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For instance, does taking the square root of a complex number and its complex conjugate create a metric that "automatically" makes it an inner product space?

Is a complex space more complete than a real space? If not, what must be done to make it complete, so that it IS a Hilbert space?

And does the modulus represent a norm that makes a complete complex space a Banach space?

t.b.
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Tom Au
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  • I'm not sure what "automatically" could really mean here. When you say "complex space", do you mean "vector space over the complex numbers"? Modulus is an operation on complex numbers, not vector spaces over the complex numbers. Also, note that complete and norm are mathematics terms with precise definitions - the scare quotes are unnecessary. – Zev Chonoles Jun 30 '11 at 13:47
  • Possibly related: http://math.stackexchange.com/questions/21792/norms-induced-by-inner-products (I'm just guessing that you mean to ask among other things if the norm obtained by $|z| = \sqrt{\langle z, z\rangle}$ determines the scalar product). The answer to that would be yes. – t.b. Jun 30 '11 at 13:48
  • Please forgive me if this is too basic for you but let me mention that the inner-product on $\mathbb{C}$ defined by the rule $\langle z,w \rangle = z\overline{w}$ for all $z,w\in\mathbb{C}$ induces the norm on $\mathbb{C}$ that you state. Moreover, $\mathbb{C}$ is complete under the norm induced by this inner product and therefore it is a Hilbert space. – Amitesh Datta Jun 30 '11 at 13:53
  • @zev: Not all real vector spaces are inner products spaces because not all of them have inner products. Does multiplication by a complex conjugate give it an inner product and ipso facto make it an inner product space? A complex number has an angle and a radius (I may have called it modulus by mistake.) Do these serve as a norm? I've edited out the quotes. – Tom Au Jun 30 '11 at 13:56
  • @amitesh: It's not too "basic," it's probably what I need. If anything, it might not be "basic" enough. But it is (in essence), the answer I was looking for. Put it in the form of a formal answer and I'll upvote it. Using a wikipedia page, I'd copy the reference from the browser, and post it on this page. – Tom Au Jun 30 '11 at 13:57
  • @Zev Although this is quite irrelevant to this discussion, I am curious: how do you link to the Wikipedia pages in your comment? If I must link to "modulus", then I need to type the entire (web) address; I am not sure how one can link to the relevant Wikipedia page otherwise. (I am happy to delete this comment afterwards.) – Amitesh Datta Jun 30 '11 at 13:58
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    @Amitesh: use markdown. Link obtained by writing [markdown](http://daringfireball.net/projects/markdown/). – t.b. Jun 30 '11 at 14:00
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    @Tom: And not all complex vector spaces are inner product spaces either. Complex conjugation is an operation that is defined only for complex numbers; and even then, a vector space is not equipped with a multiplication operation (the complex numbers are). So, the notion of multiplying an element of a vector space by its complex conjugate does not make sense. – Zev Chonoles Jun 30 '11 at 14:00
  • @Tom Are you sure? If so, I can add my comment as an answer if you would like. I thought (and indeed I think others thought) that you were asking whether or not there are other inner products on $\mathbb{C}$ that induce the norm on $\mathbb{C}$ that you stated. (Theo has already answered that question.) – Amitesh Datta Jun 30 '11 at 14:01
  • @Tom I just saw your edited comment (too many comments too fast!); I will add my comment as an answer then. – Amitesh Datta Jun 30 '11 at 14:03
  • Why are there three downvotes to this question? It'd be nice if some of the voters would care to provide some constructive comments. – Willie Wong Jun 30 '11 at 14:25
  • @Willie: I suppose my comments are not technically constructive, in the sense that they don't propose improvements to the question, but that's because I have no idea how to improve the question, or what it's really trying to ask. I feel that I've raised several valid criticisms of the question that Tom has not addressed, and that's why I downvoted. – Zev Chonoles Jun 30 '11 at 14:31
  • @zev: I have a confession to make. I'm really "in over my head." Meaning that I don't fully understand my own question or how to "tweak" it. I'm really working on an "analogies" basis, that is, if a vector space has an inner product, and a complex space has a product of z multiplied by its complex conjugate, that you can take the square root of, does that make it a 1) and inner product space and 2) a norm. I'm not a true mathematician, only a "hedgie" that uses mapping algorithms. – Tom Au Jun 30 '11 at 14:37
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    @Tom: I'm sorry if I was being harsh. I would summarize my point as simply that there are many things involving complex numbers (e.g. modulus, complex conjugate, argument, multiplication) that you are incorrectly assuming also make sense for a elements of a vector space over the complex numbers. – Zev Chonoles Jun 30 '11 at 15:58
  • @Zev: I know that I'm prone to making statements that are wrong. That's EXACTLY why I asked the question. What I'm doing is throwing out a number of statements that I believe to be true. And then asking "what do I know?" versus "what do I 'know' that ISN'T SO?" – Tom Au Jun 30 '11 at 16:02
  • I'm also not aware of any notion of one metric space being "more" complete than another. Finally, you should be aware of the following constructions for a vector space: $$\text{inner product }\Rightarrow\text{ norm }\Rightarrow\text{ metric }\Rightarrow\text{ topology}$$ That is, given an inner product, there is an induced norm, but not necessarily vice versa, and so on. – Zev Chonoles Jun 30 '11 at 16:04
  • @Zev: That's helpful, and basically the kind of answer that I'm looking for. – Tom Au Jun 30 '11 at 16:11
  • @ZevChonoles let us continue this discussion in chat – Tom Au Jun 30 '11 at 16:11
  • @Zev, Tom: In view of Zev's last comment, maybe my answer here is of interest – t.b. Jun 30 '11 at 17:39
  • @theo:another great answer. – Tom Au Jun 30 '11 at 17:44
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    Tom: Thank you, much appreciated! – t.b. Jun 30 '11 at 17:51
  • @Zev: I figured at least one of the down-votes came from someone already engaged in the discussion, which was why I wrote "some of the voters". The others, feel free to carry on with the good job! – Willie Wong Jun 30 '11 at 18:43
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    @Tom: just a suggestion. Providing just a little information about yourself in your profile (e.g. your age; are you a student or an academic; if not, what is your job, etc.) can make it a lot easier for people to give answers which are better tailored for you to understand. One of the first rules of teaching is "Know your audience". If your audience is a nearly anonymous internet entity, that makes things much harder. (I say this especially because it seems like you are willing to divulge some of this information anyway, and it would be more efficient to put it in your profile.) – Pete L. Clark Jun 30 '11 at 19:51

