I've stumbled in the german wikipedia over the question if a vector space over a finite field exists, that has a dot product.
Definition of dot product
A dot product over a $\mathbb{K}$-vector space V is a mapping $F : V \times V \rightarrow \mathbb{R}$ which is...
... bilinear: $\forall a, a_1, a_2, b, b_1, b_2 \in V \forall \lambda_1, \lambda_2, \mu_1, \mu_2: $ $\begin{align*} F(\lambda_1 \cdot a_1 + \lambda_2 \cdot a_2, b) &= \lambda_1 \cdot F(a_1, b) + \lambda_2 \cdot F(a_2, b)\\ F(a, \mu_1 \cdot b_1 + \mu_2 \cdot b_2) &= \mu_1 \cdot F(a, b_1) + \mu_2 \cdot F(a, b_2) \end{align*}$
... symmetric: $F(a, b) = F(b,a)$
... positive definite: $\forall a \in V: F(a,a) \geq 0 \land ( F(a,a) = 0 \Leftrightarrow a = 0)$
My try
Please correct my choice of words if you know what I mean and if you know how to write it in English. I am not used to write math texts in English.
$\mathbb{K} := \mathbb{Z} / 2 \mathbb{Z}$ Let $V$ be a $\mathbb{K}$-vector space over the set $\{0,1\}$.
$\mathbb{K}$ is a finite filed.
Define $\langle 0, 0 \rangle := 0$, $\langle 1, 0 \rangle := \langle 0,1 \rangle := 1$ and $\langle 1, 1 \rangle = 1$.
symmetry
$\langle, \rangle$ is symmetric per definition
bilinear form
$f(1 + 1, 1) = f(0,1) = 1 \neq 0 = 1+1 = f(1,1) + f(1,1)$
Is it possible to define a dot product for a bigger vector space / finite field?
(If the answer is yes, please give me an example)
edit: argh - $f(a,a) = 0 \Leftrightarrow a = 0$, but $f(a,b)$ can be 0. I always forget that :-/