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 :-/
Best Answer
Certainly the "obvious" inner product of $(v_1,\dots v_n)$ with $(w_1,\dots w_n)$ given by $\sum_n v_n w_n$ is defined for any (finite dimensional) vector space. It is symmetric and bilinear and fairly useful. It is used extensively, for example, when discussing linear codes over finite fields. (The codes are vector spaces!)
There isn't any reason to discouraged that the "positive definite" condition is missing. There are many interesting inner product spaces where the inner product is not positive definite... even over $\mathbb{R}$! The inner product of Minkowski space is not positive definite, for example, and this is the setting for special relativity.
Positive definiteness is kind of a "Euclidean ideal" that we would hope for, but the Minkowski space example shows that nature isn't always Euclidean, and that non-positive-definite inner products arise naturally.