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Basis and Dimension of Solution Space

  1. Mar 12, 2015 #1

    B18

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    1. The problem statement, all variables and given/known data
    Find a basis for and the dimension of the solution space of the homogenous system of equations.

    2x1+2x2-x3+x5=0
    -x1-x2+2x3-3x4+x5=0
    x1+x2-2x3-x5=0
    x3+x4+x5=0
    2. Relevant equations


    3. The attempt at a solution
    I reduced the vector reduced row echelon form. However the second row doesn't have a 1 in the x2 position which has thrown me off. Do I parametrize here like I did? Something just doesn't seem correct.
    Here is my work thus far:
     
    Last edited: Mar 12, 2015
  2. jcsd
  3. Mar 12, 2015 #2

    B18

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  4. Mar 12, 2015 #3

    Mark44

    Staff: Mentor

    Based on your final matrix, which I didn't check, you have
    x1 = -x2 - x5
    x2 = x2
    x3 = -x5
    x4 = 0
    x5 = x5
    If you stare at this long enough you should see that the above represents all possible linear combinations of two vectors.

    BTW, it's confusing to have that last column of zeroes in your matrix. It would be better to eliminate that column since it can't possibly change.
     
  5. Mar 14, 2015 #4

    B18

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    Well I got the following equations from the matrix:
    x1+x2+x4=0
    x3+x5=0
    x4=0
    Not really sure how I can parametrize these to solve for the basis of the solution space.
     
  6. Mar 14, 2015 #5

    HallsofIvy

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    If those are correct, since [itex]x_4= 0[/itex], your first equation is [itex]x_1+ x_2= 0[/itex] which means that [itex]x_2= -x_1[/itex]. And, of course, your second equation s the same as [itex]x_5= -x_3[/itex]. That means we can write [itex]<x_1, x_2, x_3, x_4, x_5>= <x_1, -x_1, x_3, 0, -x_3>= <x_1, -x_1, 0, 0, 0>+ <0, 0, x_3, 0, -x_3>= x_1<1, -1, 0, 0, 0>+ x_3<0, 0, 1, 0, -1>[/itex]. Now do you see what basis vectors for the solution space are?
     
  7. Mar 18, 2015 #6

    B18

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    This is the solution I got. However I am a bit confused on my basis. How can the x4 slot both have zeros in the basis vectors and still span the space? Did I make a mistake along the way?
    Thanks for the help everyone.

    ImageUploadedByPhysics Forums1426716928.795578.jpg
     
  8. Mar 18, 2015 #7

    Mark44

    Staff: Mentor

    There's no reason that the x4 component shouldn't be zero. You can do a quick check by multiplying each of the two vectors by your matrix. If you don't get a zero vector for each product, that's a sign that you did something wrong.
     
  9. Mar 18, 2015 #8

    HallsofIvy

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    Have you forgotten what it is spanning? Not the original space- these vectors are to span the solution set for these equations- and as you have calculated x4 is 0!
     
  10. Mar 18, 2015 #9

    B18

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    Both of your replies helped me understand what was going on here much better. Thank you.
     
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