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I know that it is possible to read the basis for $\operatorname{Null}(A)$ and $\operatorname{Null}(A^T)$ by simply looking at the reduced row-echelon form (RREF) of the matrix $A$. I have only an approximate idea of how to do it.

For example, if we $$A = \begin{bmatrix} 1 & 2 & 0 & -1 \\ 0 & 0 &1 &2 \\ 0 &0 &0 &0 \end{bmatrix},$$ then one of the basis vectors for $\operatorname{Null}(A)$ will be $\begin{bmatrix}-2 & 1 & 0 &0\end{bmatrix}^T$, and the other one will be $\begin{bmatrix}1 & 0 & -2 &1 \end{bmatrix}^T$. So, basically we're "going around" over eight digits and alternating signs. But I noticed that this method isn't perfect. Does someone have a better idea of how to read bases for these two spaces from the RREF of $A$?

user642796
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sequence
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  • Perhaps you can rewrite the RREF of $A$ as $x_1 = x_4 - 2x_2$, $x_3 = -2x_4$. Note this is the same as writing it out in terms of systems of linear equations (a matrix is a compact way to write this without the messy $x_1,x_2$, and etc.). Then you can form the vector $\begin{bmatrix} x_1 \ x_2 \ x_3 \ x_4 \end{bmatrix} = \begin{bmatrix} x_4 - 2x_2 \ x_2 \ -2x_4 \ x_4 \end{bmatrix} = x_4\left(\begin{bmatrix} 1 \ 0 \ -2 \ 1\end{bmatrix}\right) + x_2\left(\begin{bmatrix} -2 \ 1 \ 0 \ 0 \end{bmatrix}\right) $. – MathNewbie Jun 17 '15 at 04:06
  • This way we're calculating the null space, but my question is about finding the null space by inspection. – sequence Jun 17 '15 at 04:17
  • well your question wasn't clear. If you're asking for a shortcut, sure there's a way, and that's the way you just mentioned. Practice it, but be careful of trying to do too much mental arithmetic at once. You're bound to make a careless mistake. – MathNewbie Jun 17 '15 at 04:19
  • I have Java code for doing this; I've put it up as a gist in case it's of any use to you. – joriki Jul 27 '15 at 06:33

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