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Here's what I'm reading right now:

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So, the question that I have relating to this is if it needs to be proven that an inner product can be defined on every real vector space and if there are infinitely many inner products that can be defined on each real vector space, with the exception of $\{0\}$. If there is a proof for this, how would it go? I'm somewhat confused by Dr Klaus Janich's wording because it seems like this is something that can be proved.

Also, the way that it has been worded kind of gives me the impression that one can prove this by taking a real vector space and showing that there is a way to construct a set of inner products on that vector space. Then, you'd have to show that that set has infinitely many elements. Would this be a correct way of looking at it or am I just getting this entirely wrong?

Abhijeet Vats
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    Given a real vector space $V$, if $\beta$ is a basis you can define an inner product in terms of $\beta$ simply by declaring $\beta$ to be an orthonormal basis. – Arturo Magidin Apr 17 '20 at 19:37
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    I would point out that the real numbers are a vector space in and of themselves and that in that case $A=a$ is just a scalar. You can think of this change in inner product like a change in units or basis. – CyclotomicField Apr 17 '20 at 19:37
  • Oh okay, thank you for your comments. I will take note of all of this. – Abhijeet Vats Apr 17 '20 at 19:40

1 Answers1

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Suppose $V$ is a real vector space with inner product $(\cdot, \cdot)$. Then for any $c >0$ define inner product $\langle \cdot, \cdot \rangle$ by $\langle x,y\rangle = c(x,y)$. Show that this defines a perfectly good inner product.

Thus if we have inner product (not on the zero space, the only space for which the inner product is just the constant function to 0), we have infinitely many, just given by scaling as above.

Now what's left to be shown is any real vector space admits an inner product. Let $V$ a real vector space. By a standard result, every real vector space has a basis (note this requires the Axiom of Choice in the infinite case and this is necessary for this proof; this is really a technical detail you don't have to worry about i you don't want to/don't understand though). Let $\{b_i\}_{i \in I}$ be a basis for $V$. Then define an inner product $(x,y) = \sum_i x_i y_i$ where $x_i, y_i$ denote the coefficient on $b_i$ of $x$ (resp. $y$) represented in the basis. Note that only finitely many of the $x_i$ and finitely many of the $y_i$ will be nonzero, thus that sum makes sense and is finite. This is from the definition of a basis. You can check this really defines an inner product.

Physical Mathematics
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    Thanks so much for your answer. Unfortunately, I haven't gone as far as the Axiom of Choice yet so I can't really use it to prove things. I'll prove that first statement that you asked me to the prove and the last statement in your answer. I'll probably take note of your answer so that I may use it as a reference once I have learned the relevant material. – Abhijeet Vats Apr 17 '20 at 19:41
  • You do need the Axiom of Choice (in the form of Zorn's Lemma) to prove every vector space has a basis, but if you assume that, then my answer works without having to actually use Choice. Thus you should be able to understand my answer as long as you assume that every vector space has a basis (which certainly sounds like a reasonable enough thing to assume). – Physical Mathematics Apr 17 '20 at 19:43
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    Yeap, I don't mind assuming that till I can prove it at a later point. – Abhijeet Vats Apr 17 '20 at 19:44
  • If the vector space is a topological vector space, then the basis is going to be a topological basis. $L^p$ spaces for $0<p<1$ do not have a norm. I'm not an expert, it'll be great if you clear my doubt! – Offlaw Apr 17 '20 at 19:45
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    We are not assuming that there is already a norm. Every norm does not come from an inner product but every vector space (without any additional structure) can be given an inner product. – Physical Mathematics Apr 17 '20 at 19:46
  • +1. But I somewhat disagree with "this is really a technical detail you don't have to worry about if you don't want to/don't understand though". The bases given by Zorn's Lemma are potentially huge and a bit exotic. So while an inner product does exist, you won't be able to construct it, and it's not going to be natural in any sense. I think this should be kept in mind. Regardless, this is a solid answer. – Jair Taylor May 21 '20 at 19:38