Hermitian dot product (finite fields)

Let $q$ be an http://planetmath.org/node/4703even http://planetmath.org/node/438prime power (in particular, $q$ is a square) and $\\mathbb{F}_{q}$ the finite field with $q$ elements. Then $\\mathbb{F}_{\\sqrt{q}}$ is a subfield of $\\mathbb{F}_{q}$ . The $\\overline{k}$ of an element $k\\in\\mathbb{F}_{q}$ is defined bythe $\\sqrt{q}$ -th power Frobenius map

\\overline{k}:=\\operatorname{Frob}_{\\sqrt{q}}(k)=k^{\\sqrt{q}}.

The has properties similar to the complex conjugate. Let $k_{1},k_{2}\\in\\mathbb{F}_{q}$ , then

1.
$\\overline{k_{1}+k_{2}}=\\overline{k_{1}}+\\overline{k_{2}}$ ,
2.
$\\overline{k_{1}k_{2}}=\\overline{k_{1}}\\,\\overline{k_{2}}$ ,
3.
$\\overline{\\overline{k_{1}}}=k_{1}$ .

Properties 1 and 2 hold because the Frobenius map is ahttp://planetmath.org/node/1011homomorphism.Property 3 holds because of the identity $k^{q}=k$ whichholds for any $k$ in any finite field with $q$ elements.See also http://planetmath.org/node/2893finite field.

Now let $\\mathbb{F}_{q}^{n}$ be the $n$ -dimensional vector space over $\\mathbb{F}_{q}$ , then the Hermitian dot product of two vectors $(u_{1},\\ldots,u_{n}),(v_{1},\\ldots,v_{n})\\in\\mathbb{F}_{q}^{n}$ is

(u_{1},\\ldots,u_{n})\\cdot(v_{1},\\ldots,v_{n}):=\\sum\\limits_{i=1}^{n}u_{i}%\\overline{v_{i}}.

Again, this kind of Hermitian dot product has properties similar toHermitian inner products on complex vector spaces. Let $k_{1},k_{2}\\in\\mathbb{F}_{q}$ and $v_{1},v_{2},v,,w\\in\\mathbb{F}_{q}^{n}$ , then

1.
$(k_{1}v_{1}+k_{2}v_{2})\\cdot w=k_{1}(v_{1}\\cdot w)+k_{2}(v_{2}\\cdot w)$ (linearity)
2.
$v\\cdot w=\\overline{w\\cdot v}$
3.
$v\\cdot v\\in\\mathbb{F}_{\\sqrt{q}}$

Property 3 follows since $\\sqrt{q}+1$ divides $q-1$ (see http://planetmath.org/node/2893finite field).