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about orthogonality and nullspaces |
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| May30-12, 08:38 AM | #1 |
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about orthogonality and nullspaces
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. |
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| May30-12, 11:32 AM | #2 |
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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.
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| May30-12, 11:47 AM | #3 |
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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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