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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?

Garmekain
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  • Almost a duplicate of https://math.stackexchange.com/questions/49348/inner-product-spaces-over-finite-fields – Travis Willse Nov 08 '18 at 19:24
  • At you asking what an ordered field is? – rschwieb Nov 08 '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 08 '18 at 22:32
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    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 09 '18 at 14:13

1 Answers1

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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, $$\langle(a+bi,c+di),(a+bi,c+di)\rangle=\\a^2+b^2+c^2+d^2.$$ )

Akiva Weinberger
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