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Finding the basis of a null space

  1. Nov 20, 2011 #1
    1. The problem statement, all variables and given/known data
    The matrix is:
    -2 -2 -4 4
    -1 1 2 -2
    -1 0 -3 0
    -4 1 -7 -2

    I know the dimensions for the null space are 2

    2. Relevant equations
    I know that to find the basis for a null space Ax=0, so I row reduced it and I got
    1 0 3 0
    0 1 5 -2
    0 0 0 0
    0 0 0 0

    3. The attempt at a solution
    Since I got this matrix, I thought the basis for the null space would be
    -3 and 0
    -5 2
    1 0
    0 1

    but my computer keeps saying it is wrong. Can someone tell me what I'm doing wrong?
     
  2. jcsd
  3. Nov 20, 2011 #2

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    Hi angiep410! :smile:

    Did you try to multiply your original matrix with your solutions?
    Does the product come out as the zero vector?
     
  4. Nov 20, 2011 #3
    No it doesn't come out to zero, but I don't know how to get it to come out to zero. Do you have any suggestions?
     
  5. Nov 20, 2011 #4

    HallsofIvy

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    This is not the row reduction I get.

     
  6. Nov 20, 2011 #5
    What is the row reduction you get?
     
  7. Nov 20, 2011 #6

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    What HoI said!

    Redo your row reduction.
    If you get the same row reduction, perhaps you can show a couple of the steps you did?
     
  8. Nov 20, 2011 #7
    Row Operation 1:
    multiply the 1st row by 1/2
    1 -1 -2 2
    -1 1 2 -2
    -1 0 -3 0
    -4 1 -7 -2
    Row Operation 2:
    add 1 times the 1st row to the 2nd row
    1 -1 -2 2
    0 0 0 0
    -1 0 -3 0
    -4 1 -7 -2
    Row Operation 3:
    add 1 times the 1st row to the 3rd row
    1 -1 -2 2
    0 0 0 0
    0 -1 -5 2
    -4 1 -7 -2
    Row Operation 4:
    add 4 times the 1st row to the 4th row
    1 -1 -2 2
    0 0 0 0
    0 -1 -5 2
    0 -3 -15 6
    Row Operation 5:
    interchange the 2nd row and the 3rd row
    1 -1 -2 2
    0 -1 -5 2
    0 0 0 0
    0 -3 -15 6
    Row Operation 6:
    multiply the 2nd row by -1
    1 -1 -2 2
    0 1 5 -2
    0 0 0 0
    0 -3 -15 6
    Row Operation 7:
    add 3 times the 2nd row to the 4th row
    1 -1 -2 2
    0 1 5 -2
    0 0 0 0
    0 0 0 0
    Row Operation 8:
    add 1 times the 2nd row to the 1st row
    1 0 3 0
    0 1 5 -2
    0 0 0 0
    0 0 0 0
     
  9. Nov 20, 2011 #8

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    Right at the first step you seem to have lost a minus sign.

    The rest looks just fine, except that.
     
  10. Nov 20, 2011 #9
    where? I don't see it
     
  11. Nov 20, 2011 #10
    sorry. i see what you're talking about. i typed the matrix in wrong on this forum. the actual matrix is:
    2 -2 -4 4
    -1 1 2 -2
    -1 0 -3 0
    -4 1 -7 -2
     
  12. Nov 20, 2011 #11

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    In that case you have a correct basis that spans the null space.
    Your matrix multiplied by each of your 2 solutions yields the zero vector, and there is no other independent vector with that property.

    Why would you say that your computer keeps saying that your answer is wrong?
     
  13. Nov 20, 2011 #12
    I am asked to find the basis for the row space, which I said was:
    (2,-2,-4,4), (-1,1,2,-2)

    the column space, which I said was:
    2 -2
    -1 1
    -1 0
    -4 1

    and the null space, which I said was:
    -3 0
    -5 -2
    1 0
    0 1

    and my computer says that at least 1 thing is wrong, but it doesn't tell me what it is.
     
  14. Nov 20, 2011 #13

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    Your row space vectors are dependent on each other.
     
  15. Nov 20, 2011 #14
    so what does that mean in terms of my answer?
     
  16. Nov 20, 2011 #15

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    You need to pick another row vector.
    One that is not a scalar multiple of the other.
     
  17. Nov 20, 2011 #16
    but how do i pick another row vector?
     
  18. Nov 20, 2011 #17

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    Keep one row and replace the other one by any other row you like.

    Actually, you should check that if you left-multiply the row with the matrix, that it is not zero.
     
  19. Nov 20, 2011 #18
    I tried keeping (2,-2,-4,4) and putting in the 2 other rows, but my computer still says it's wrong
     
  20. Nov 20, 2011 #19

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    In that case, I believe you're having trouble with entering your data in a format that your computer accepts.

    Did you type in your vectors the way you showed them in your post?
     
  21. Nov 20, 2011 #20
    yes, I entered it exactly as shown. The program gives me spaces to put the numbers in.
     
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