Definition talk:Inner Product

Maybe there is need for a proof that $<Ax,y>=<x,A^Ty>$? --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: [1] , [2], [3]. 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)

Can the above be extracted into whatever theorem pages are necessary? That would give us an edge. --prime mover (talk) 17:26, 28 January 2023 (UTC)

Just double-checked my sources, and my above comments are not entirely correct. The Completion Theorem (Inner Product Space) does work for non-complete fields, but the completion will be equal to an inner product space over a complete field. For instance, if you complete an inner vector space over $\Q$, the result will be equal to an inner product space over $\R$, though they will not be isomorphic as vector spaces.

The main theorem where the inner product space $V$ needs to be over $\R$ or $\C$ is the Gram-Schmidt Orthogonalization, where we use unit vectors of the type:

$\dfrac v {\norm v} = \dfrac v {\sqrt{\innerprod v v} }$

Here, we need to require that the field is closed under the norm + square root operation. I will add the requirement to the Gram-Schmidt Orthogonalization theorem. This is, as far I can se after a quick skim, the only theorem we have where we need to add a requirement.

So how about we explain the issue on the Definition:Inner Product page by adding an "Also defined as" section, and write:

"Some texts define an inner product only for vector spaces over $\R$ or $\C$.

This ensures that for all $v \in V$, the inner product norm $\norm v = \sqrt {\innerprod v v}$ is an element of the field $\Bbb F$.

$\mathsf{Pr} \infty \mathsf{fWiki}$ prefers the more general definition, and lists additional requirement on $\Bbb F$ in theorems where it is needed, such as the Gram-Schmidt Orthogonalization theorem."

--Anghel (talk) 21:02, 28 January 2023 (UTC)

Go for it --prime mover (talk) 22:33, 28 January 2023 (UTC)

Good discussion, fully agree with the outcome. — Lord_Farin (talk) 17:26, 31 January 2023 (UTC)

Not sure if you have already mentioned but in some sense the inner product only makes sense for subfields $F \subseteq \C$ because of complex conjugate. If you define it for a more general field, then it is no longer compatible with the usual one on $\C^n$ --Usagiop (talk) 22:07, 31 January 2023 (UTC)

What you mean is probably called Definition:Symmetric Bilinear Form. This is not called the inner product, usually. --Usagiop (talk) 22:11, 31 January 2023 (UTC)

Usagiop, is it not sufficient to introduce a certain involution on the field which generalizes the conjugation? And then still impose sesquilinearity with the general involution? We might be getting overly generic here but I don't immediately see a downside to that. Worst case "it so happens" that subfields of $\C$ are the only ones, but then this could be a theorem putting all the above on a solid foundation. — Lord_Farin (talk) 09:08, 1 February 2023 (UTC)

Do you want to generalize Definition:Sesquilinear Form using this abstract conjugation, and call that an inner product on a algebra? I am not sure if I understood it correctly but I mean the inner product is usually defined for $\R$ and $\C$ and has the well-known geometric meaning. For example, a 'sesquilinear form' on $\paren {\Z \sqbrk i}^n$ is like a inner product but is usually not called so. Anyway this is just a minor comment from me, I have neither objections nor deeper knowledge. --Usagiop (talk) 20:38, 1 February 2023 (UTC)