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About orthogonality and nullspaces

  1. May 30, 2012 #1
    Hi:
    I see an example about nullspace and orthogonality, the example is following:

    $$Ax=\begin{bmatrix} 1 & 3 &4\\ 5 & 2& 7 \end{bmatrix} \times \left[ \begin{array}{c} 1 \\ 1\\-1 \end{array} \right]=\begin{bmatrix} 0\\0\end{bmatrix}$$

    The conclusion says the nullspace of [itex]A^T[/itex] is only the zero vector(orthogonal to every vector). I don't know why the columns of A and nullspace of [itex]A^T[/itex] are orthogonal spaces.
    I know nullspace is the solution of Ax=0; but in this theorem, why columns of A is related
    to nullsapce of [itex]A^T[/itex].
    Thanks.
     
    Last edited: May 30, 2012
  2. jcsd
  3. May 30, 2012 #2
    Notice the null space of A^T is orthogonal to (1,5), (3,2) and (4,7). So it is orthogonal to the space spanned by the vectors, which is the column space of A.
     
  4. May 30, 2012 #3

    HallsofIvy

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    [tex]A^T= \begin{bmatrix}1 & 5 \\ 3 & 2 \\ 4 & 7\end{bmatrix}[/tex]
    so that the condition that a vector be in the null space of [itex]A^T[/itex] is
    [tex]\begin{bmatrix}1 & 5 \\ 3 & 2 \\ 4 & 7\end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}= \begin{bmatrix}0 \\ 0 \\ 0 \end{bmatrix}[/tex]
    which is the same as
    [tex]\begin{bmatrix}x+ 5y \\ 3x+ 2y \\ 4x+ 7y\end{bmatrix}= \begin{bmatrix}0 \\ 0 \\ 0 \end{bmatrix}[/tex]
    Do you see now, how the columns of A, which become the rows of AT, are relevant here?
     
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