Definition talk:Inner Product

Maybe there is need for a proof that $=$? --Espen180 22:25, 22 October 2009 (UTC)

I would say so, definitely. Feel free to write such a page. --Prime.mover 05:28, 23 October 2009 (UTC)

It's done. I didn't know what to call it though. Espen180 05:51, 23 October 2009 (UTC)

Complex and reals?
How important is it to specify "or $\R$" in that first sentence? A subfield of $\R$ is by definition also a subfield of $\C$ is it not? And if you have the language of abstract algebra under your belt enough to understand what a subfield is, you'll know that. Furthermore, the concept of an "inner product" is fairly well advanced down the route of abstract algebra (coming as it does after vector spaces) that such an interpolation would seem clumsy. --prime mover 04:43, 29 April 2012 (EDT)
 * Well, in my class we're doing inner product spaces, and we haven't been introduced to fields yet. But maybe my class's curriculum is pathological. Fraleigh has inner product spaces as chapter $3.5$ and fields as $9.2$. We're introduced to it as a specific type of a regular vector space, and answering questions like "is $\left \langle {\cdot, \cdot} \right \rangle : \mathbf M_2 \left({\R}\right) \times \mathbf M_2 \left({\R}\right) \to \mathbb \R, \left \langle {\begin{bmatrix} a_1 & a_2 \\ a_3 & a_4 \end{bmatrix}, \begin{bmatrix} b_1 & b_2 \\ b_3 & b_4 \end{bmatrix}} \right \rangle = a_1b_1 + a_2b_2 + a_3b_3 + a_4b_4$ an inner product?" --GFauxPas 07:46, 29 April 2012 (EDT)


 * IMO Fraleigh *is* pathological. OK so what's anyone else think? --prime mover 09:51, 29 April 2012 (EDT)

I would deem the addition 'or $\R$' unnecessary and indeed a bit clumsy; introducing inner product spaces before fields is sort of possible, as its usually practically restricted to $\Bbb F$ being $\R$ or $\C$ anyway. It's not my preferred route, though. --Lord_Farin 17:36, 29 April 2012 (EDT)

Expansion of definition to other subfields (answer to User:Prime.mover)
The question of whether inner product can (or should be defined) for other fields that $\R$ and $\C$ has been posed multiple times at math.stackexchange.com:, ,. Here is my summation of the answers:

1) Let $F$ be a field with multiplicative identity $1$. By a combination of the axioms for linearity, non-negative definiteness and positiveness, we have:
 * $\innerprod 1 1 < \innerprod {1+1}{1+1} < \innerprod {1+1+1}{1+1+1} < \ldots$

This implies that $F$ is an infinite field, and $\Q$ will be a subfield of $F$.

2) Since we want an inner product space to be a normed vector space, we want $\sqrt{ \innerprod r r } \in F$ for all $r \in F$. This rules out $\Q$, and various subfield of the type $\Q [ \sqrt 2]$. Of course, a definition for inner product of subfields of $\R, \C$ is valid, and this is indeed what we do now.

3) More technical: Most applications of inner product spaces treat it as a pre-Hilbert Space, which can be completed to form a Hilbert Space: Completion Theorem (Inner Product Space). For this, the field $R$ needs to be complete. You can define an inner product space over, say the algebraic numbers, but that field is not complete. The answers on StackExchange imply that there is no applications for inner product spaces over fields that are not complete, and this is the main reason why almost all sources define these spaces over $\R$ and $\C$ only. --Anghel (talk) 16:25, 28 January 2023 (UTC)