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Basis for matrix

  1. Oct 14, 2012 #1
    When you have a matrix like:
    3 1
    0 1

    The RREF is
    1 0
    0 1
    the identity matrix.

    Is the basis always [0 0]^t?
     
  2. jcsd
  3. Oct 14, 2012 #2

    Ray Vickson

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    No, a basis is never [0 0]^t. A basis always consists of nonzero vectors, and for a 2-D space you need two of them. Anyway, what is a "basis of a matrix"? I have never heard of that term.

    RGV
     
  4. Oct 15, 2012 #3
    I was looking at the answers in the back of the book for the attached problem.
    The problem statement is "find a basis for the null space of the linear transformation T.

    Their answer was {[0 0]^t}. I don't understand how they got this.
     

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  5. Oct 15, 2012 #4

    Ray Vickson

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    OK, that seems to be a different question from what you first asked. The question is: what is the null space of the matrix? For the given matrix, the null space consists of a single point (x,y) = (0,0); do you know why?

    RGV
     
  6. Oct 15, 2012 #5
    Isn't it asking for the basis of the nullspace or is the nullspace of the matrix the same thing?

    Hmm, I'm not sure why, it seems if the RREF form has a pivot in every column then this is the case.
    Is it cause the augmented matrix would look something like
    1 0 0
    0 1 0
    thus x=0, y=0?
     
  7. Oct 17, 2012 #6
    Can someone answer this?
     
  8. Oct 17, 2012 #7

    Mark44

    Staff: Mentor

    The problem is asking for a basis of the nullspace of the transformation. A vector space is different from a basis for that vector space. You can get all of the vectors in some vector space by taking linear combinations of the vectors in a basis.

    Here's an example. Suppose that the nullspace happened to be all of the vectors in R2 that lie along the x-axis. A basis for this space is the vector <1, 0>T. This nullspace has infinitely many vectors in it, but all of them are a linear combination (multiples of in the case) the vector <1, 0>T.
    Yes.
     
  9. Oct 17, 2012 #8

    HallsofIvy

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    The question did not ask for a basis of the nullspace, it asked for the nullspace itself. The entire null space consists of the 0 vector. That does not have a basis.
     
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