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Nullspace matrix

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

    Let A be the matrix:


    a) An example of a vector in the nullspace of A is
    b) An example of a vector NOT in the nullspace of A is

    Sorry guy but I'm really STRUGGLING

    3. The attempt at a solution

    a) I found x1 ,x2,x3,x4 = -x2+4/3x4, x2, 2x4, x4

    =x2( -1,1,0,0) + x4(4/3,0,2,1)

    For more than an hour, the answer sheet is telling that my vectors are wrong.
    I don't see how it can be wrong.

    Ax = 0
    I reduced A to row echelon form, then solve in terms of the free variables x2 and x4.
  2. jcsd
  3. Mar 20, 2014 #2


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    Science Advisor

    That answer is correct as you can see by apply A to each of (-1,-, 0, 0) and (4/3, 0, 2, 1). The nullspace is two dimensional and there are an infinite number of bases for it. How is the answer sheet telling you that you are wrong? If it gives a different pair of vectors, that may just be a different basis for the same space. For example, if (4/3, 0, 2, 1) is a basis vector, so is (4, 0, 6, 3).
  4. Mar 20, 2014 #3
    The answer sheet is an online portal prompting me of my error whenever I submit the answer.
    I know my answer has to be correct. I submit ( -1,1,0,0),(4/3,0,2,1) in vector form. and I know my answer cannot be wrong because 1) I perform RREF on the matrix A then, 2) express the solution as free variables.

    I do not know what is going on with the portal prompting me of my error.

    edit: Message: "What you have given as vector in the nullspace of A is not of the correct dimension."
  5. Mar 20, 2014 #4
    I'll solve part (b) first:

    so, a vector that is not in the nullspace implies ax =/= 0

    [3,3,-2,0;-3,-3,3,-2] [x1,x2,x3,x4] = [0;1]

    I get

    x1 = 2/3 - x2 + 4/3x4
    x3 = 1 + 2x4
    free : x2 , x4

    x1,x2,x3,x4 = ( 2/3 - x2 + 4/3x4, x2, 1+2x4, x4)
  6. Mar 20, 2014 #5


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    A has four columns so acts on four dimensional vectors. If you gave (-1, 1, 0, 0) and (4/3, 0, 2, 1) they have the correct dimension. How did you enter the fraction 4/3? The answer "sheet" might be interpreting that as (4, 3, 0, 2, 1). Try (4, 0, 6, 3), as I suggested before, instead.
  7. Mar 20, 2014 #6
    I entered the fraction as it is, that is- 4/3. There is a button that shows my answer in some kind of tex format and it corresponds to 4/3 as how one would normally write a value as a fraction form.

  8. Mar 20, 2014 #7
    It isn't working. There isn't any reason for me to enter (4,0,..) because the system does in fact recognize 4/3 as an input language for the answer.
  9. Mar 20, 2014 #8

    I might be wrong, but the way you are inputting the answer might be incorrect.

    Notice the vectors you are putting in are 1x4, and you know you should be inputting 4x1.

    Maybe try putting a semicolon after your numbers instead?

    If that doesn't work.. well, maybe there is an error in the program.
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