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Range & Null space of A matrix

  1. Aug 4, 2011 #1
    1. The problem statement, all variables and given/known data
    Let x [itex]\in[/itex] RN, y [itex]\in[/itex] RM & A [itex]\in[/itex] RMxN be a matrix. Denote the columns of A by Ak, k = 1,...,N. Let R(A) & N(A) be the range & null space of A respectively.
    a) How do the colmuns of A relate to the range of A?
    b) Your task is to find the solution to the problem y = Ax, where y & A are known & M = N. What role do R(A) & N(A) play?
    c) Let RN [itex]\ni[/itex] x = (x1,x2), x1 [itex]\in[/itex] RN1, x2 [itex]\in[/itex] RN2 & N1 + N2 = N. Let A [itex]\in[/itex] RN1xN & consider the problem Ax = 0. Assume that you know x2. Solve for x


    2. Relevant equations



    3. The attempt at a solution
    a) This is easy, the column space of A is just the range of A.
    b) Do we just use the definitions of R(A) and N(A)?
    c) I have no idea on this one.
    Any help please.
     
  2. jcsd
  3. Aug 5, 2011 #2

    lanedance

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    be careful mixing N(A)=Nullspace and N=number of columns - I would use n instead for the number of columns, m for rows
     
  4. Aug 5, 2011 #3

    lanedance

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    for b), what would happen if y is not in the range of A?
     
  5. Aug 5, 2011 #4

    lanedance

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    for c) consider writing a as 4 matrices to understand how it works
    [tex] \begin{pmatrix}
    B & C \\
    D & E
    \end{pmatrix}[/tex]

    with
    B - n1 x n1
    C - n1 x n2
    D - n2 x n1
    E - n2 x n2

    then consider the product
    [tex] Ax = \begin{pmatrix}
    B & C \\
    D & E
    \end{pmatrix}x = \begin{pmatrix}
    B & C \\
    D & E
    \end{pmatrix}\begin{pmatrix}
    x_1 \\
    x_2
    \end{pmatrix}[/tex]

    bit of an abuse of notation, but hopefully its clear what we're trying to do
     
  6. Aug 6, 2011 #5
    Bx1 + Cx2 = 0
    Dx1 + Ex2 = 0

    Is that correct mate, if so I'm lost on what to do next?
     
  7. Aug 14, 2011 #6
    Still lost on this question.
     
  8. Aug 15, 2011 #7
    If x2 is known what does that mean for x1 and in turn x.
     
  9. Aug 22, 2011 #8

    lanedance

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    well solving for x is essentially solving for x_1 as x_2 is known
     
  10. Aug 22, 2011 #9

    lanedance

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    Now re-arranging the equation we get

    [tex] Ax = \begin{pmatrix}
    B & C \\
    D & E
    \end{pmatrix}x = \begin{pmatrix}
    B & C \\
    D & E
    \end{pmatrix}\begin{pmatrix}
    x_1 \\
    x_2
    \end{pmatrix} = 0
    [/tex]

    [tex] Ax = \begin{pmatrix}
    B \\
    D
    \end{pmatrix}x_1 = - \begin{pmatrix}
    C \\
    E
    \end{pmatrix}
    x_2
    [/tex]

    this is a system of N equations with N1 unknowns
     
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