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Orthogonal vectors

  1. Mar 20, 2014 #1
    1. The problem statement, all variables and given/known data

    Let S be the subspace of all vectors in R4 that are orthogonal to each of the vectors
    (0, 4, 4, 2), (3, 4, -2, -4)
    What is an example of a matrix for which S is the nullspace?




    3. The attempt at a solution

    I'm not sure how I should be intepreting the question:

    [ 0 ,4 ,4 ,2 ;3, 4, -2, -4] = [ x , y ,z , t] = [ 0 , 0 ]

    0x + 4y + 4z + 2t = 0

    3x + 4y -2 z -4t = 0

    from here I set up an augmented matrix and solve for the set of vectors x ,y ,z ,t? By definition if x, y, z, t results in each linear equation = 0, then, x ,y ,z ,t are vectors with properties of orthogonality?
     
  2. jcsd
  3. Mar 20, 2014 #2

    ehild

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    There is multiplication instead of equality in the first line? If yes, it looks all right, do it.

    ehild
     
  4. Mar 20, 2014 #3
    [ 0 ,4 ,4 ,2 ;3, 4, -2, -4] .[ x , y ,z , t] = [ 0 , 0 ]

    Yes, it should be a multiplicative operation-dot product.
    But I aren't exactly sure what should I be doing with the 2 linear equation. Do I perform reduce row echelon?
     
  5. Mar 20, 2014 #4

    ehild

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    Yes. But it is of that form already... You will have two free parameters.


    ehild
     
  6. Mar 20, 2014 #5
    I'm still unclear. Could you explain?
     
  7. Mar 20, 2014 #6

    ehild

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    There is the system of equations:

    0x + 4y + 4z + 2t = 0

    3x + 4y -2 z -4t = 0

    What is the general solution?

    ehild
     
  8. Mar 20, 2014 #7
    It is done by setting up an augmented matrix:

    0 4 4 2 | 0
    3 4 -2 -4 | 0

    Perform REF(Not RREF), then solve for the variables.


    Edit: x1 = 2x3 + 2x4, x2 = -x3-0.5x4, x3, x4
     
    Last edited: Mar 20, 2014
  9. Mar 20, 2014 #8

    ehild

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    Well, solve for the variables....

    ehild
     
  10. Mar 20, 2014 #9
    Post #7
     
  11. Mar 20, 2014 #10

    ehild

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    X3=a, x4=b a,b free parameters. So the S subspace consists of vectors ((2a+2b); (-a-0.5b); a; b).
    And a matrix for which S is the nullspace?

    ehild
     
  12. Mar 20, 2014 #11
    Relooking at the question, I'm not sure if we interpreted it correctly.

    I think the question is asking for the {SX=0|x1=0,4,4,2, x2=(3,4,-2,-4),S[itex]\in R4[/itex]}

    Edit: I think we were on the right track.

    But this brings up a question. What if we're asked to solve S, given X where SX = 0?
     
    Last edited: Mar 20, 2014
  13. Mar 20, 2014 #12

    Mark44

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    Let's untangle your notation. S is a subspace of R4 such that if v ##\in## S, then v ##\cdot## <0, 4, 4, 2> = 0 and v ##\cdot## <3, 4, -2, -4> = 0.

    The problem asks that you find a matrix (call it A) for which S is the nullspace of A. IOW, if v ##\in## S, then Av = 0.
     
  14. Mar 20, 2014 #13


    Isn't the vector [2z+2t; -z-0.5t;z;t] the answer?
     
  15. Mar 20, 2014 #14

    ehild

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    S is the nullspace of a linear transformation T, that is, Tv =0 for all vectors v[itex]\in[/itex]S. Show a matrix of that transformation. (You have shown one in Post #3) .

    ehild
     
  16. Mar 20, 2014 #15

    Mark44

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    See post #12. The "answer" should be a matrix, not a vector.

    In addition, what you have above would be better written as two vectors - a basis for S.
     
  17. Mar 20, 2014 #16
    Am I suppose to perform another one? I thought we have had all the solutions already. 2 are free variables and the corollary is that there are infinite solutions.
     
  18. Mar 20, 2014 #17

    ehild

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    The subspace S contains infinite number of vectors. You have found these vectors already.

    The next question is
    ehild
     
  19. Mar 20, 2014 #18
    any matrix with coefficient [itex]\in R[/itex]?
     
  20. Mar 20, 2014 #19
    basis for S : z (2,-1,1,0) + t (2,-0.5,0,1)
     
  21. Mar 20, 2014 #20

    HallsofIvy

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    Your wording is peculiar. z (2,-1,1,0) + t (2,-0.5,0,1) is a general vector in S. The set of vectors {(2, -1, 1, 0), (2, -0.5, 0, 1)} is a basis.

    The answer to this question is any matrix, A, such that A applied to each of those two vectors is 0.
     
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