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Seemingly simple linear algebra

  1. May 24, 2013 #1
    Why is it that for arbitrary z, A[itex]^{T}[/itex]z = 0 and b[itex]^{T}[/itex]z ≠ 0 when there does not exist an x such that Ax = b, i.e. that b is not in the range space of A, where A is an n x m matrix?
     
    Last edited: May 24, 2013
  2. jcsd
  3. May 24, 2013 #2
    "Range space" ?? Do you mean "row space" ??
     
  4. May 24, 2013 #3
    Actually, it means column space. This means that there is no linear combination of A's columns that gives b. This must mean that there is no x for which Ax = b, right?
     
  5. May 24, 2013 #4

    Ray Vickson

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    Huh? How can 'b' not be in the range space of A when AX=b? That is contradictory!
     
  6. May 24, 2013 #5
    Edited the post so that it makes more sense. Does it now?
     
  7. May 24, 2013 #6

    Ray Vickson

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    Yes.
     
  8. May 24, 2013 #7
    Okay, sorry. I am just not used to using that term. I usually use the following for the 4 fundamental spaces

    C(A) - column space (of A)
    [itex] C(A^T) [/itex] - row space (of A)
    N(A) - null space (of A)
    [itex] C(A^T) [/itex] - left handed nullspace of A (i.e nullspace of A transpose)
     
    Last edited: May 24, 2013
  9. May 25, 2013 #8
    So the column space of A transpose is the null space of A?
     
  10. May 25, 2013 #9

    Ray Vickson

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    What have you tried so far? Show your work.
     
  11. May 25, 2013 #10
    Whoops... My error. No it is not. The nullspace of A transpose is NOT the column space of A transpose I meant to write

    [tex] N(A^T) [/tex]

    The null space of A transpose is of course the set of all vectors y such that

    [tex] A^T y \, = \, 0 [/tex]

    We sometimes call it the "left null space" because is we take the transpose of both sides of the above equation.

    [tex] (A^Ty)^T \, = \, 0^T [/tex]

    we get

    [tex] y^T A \, = \, 0^T [/tex]

    Sorry for my mistake. I'm going to try to edit it and correct my typo before it causes confusion. Unfortunately I don't see an edit function available :(
     
  12. May 26, 2013 #11
    This statement came from the book Convex Optimization by Boyd and Vandenberghe. I forget which page now, but the idea has come up again in Farkas' lemma, but in a different form.

    http://en.wikipedia.org/wiki/Farkas'_lemma

    Let A be an n × m matrix and b an n-dimensional vector. Then, exactly one of the following two statements is true:

    (1) There exists an x ∈ Rm such that Ax = b and x ≥ 0.
    (2) There exists a y ∈ Rn such that A[itex]^{T}[/itex]y ≥ 0 and b[itex]^{T}[/itex]y < 0.

    This is close but the original statement involves equalities similar to (2).
     
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