2 Answers2

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We can define an inner product on $\mathbb{C}$ by the rule $\langle z,w \rangle = z\overline{w}$ for all $z,w\in\mathbb{C}$. The norm on $\mathbb{C}$ induced by this inner product is then the map $z\to \left|z\right|$ where $\left|z\right|$ denotes the modulus of the complex number $z$. Finally, $\mathbb{C}$ is complete under the norm induced by this inner product and is therefore a Hilbert space.

Additional Details:

A vector space $V$ over a field $F$ ($F=\mathbb{R}$ or $F=\mathbb{C}$) is an inner-product space if there is a map $V\times V\to F$ (for notational convenience we denote the image of $(z,w)$ under this map by $\langle z,w \rangle$) that satisfies the following axioms:

(1) $\langle v,v \rangle \geq 0$ for all $v\in V$ and $\langle v,v \rangle =0$ if and only if $v=0$.

(2) If $u,v,w\in V$, then $\langle u+v,w \rangle =\langle u, w\rangle + \langle v,w \rangle$.

(3) If $a\in F$ and if $u,v\in V$, then $\langle au,v\rangle = a\langle u,v\rangle$.

(4) If $u,v\in V$, then $\langle u,v\rangle =\overline{\langle v,u\rangle}$.

Exercise 1: Prove that the inner product on $\mathbb{C}$ given by the rule described at the very beginning of this answer is indeed an inner product, that is, it satisfies axioms (1)-(4) above.

Note that axiom 4 can be removed if $F=\mathbb{R}$. If $V$ is an inner product space, then the norm induced by the inner product on $V$ is the map $v\to \sqrt{\langle v,v\rangle}$. (The image of $v\in V$ under this map is denoted by $\left\|v\right\|$ for notational convenience.)

Exercise 2: Prove that the norm induced by the inner product on $\mathbb{C}$ given by the rule described at the very beginning of this answer is indeed the map $z\to \left|z\right|$ where $\left|z\right|$ denote the modulus of the complex number $z$.

If $V$ is an inner product space, then we can define a metric $d:V\times V\to [0,\infty)$ by the rule $d(u,v)=\left\|u-v\right\|$ for all $u,v\in V$.

Exercise 3: Prove that $d$ is indeed a metric on $V$.

Finally, a Hilbert space is an inner-product space $V$ that is complete under the metric induced by its norm.

Exercise 4: Prove that $\mathbb{C}$ is a Hilbert space under the inner product described at the very beginning of this answer.

I hope this helps!

Amitesh Datta
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Here's a compendium of answers by others, from the comments:

Connections between metrics, norms and scalar products (for understanding e.g. Banach and Hilbert spaces)

Norms Induced by Inner Products and the Parallelogram Law

You should be aware of the following constructions for a vector space: inner product ⇒ norm ⇒ metric ⇒ topology That is, given an inner product, there is an induced norm, but not necessarily vice versa, and so on. –

Community
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Tom Au
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