may have done wrong.


    2x+4y-1z-6k+2j=-1
    3x+6y-3z-8k+4j=1
    1x+3y+4z-2k+4j=47
    4x-2y-2z+6k+2j=28
    5x+3y+4z+3k+3j=69

    (The values are X=2, Y=3, Z=5, K=4, J=6).
     
  2. jcsd
  3. Dec 12, 2007 #2
    I think you have to first precise the number of dimension(s) of the space where (in which) you are working. For example, if you write x = 0 in a 2-D space: this is describing a line. If you write x = 0 in a 3-D space: this is a plane because the fact you are giving a precise value to only one of the dimension reduce your freedom to move into this space of only one degree. And so and... Hope it was clear enought for you.
     
  4. Dec 12, 2007 #3
    Okay, that makes sense, but since I don't have any zeros in my equations, doesn't that mean that I'm dealing with graphs in five dimensional space? If so, is their any way to tell what shape these lines are? Is there any way I can manipulate these equations so that they represent, say a definitive shape? I'm sorry, but could someone please explain in depth what is going on here?
     
  5. Dec 12, 2007 #4

    Ben Niehoff

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    In two-dimensional space, an equation like

    2x + 3y = 11

    represents a line. A system of two equations

    2x + 3y = 11
    8x - 6y = 7

    represents two lines, and the solution (x,y) represents the point where those lines intersect.

    Now, carry this to three dimensions:

    A single linear equation like

    4x + 3y - 6z = 8

    represents a plane. Two equations

    4x + 3y - 6z = 8
    2x - 5y + 2z = 3

    would then represent two planes. Since there are fewer equations than variables, you can't solve for a unique (x,y,z); however, what you can do is find the equation of a line that is the intersection of the two planes.

    Three equations

    4x + 3y - 6z = 8
    2x - 5y + 2z = 3
    3x - 8y - 7z = 10

    give you three planes, and they will intersect in a single point (x,y,z).

    Now, you can extend this upward. A linear equation with five variables

    3x - 3y + 2z - 6u - 5v = 7

    represents a linear 4-dimensional hyperplane; that is, it represents the analogue of a plane in 5 dimensions. If you have five such equations, then they will intersect in a single point (x,y,z,u,v), as long as no two of the hyperplanes are parallel.
     
  6. Dec 13, 2007 #5
    And how do you tell if the planes are parallel? And how do I find the equation for the intersection of the line? Also, Why is it a four dimensional hyperplane if I have five variables?
     
  7. Dec 13, 2007 #6

    HallsofIvy

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    The planes are parallel if their normal vectors are parallel. Fortunately, for planes, finding the normal vector is easy: if the plane is given by Ax+ By+ Cz= D then the normal vector is Ai+ Bj+ Ck.

    By "intersection of the line" do you mean the line of intersection of two planes? If the planes are Ax+ By+ Cz= D and Px+ Qy+ Rx= S you can solve those two equations for two of the variables, say x or y, in terms of the third, z. If x= f(z), y= g(z), you can then write the line in parametric equations x= f(t), y= g(t), z= t. (Since a line is one dimensional, any point on it can be written in terms of a single variable. Parametric equations is a good way to do that.)

    If you have one equation in n variables, you can solve for one of the variables in terms of the other n-1. That is you are free to choose any values for n-1 variables and the last is given by the equation so your surface is n-1 dimensional. If the equation is linear, then the surface is a hyperplane.
     
  8. Dec 13, 2007 #7

    Ben Niehoff

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    Actually, my earlier statement was slightly incorrect. What really matters, is whether all of the equations are linearly independent of each other. In three dimensions, if your equations are:

    [tex]\begin{array}{rcl} a_1x + a_2y + a_3z & = & A \\ b1_x + b_2y + b_3z & = & B \\ c_1x + c_2y + c_3z & = & C \end{array}[/tex]

    then you can find a unique solution (x,y,z) as long as the determinant

    [tex]\left| \begin{array}{ccc} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{array} \right| \neq 0[/tex]

    The same idea carries over into higher dimensions; if the determinant of all the coefficients is nonzero, then the solution will be a single point.
     
  9. Dec 13, 2007 #8
    So, then is there really any way to tell mathematically what these shapes look like? Also, what is a parametric equation? I'm not familiar with that.
     
  10. Dec 13, 2007 #9
    Also, once you've found that the determinant doesn't equal zero, how do you solve for a single variable? When you've solved for a single variable, doe that mean you have the point where each plane intersects? Also, someone said that the equations represent a four dimensional hyperplane. But why wouldn't it be five dimensional since I have five variables?
     
  11. Dec 13, 2007 #10

    rcgldr

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    A paremetic equation relates variables to functions of another variable, such as:

    x = cos(t)
    y = sin(t)

    An equation with "n" multiple variables don't have to imply "n" dimensions , just "n" independent components, such as X, Y, Z, temperature, density, pressure, which has 6 components but is only 3 dimensional.
     
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