Orthogonalize method, keeping to binary vectors

Question

The built in function:

Orthogonalize[m]

where m is a list of vectors returns a orthogonal basis for m. However, I want to stick to the binary field. However, I tried this:

Orthogonalize[m,Mod[#1.#2,2]&]

But this just still gives a set of vectors with negative coordinates.

Can anyone please help.

Thanks.

Can you give an example of a set of vectors where your proposal fails, and what the answer ought to be? — J. M.'s eventual burnout, Aug 14, 2012 at 5:29
Will vectors that are self-orthogonal be problrmatic for your purposes? if not, could just do Gram-Schmidt without the normalization part. — Daniel Lichtblau, Aug 14, 2012 at 15:33

whuber · Accepted Answer · 2012-08-14 18:07:37Z

9

Define the inner product modulo $2$:

orthogonalize[a_] := Mod[Orthogonalize[a, Mod[#1.#2, 2] &], 2]

Example:

(m = Union[orthogonalize[RandomInteger[{0, 1}, {500, 64}]] ]) // ArrayPlot

Orthogonal basis

Check orthogonality:

Mod[m . Transpose[m], 2] // ArrayPlot

enter image description here

answered Aug 14, 2012 at 18:07

whuber

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