# Is a complex space more “advanced” than a “generic” real space?

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?

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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
@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: 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
Tom: Thank you, much appreciated! – t.b. Jun 30 '11 at 17:51
@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

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.

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!

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