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Basis for ROW(A), COL(A) and NUL of a square matrix

  1. Feb 4, 2010 #1

    No one answered
    Last edited: Feb 4, 2010
  2. jcsd
  3. Feb 4, 2010 #2


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    Looks fine up until the null space. Ax=0 yields the equations

    [tex]x_1+(1+i)x_2+x_4 = 0[/tex]
    [tex]x_3+5ix_4 = 0[/tex]

    You have four unknowns and only two equations, so you can solve for two of the variables in terms of the other two.

    [tex]x_1 = -(1+i)x_2-x_4[/tex]
    [tex]x_3 = -5ix_4[/tex]

    In vector form, this would be

    [tex]\begin{pmatrix}x_1\\x_2\\x_3\\x_4\end{pmatrix} = x_2\begin{pmatrix}{-1-i\\1\\0\\0}\end{pmatrix} + x_4\begin{pmatrix}-1\\0\\-5i\\1\end{pmatrix}[/tex]

    The two constant vectors are a basis of the null space.
  4. Feb 4, 2010 #3
    Sorry I deleted it. I thought no would answer and got frustrated. But thanks.
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