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We know that the standard inner product of a vector space over a finite field might not be positive definite.

e.g. $$x = (1, 1) \in \mathbb{Z_2}\times\mathbb{Z_2}$$ $$x \cdot x = 1 + 1 = 2 = 0$$

Also we defined two vectors $x,y$ to be orthogonal if $x \cdot y = 0$

So my question is,

Let $V$ be an $n$ dimensional vector space over a finite field $F$,

Let $B = \{x_1, x_2,...x_n\}$ be a basis such that

$$x_i \cdot x_j = 0 \text{ for } i \neq j \ \ (1)$$ $$x_1 \cdot \ x_1 = 0 \ \ (2)$$

Then any $1 \times n$ matrix with $x_1$ being its row would fail the rank-nullity theorem.

I know that this is impossible because you cannot have a non-zero vector being orthogonal to everything in the space. Therefore the construction of $B$ is impossible.

But you take a basis $C = \{x_1, v_2, v_3, ... v_n\}$, apply gram-schmidt to that without normalizing the vectors, then won't you get an orthogonal basis that satisfy the condition $(1), (2)$

So I'm just wondering what is wrong here?

Nevermind I figured out.... because $x_1 \cdot x_1 = 0$, gram-schmidt would blow up as soon as I try to work out the second vector.

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    Orthogonality does not make any sense over finite fields. Grahm Schmidt does not apply over finite fields. – 2017-01-26
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    Why would it satisfy (2)? Just because you did not normalize does not mean that it might not still have at least some vectors of non-zero "length". – 2017-01-26
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    by gram schmidt, you keep the first vector in your final set, therefore (2) is satisfied – 2017-01-26
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    Rank nullity works independent of what field you're thinking about, and independent of any inner product. – 2017-01-26
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    @Crostul Orthogonality makes sense, we just use reflexive bilinear forms instead of inner products (so you are correct that Gram-Schmidt does not work without some modification, like skipping the "scaling" step) – 2017-01-27

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The rank nullity theorem just says something about the dimensions of a pair of related subspaces. Whether or not you can find orthogonal bases of those subspaces using a given inner product is a different question. Your example shows the answer may be "no".

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    yes but what I'm saying is that suppose such $B$ exist, then rank-nullity would fail for that $1 \times n$ matrix because rank is 1, but kernel is $n$. But how do you prove that such $B$ cannot exist – 2017-01-26
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    I'm puzzled. You've accepted my answer (thanks) but still ask your question. You have a basis $B$, What map from $n$-space to $1$-space are you thinking about? It can't have "$x_1$" in it. It's just a column vector of scalars that represent the action of the map on vectors written with respect to the coordinate system $B$. Perhaps it maps $x_1$ to $1$ and the other basis vectors to $0$. Then its rank is ... and its nullity is ... – 2017-01-26