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In answer to the question here:

Lights out game on hexagonal grid

The following argument is given:

"Thus if v is in the null space of A, then d is orthogonal to v and as a consequence, d is in the row space of A."

Here everything is over $\mathbb{F}_2$, and the "inner product" is the standard dot product $x\cdot y=\sum_{i=1}^nx_iy_i$.

However, this is not an inner product at all! It may very well be the case that $x\cdot x=0$ but $x\ne 0$.

I'm sure that a certain amount of linear algebra can still be preserved here. In particular, if I can prove that $$\dim W + \dim W^\perp = \dim V$$ it will suffice. So my main question here is how can I prove the above equality in this setting.

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    Dear Gadi: The equality $\dim W + \dim W^\perp = \dim V$ holds whenever the bilinear form is non-degenerate. The proof is the obvious one. (I assume $\dim V < \infty$.)2012-01-24

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