Classification of anisotropic spaces over finite fields

Statement
Suppose $$K$$ is a finite field of characteristic not equal to two and $$b$$ is a symmetric bilinear form on a finite-dimensional vector space $$V$$ over $$K$$. Suppose, further, that $$V$$ is anisotropic with respect to $$b$$, i.e., for any nonzero vector $$v \in V$$, $$b(v,v) \ne 0$$. Then, the dimension of $$V$$ is either 1 or 2. The complete classification is given below.

For describing the classification, we denote by $$q$$ the size of the field. $$q$$ is a prime power (specifically, a power of the field characteristic) and completely determines the field up to isomorphism. Since the characteristic is not two, $$q$$ is odd.

One-dimensional spaces
If $$V$$ is one-dimensional, there are two possibilities for the form $$b$$ up to equivalence:

Two-dimensional spaces when field size is 1 mod 4
If $$q$$ is 1 mod 4, then, up to equivalence, there is a unique anisotropic two-dimensional space. A representative matrix is:

$$\begin{pmatrix} 1 & 0 \\ 0 & u \\\end{pmatrix}$$

where $$u$$ is a quadratic nonresidue. More generally, we can choose any diagonal matrix where one of the diagonal entries is a quadratic residue and the other is a quadratic nonresidue.

Two-dimensional spaces when field size is 3 mod 4
If $$q$$ is 1 mod 4, then, up to equivalence, there is a unique anisotropic two-dimensional spaces. Representatives for these are:

$$\begin{pmatrix} 1 & 0 \\ 0 & 1 \\\end{pmatrix}$$

Facts used

 * 1) uses::Anisotropic form can be represented by diagonal matrix
 * 2) uses::Product of subsets whose total size exceeds size of group equals whole group (the way we use this generalizes the idea behind every element of a finite field is expressible as a sum of two squares)
 * 3) uses::Quotient of determinants of equivalent symmetric bilinear forms is a square

Proof
Given: A finite field $$K$$ of size $$q$$, a $$n$$-dimensional vector space $$V$$ over $$K$$, a symmetric bilinear form $$b$$ on $$V$$ that is anisotropic.

To prove: $$n \le 2$$ and the classification above holds.

Proof: