Rank Nullity for Vector Space over Finite Field 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.

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".
