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Linear Sys. Solution Space of Homogenous Sys.

  1. Apr 30, 2012 #1
    1. The problem statement, all variables and given/known data
    Alright, so the question is:
    Determine a basis for, and solution space of, the following homogenous system:
    1 -2 0 -1 = 0
    2 -4 0 -2 = 0
    -4 8 0 4 = 0
    -2 4 0 2 = 0

    2. Relevant equations

    3. The attempt at a solution
    Okay. This question is posing some issues for me. First is the fact that all the equations are the same. From looking at a similar problem from the book
    |1 -2 3||x|=|0|
    |2 -4 6||y|=|0|
    |3 -6 9||z|=|0|
    it appears to me that the professor might have not written it out properly. But anyway, I just assumed that each row represents a variable, x,y,z, and t.
    Therefore, the solution space is x-y-t=0?
    If I multiply by the nonexistent 4x1 matrix of x,y,z, and t, then I end up with, once again, four equivalent equations, and the solution space is
    -2x-4y+2z+t=0.
    I really just have no understanding of how to do this when all the equations are the same. Help?
     
  2. jcsd
  3. Apr 30, 2012 #2

    sharks

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    Start by converting the matrix into row echelon form.
     
  4. Apr 30, 2012 #3
    When you reduce to row echelon form you get
    1 -2 0 -1 0
    0 0 0 0 0
    etc.
    At least that is my understanding.
     
  5. Apr 30, 2012 #4

    sharks

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    What are your pivots and free variables?

    Then, use the definition: A basis is a linearly independent spanning set.
     
  6. Apr 30, 2012 #5

    LCKurtz

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    Here's an example for you. Say, like in your example, you have 4 equations in 4 unknowns but when you row reduce it you are left with only one row being nonzeros. For example, (making it up) say

    ##x -5y + 2z -3w = 0##

    You could solve for any variable in terms of the others. In my example, ##x## is easiest to solve for. ##x=5y-2z+3w##, where the ##y,\ z,\ w## can be anything. You could then write the solution like this:$$
    \left[\begin{array}{c}x\\y\\z\\w \end{array}\right ]=
    \left[\begin{array}{c}5y-2z+3w\\y\\z\\w \end{array}\right ]=
    y\left[\begin{array}{c}5\\1\\0\\0 \end{array}\right ]+
    z\left[\begin{array}{c}-2\\0\\1\\0 \end{array}\right ]+
    w\left[\begin{array}{c}3\\0\\0\\1 \end{array}\right ]
    $$From writing it like this a basis for the solution space is obvious.
     
  7. Apr 30, 2012 #6

    sharks

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    If i understood correctly, the question by the OP, is in two parts:
    1. Find the basis.
    2. Find the solution space.

    Maybe i'm getting a little confused, but those are not the same as "the basis for the solution space". Or is it?
     
  8. Apr 30, 2012 #7
    To be quite honest I have no idea. I don't have much faith in my professor.

    My understanding of the question is
    Find a basis for the homogenous system.
    Find the solution space of the homogenous system.

    Half the test is vocabulary too. Kind of frustrating to be completely stumped here after doing fine in Calc 1-3.

    Kurtz, thank you very much for the effort in your response. I think I'm lacking some kind of fundamental understanding of 'basis'.
     
  9. Apr 30, 2012 #8

    sharks

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    I found another post to be very helpful for this problem. Read the thread, especially HallsofIvy's reply.
    https://www.physicsforums.com/showthread.php?t=229074
    However, in that other post, it deals with finding the basis for the solution space, which is not exactly the wording in this problem here.

    Another (maybe easier) definition for basis: A basis is a set of linearly independent vectors that, in a linear combination, can represent every vector in a given vector space.
     
  10. Apr 30, 2012 #9
    So then from
    x-2y+0z-t=0
    I would get x=2y+t, plug in for <x, 1,0> and <x, 0,1>, get the values of x, and then those would be the basis vectors of the solution space, assuming that's what the question is asking?
    Of course, I cut out z. If z has to be included then I assume it would be <x, 1, 0, 0> and <x, 0, 0, 1>. Seeing as any value could be used for z and it would be of no consequence.

