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(LINALG) : Nullspace of transpose : N(A^T)

  1. Apr 19, 2005 #1
    I'm not sure if I am making a mistake, or my book is wrong, or if both answers are correct. But, it is confusing me, and I would like to know why. We are asked to find the basis of the following subspaces on the matrix A.

    Find: [tex] R(A^T),\,\,N(A),\,\,\,R(A),\,\,N(A^T)[/tex]

    I'm having trouble finding N(A^T). Here is how I'm doing it.

    [tex]A = \left[ \begin{array}{cccc}
    1 & 0 & 0 & 0 \\
    0 & 1 & 1 & 1 \\
    0 & 0 & 1 & 1 \\
    1 & 1 & 2 & 2
    \end{array} \right]
    [/tex]


    thus:
    [tex]
    A^T = \left[ \begin{array}{cccc}
    1 & 0 & 0 & 1 \\
    0 & 1 & 0 & 1 \\
    0 & 1 & 1 & 2 \\
    0 & 1 & 1 & 2
    \end{array} \right]
    [/tex]

    so...

    [tex]
    rref(A^T) = \left[ \begin{array}{cccc}
    1 & 0 & 0 & 1 \\
    0 & 1 & 0 & 1 \\
    0 & 0 & 1 & 1 \\
    0 & 0 & 0 & 0
    \end{array} \right]
    [/tex]

    then we are left with...

    [tex]
    \begin{array}{c}
    x_1+\alpha = 0 \\
    x_2 + \alpha = 0 \\
    x_3 + \alpha = 0 \\
    x_4 = \alpha
    \end{array}
    [/tex]

    which gives:

    [tex]
    \alpha\left[
    \begin{array}{c}
    -1\\
    -1\\
    -1\\
    1
    \end{array} \right]
    [/tex]

    The book gives:
    [tex] \left[
    \begin{array}{c}
    1 \\
    1 \\
    1 \\
    -1
    \end{array} \right]
    [/tex]

    as the basis for [tex]N(A^T) [/tex]

    Is this the same? And why?

    I mean [tex] \alpha [/tex] can be anything, so if [tex] \alpha = -1 [/tex] then I get the same answer as the book. So spanning the set with either "my" vector, or the books accomplishes the same thing. It's just confusing to me why the book would not follow the algorithm to get the answer. Thanks in advance.
     
  2. jcsd
  3. Apr 19, 2005 #2
    As you noted, using alpha = -1 makes your answer the same. They are the same answer.
     
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