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How they found the left nullspace in each of these examples

  1. Jan 22, 2015 #1
    1. The problem statement, all variables and given/known data

    Part b)

    http://www.math.utah.edu/~zwick/Classes/Fall2012_2270/Lectures/Lecture19_with_Examples.pdf
    upload_2015-1-22_11-56-16.png


    For B
    upload_2015-1-22_11-57-0.png

    Left nullspace is solution to A ^ T times Y =0
    So we have a free variable for the third row so don't we have infinitely many solutions as x3 could be anything?
    upload_2015-1-22_11-51-3.png

    In this problem's part b) I don't think that they took the transpose of the matrix A.

    http://staff.imsa.edu/~fogel/LinAlg/PDF/29 Fundamental Subspaces.pdf
    Go the bottom of page 1 under 4) and you'll see "
    Perform Gaussian (or better, Gauss-Jordan) elimination on [A | I] to produce [U | B] (or
    [R | C]). Claim: the last m – r rows of B (which equal those of C because once we get zeroes
    there’s no more work to do in those rows) form ...."

    So why isn't the last three rows of 0's in A just the left nullspace?

    I am trying to figure out why

    2. Relevant equations


    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Jan 22, 2015 #2

    vela

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    B is a 2x3 matrix, so it maps vectors from ##\mathbb{R}^3## to ##\mathbb{R}^2##. In which of these two vector spaces does the left nullspace reside?

    You start with the (untransposed) A and form the augmented matrix [A | I] and perform row operations to produce [U | B]. The basis vectors are the last m-r rows of B. So why are you thinking the rows of A have anything to do with the basis of the left nullspace? It should be clear the last three rows of A can't form basis because they're all 0s.
     
  4. Jan 28, 2015 #3
    I don't think so.

    But then what does this mean?
    https://api.viglink.com/api/click?f...fogel/LinAlg/PDF/29 Fundamental Subspaces.pdf
    Perform Gaussian (or better, Gauss-Jordan) elimination on [A | I] to produce [U | B] (or
    [R | C]). Claim: the last m – r rows of B (which equal those of C because once we get zeroes
    there’s no more work to do in those rows) form ...."
     
  5. Jan 30, 2015 #4

    vela

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    Sorry for the late reply. I've been kinda busy this week.

    I simply paraphrased the passage you quoted from the PDF, so I'm not sure what you're disagreeing with. As a result of the row operations, matrix A turns into U, and the identity matrix turns into B. The passage says the last rows of B then form a basis for the left nullspace.
     
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