# Does any field induce a partial order?

I was reading about inner product spaces, and came accross the definition of positive definiteness, which says that if $$V$$ is a vector space over a field $$F$$ and $$\langle\cdot,\cdot\rangle:V\times V\to F$$ is the inner-product, then

$$\langle v,v\rangle\geq 0$$

Which means that there must be some ordering in that field, how is this handled formally?

• Almost a duplicate of math.stackexchange.com/questions/49348/… – Travis Nov 8 '18 at 19:24
• At you asking what an ordered field is? – rschwieb Nov 8 '18 at 22:26
• No, I'm asking why, in the definition, the field need to have an ordered structure, or if I'm missing something. – Garmekain Nov 8 '18 at 22:32
• Where were you reading it? That would be a very bad book if it defines positive definite inner product over a field $F$ without first specifying that $F$ has an order! – GEdgar Nov 9 '18 at 14:13

Usually, inner product spaces are only defined over $$\Bbb R$$ or $$\Bbb C$$.
If it's $$\Bbb R$$, then $$\langle v,v\rangle\ge0$$ makes sense since we have an ordering of reals.
If it's $$\Bbb C$$, it's defined to satisfy $$\langle v,w\rangle=\overline{\langle w,v\rangle}$$. An example of an inner product space over $$\Bbb C^2$$ would be: $$\langle(w_1,w_2),(z_1,z_2)\rangle= w_1\overline{z_1}+w_2\overline{z_2}.$$ Because of this, $$\langle v,v\rangle$$ will equal its own conjugate, and thus be real. Since it's real, $$\langle v,v\rangle\ge0$$ makes sense.
(Note that, in the inner product given for $$\Bbb C^2$$ above, $$\langle(a+bi,c+di),(a+bi,c+di)\rangle=\\a^2+b^2+c^2+d^2.$$ )