If $B:V \times V \rightarrow K$ is a symmetric bilinear form over a finite-dimensional vector space, where the characteristic of the field is not 2, then we may prove that there is an orthogonal basis such that $B$ is represented by

$$ \bordermatrix{& \cr & a_{1} & 0 & \ldots & 0\cr & 0 & a_{2} & \ldots & 0\cr & \vdots & \vdots & \ddots & \vdots\cr & 0 & 0 &\ldots & a_{n}\cr } $$

Recall that a bilinear form has a well-defined rank, and denote this by $r$

If $K = \mathbb{R}$ we may choose a basis such that $a_1 = \cdots = a_t = 1$ $a_{t+1} = \cdots = a_{t+p} = -1$ and $a_{t+p+j} = 0$ for some integers $p$ and $t$ where $1 \le j \le n-t-p$ Furthermore, these integers are invariants of the bilinear form. This is known as Sylvester's Law of Inertia. $B$ is positive definite if and only if $t = n$ $p = 0$ Such a form constitutes a real inner product space.

If $K = \mathbb{C}$ we may go further and choose a basis such that $a_1 = \cdots = a_r = 1$ and $a_{r + j} = 0$ where $1 \le j \le n-r$

If $K = F_p$ we may choose a basis such that $a_1 = \cdots = a_{r-1} = 1$

$a_r = n$ or $a_r = 1$ and $a_{r+j} = 0$ where $1 \le j \le n-r$ and $n$ is the least positive quadratic non-residue.