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After applying Gaussian reduction mod 2, I've ended up with this matrix:

$$\left(\begin{array}{ccccccccc} 1 & 1 & 1 & 1 & 0 & 0 & 1 & 1 & 0 \\ 0 & 1 & 1 & 1 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \end{array}\right)$$

This seems like it's currently in echelon form. Is it possible to back solve this (using operations mod 2) to get basic variables so that the null space of my original matrix can be found, and if so, how?

My idea: Treat the 7th and 8th columns free variables. Then, starting with the 6th row, simplify each row, bottom to top, until each row is independent of the others.

This seems like it will work, but will a method like that always work for matrices similar to the above?

qwr
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  • I've seen that you've edited your question, now my comment doesn't make sense anymore, so i'm deleting it :) – Bman72 Aug 24 '14 at 07:52
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    Yes, it will work - indeed this is the standard method. Every column without a pivot corresponds to making the corresponding variable a free variable. Then, going right to left (equivalently, bottom to top), solve for each pivot (non-free) variable in terms of the free variables. Note that the 9th column has no pivot, and thus the 9th variable is a free variable as well! – Greg Martin Aug 26 '14 at 07:02

1 Answers1

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My idea ended up actually working.

I treated the 7th and 8th columns (and the 9th) as free variables since they did not have a corresponding row. Then through back substitution I reduced the left side until it had a nice representation.

$$\left(\begin{array}{ccccccccc} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \end{array}\right)$$

Thanks to Greg Martin for confirmation.

qwr
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