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I have been trying to construct a orthogonal basis of $\mathbb{F}_{2}^{n}$ for an odd value of $n$ which is comprised of vectors which are not $1$ (to avoid the standard basis). In particular, to mirror the assumption on $n$, I want to ensure that the basis has only vectors with an odd number of $1$'s.

Is this possible? I have tried to prove otherwise, however have run out of ideas.

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    How do you define orthogonality in $\mathbb F_2^n$?2012-11-23
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    What is $\mathbb{F}_{2}^{n}$?2012-11-23
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    Sorry about that. I define orthogonality to be with respect to the inner product $\sum x_{i} y_{i} \pmod{2}$. $\mathbb{F}_{2}^{n} $ is just the field of 2 elements.2012-11-23
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    Thanks. What does "magnitude" mean?2012-11-23
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    Here I am really just counting the number of $1$s in the vector as the magnitude (so not by the inner product).2012-11-23
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    The definition of inner product requires $\langle x,x\rangle \geq 0$, so you need an ordering on the scalars. And an ordered field has (a subfield isomorphic to) $\mathbb Q$ as its prime subfield. In particular no vector space over a finite field admits an inner product. What you have is a symmetric bilinear form on $\mathbb F_2^n$, but it's not an inner product.2012-11-23
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    Right, I asked because $\sum x_i^2$ is either $0$ or $1$ mod 2. Why call the number of 1s the magnitude? Is this standard terminology somewhere? Thanks for clarifying.2012-11-23
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    Sorry about the confusion.2012-11-23
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    Maybe you mean [Hamming weight](http://en.wikipedia.org/wiki/Hamming_weight)?2012-11-23

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