    I wish I knew what the professor was asking. Especially considering this is the pretest for the final.
     
  11. Apr 30, 2012 #10

    LCKurtz

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    They aren't quite the same thing. The solution space is the subspace containing all solutions. The basis for the solution space is a set of linearly independent solution vectors that span the space. In my example, the 3 constant vectors form the basis and as you vary the free variables that generates all linear combinations of the basis vectors which generates the solution space itself.
     
  12. Apr 30, 2012 #11
    Thanks for your help guys! I'm still not quite sure if I get it, but I think that's a mix of not going to class and possibly bad wording on the test. Not too worried about it though. Won't be needing this for my major lol.

    Thanks!
    Robert
     
  13. Apr 30, 2012 #12

    sharks

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    Thank you for clarifying, LCKurtz. :smile:

    I have a question though: Is the solution space the same as the nullspace?
    From what i understand, the nullspace is the sum of all the special solutions, so it looks like it's exactly what you've written on the R.H.S. of your example.
     
    Last edited: Apr 30, 2012
  14. Apr 30, 2012 #13

    LCKurtz

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    We talk about the nullspace of a matrix and the solution space of a system of equations. The null space of a matrix ##A## is the solution space of the system ##A\vec x = \vec 0##.
     
  15. Apr 30, 2012 #14
    Okay, yeah, check out these two problems.
    Determine the basis for, and the solution space of, the following homogenous systems:
    1.
    -1 + 2 + 0 + 1 = 0
    +2 - 4 + 0 - 2 = 0
    +3 - 6 + 0 - 3 = 0
    -8 + 16 +0 + 8 = 0
    2.
    -1 + 2 + 0 + 1 = 0
    +2 - 4 + 0 - 2 = 0
    +3 - 6 + 0 - 3 = 0
    -4 + 8 + 0 + 4 = 0

    Are these not the same question???
     
  16. Apr 30, 2012 #15

    LCKurtz

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    As written you don't have a homogeneous system of equations. You just have a bunch of false statements.

    I explained the difference between a solution space and its basis in post #10.
     
  17. May 3, 2012 #16
    Another student told me it was a 'dummy question'. So that's fun.

    And that's exactly how the question was written. I believe the variables are implied.
     
  18. May 3, 2012 #17

    HallsofIvy

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    1 -2 0 -1 = 0
    2 -4 0 -2 = 0
    -4 8 0 4 = 0
    -2 4 0 2 = 0
    Perhaps the "dummy" part is that this is a "skeleton" of the equations.

    Isn't really a "system of equations". I assume you mean something like
    [tex]x_1- 2x_2+ 0x_3- x_4= 0[/tex]
    [tex]2x_1- 4x_2+ 0x_3- 2x_4= 0[/tex]
    [tex]-4x_1+ 8x_2+ 0x_3+ 4x_4= 0[/tex]
    [tex]-2x_1+ 4x_2+ 0x_3+ 2x_4= 0[/tex]
    which is the same as the matrix equation
    [tex]\begin{bmatrix}-2 & -4 & 0 & -2 \\ -2 & 4 & 0 & -2 \\ -4 & 8 & 0 & 4 \\ -2 & 4 & 0 & 2\end{bmatrix}\begin{bmatrix}x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix}= \begin{bmatrix}0 \\ 0 \\ 0 \\ 0 \end{bmatrix}[/tex]

    One obvious point about this is all those "0"s in the third column- [itex]x_3[/itex] does not appear in the equations and so can be anything.

    Row reducing the matrix eventually gives
    [tex]\begin{bmatrix}1 & -2 & 0 & -1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\end{bmatrix}[/tex]
    which is equivalent to the single equation [itex]x_1- 2x_2- x_4= 0[/itex] which we can solve for [itex]x_1[/itex]: [itex]x_1= 2x_2+ x_4[/itex].

    That is, any vector satisfying that equation, any vector in the null space, is of the form [itex]<x_1, x_2, x_3, x_4>= <2x_2+ x_4, x_2, x_3, x_4>= x_2<2, 1, 0, 0>+ x_3<0, 0, 1, 0>+ x_4<1, 0, 0, 1>[/itex].
     
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