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There are a lot of methods I found online pertaining to computation of image of a matrix.

Suppose I had $$A = \begin{bmatrix} 0 & 1 &1 \\ 1 & -1 & 0 \\ -1 & 2 &1 \end{bmatrix}$$

Is there a clever way to compute the kernel space of the matrix without resorting to brute force i.e. compute $Ax = 0$?

Or am I asking for too much?

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    $\begingroup$ There are no generally applicable tricks other than solving $Ax=0$. In the above case you can see that there are two linearly independent rows and the last can be easily written as the difference of the first two. But this depends on the specific nature of $A$. $\endgroup$
    – copper.hat
    Dec 14, 2015 at 20:44
  • $\begingroup$ @copper.hat Good to know. Too many references computes the kernel without showing the step making me wonder if there is some unknown insight involved $\endgroup$
    – Fraïssé
    Dec 14, 2015 at 20:45
  • $\begingroup$ Still worthwhile learning tricks such as the above :-). $\endgroup$
    – copper.hat
    Dec 14, 2015 at 20:46

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In this very special case one can do it by inspection. Suppose that $\langle x,y,z\rangle$ is in the kernel. The first row of $A$ tells you that $y=-z$, and the second that $x=y$. The last row is the difference of the first two, so it adds no information. Thus, the kernel is the set of multiples of $\langle 1,1,-1\rangle$.

That, however, is just a fast solution of $Ax=0$ that happens to be possible in this case; in general there’s nothing better.

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  • $\begingroup$ Indeed it is!!! $\endgroup$
    – Fraïssé
    Dec 14, 2015 at 20:46

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