# A question about inner product

1. Nov 18, 2012

### Lajka

Hey, here's a simple question.

I have been reading some materials and, for the n-th time in my life, there was a definition of an inner product as a function $V \times V \rightarrow F$, where $V$ is an abstract vector space and $F$ is an underlying scalar field.

However, it got me thinking. Inner product is this special function which gives us some number, right? In order to get a number, you must work with numbers.

Now, in general, our abstract vectors are not sequences of numbers, be it matrices, ordered pairs, polynomials or whatever, they are just that, abstract vectors. The only thing that connects these vectors with some numbers are their coordinates with respect to some basis. And in that case, we performed an isomorphism to the vector space $R^n$ (if n is the dimension of the vector space), so we effectively consider only $R^n$ in that case, so let's not do that.

What I am getting here, is how is it possible to construct an inner product on an abstract vector space? How do we take two ordinary abstract vectors and get a number out of it? You can't sum these vectors, you can't multiply them etc., they are not numbers. You can only do that with their coordinates. And if you do that, then you are not defining an inner product on a general vector space, you have defined an inner product on $R^n$ and you are using isomorphism to indirectly define inner product on other vector spaces of the same dimension. That can't be right, can it?

So, what do you think od my dilemma?

Last edited: Nov 18, 2012
2. Nov 18, 2012

### I like Serena

Hi Lajka!

First off, the definition of a vector space also includes two binary operations: addition and multiplication with a scalar.

How?
That is something you simply have to define.
Say your vector space is the set V={apple,pear} with a field F={0,1}.
Then you also have to define what apple+pear is and what 0×apple is.
Note that these definitions need to comply with the axioms for addition and scalar multiplication.

And if you want to define an inner product, you need to define what (apple, apple) is and what (apple, pear) is, and these have to be elements of F.
To be a proper inner product, it also needs to fulfill the axioms of an inner product.

Btw, there won't always be an isomorphism between V and ℝn (like in the example I just described).

3. Nov 18, 2012

### lavinia

If you have a basis for the vector space then constucting an inner product is easy. For instance you can declare the basis vectors fo be ortonormal.

Without a basis I don't know of a general method.

If you relax the requirement that the inner product be positive definite then the trivial bilinear form works naturally.

You can also define inner products on infinite dimensional spaces

4. Nov 19, 2012

### Erland

An important example of an inner product space where the inner product is not defined through coordinates in a particular basis is the following:

Let $C[a,b]$ be the set of continuous real valued functions on the interval $[a,b]$, (where $a$ and $b$ are real numbers with $a<b$). This is a real vector space if addition of functions and multiplication of functions by scalars are defined in the obvious way.

Now, $C[a,b]$ is an inner product space with an inner product defined by:

$\langle f,g\rangle= \int_a^b f(x)g(x)\,dx$, for $f,g\in C[a,b]$.

It can be verified that this is an inner product on $C[a,b]$.

Especially important is the case $C[-\pi,\pi]$, where the functions $1$, $\cos x$, $\sin x$, $\cos 2x$, $\sin 2x$, $\cos 3x$, $\sin 3x$, ... are pairwise orthogonal w.r.t. this inner product.

Last edited: Nov 19, 2012
5. Nov 19, 2012

### HallsofIvy

Staff Emeritus
Note that Erland's example is an infinite dimensional vector space. Every vector space, V, of finite dimension, n, is isomorphic to $R^n$ and, given a basis for V, an inner product on it can written in terms of the "standard" inner product on $R^n$. (But different bases will give different inner products.)

6. Nov 19, 2012

This only applies to real vector spaces. Vector space over $F$ is generally isomorphic to $F^n$

7. Nov 19, 2012

### Number Nine

It gives us a scalar.
All vector spaces have an underlying scalar field by definition. The function maps an ordered pair of vectors to some scalar. There's no need to start talking about "numbers".

8. Nov 19, 2012

### lavinia

In the infinite dimensional cases I have always been a little unclear what a basis is.

I always thought is was a linearly independent set so that any vector could be expressed as a finite sum of basis vectors. Note finite sum.

So the orthonormal trigonometric functions described above would not be a basis since in general infinitely many of them are required to span a function.

To say that any vector space even has a basis is equivalent to assuming the Axiom of Choice. But this gives no procedure for finding one.

In this sense it is true that the inner product using an integral is defined without a basis.
But we do know that value of the function at every point and this in some sense is like knowing it on a basis After all in the finite dimensional case all a vector is with respect to a basis is a n-tuple of numbers. A function is just an infinite tuple of numbers. Same idea.

Here is another example of an inner product not defined on a basis.

On the real numbers define the usual inner product - just multiply the two numbers together. Now consider the real numbers to be a vector space over the rational numbers. The inner product is not defined in teems of a basis for this vector space. In fact, I don't think it is possible to describe a basis for this vector space.

Last edited: Nov 19, 2012
9. Nov 20, 2012

### Lajka

Yes, of course, thanks for the correction. You still get a vector, tho, that's what I was trying to say, poorly.

I have to say it never occured me it could be like that. Here I was trying to figure out how could I produce a number(scalar) out of these abstract vectors, whilst I could just define <apple,pear> = 1. It's a very interesting thought!

Ah, yes. But I was kinda trying to avoid that. After all, maybe I don't want the basis of my choice to be orthogonal/orthonormal. Actually, I was trying to understand how to construct an inner product without involving any kind of a basis. After all, in the definition of an inner product, you don't see the concept of a basis as a requirement anywhere.

@Erland Thanks, that is a good example of an inner product without a mention of any basis.
Here is the thing, though.
Functions, vectors, matrices, polynomials etc., those are all "constructions", to put it that way, that involve numbers in them at their very core. So extracting a number out of them is not an impossible task to imagine.

What I was trying to explain to myself, is how would you do that for some general, abstract vector space, which, by itself, has no numbers in them whatsoever, aside from those numbers being an underlying scalar field.

Yeah, we don't even have to use standard product on Rn, but that only complicates things, I guess.

However, must we do it like this? This way, we depend on Rn, if you get what I'm trying to say. In fact, the moment we define inner product this way, we only work with Rn and our former vector space doesn't really matter anymore, as far as computations go.

Don't get me wrong, I have no problem with this. I was just wondering if we could properly define an inner product without invoking Rn. It seems to me that, if we cannot do this, then the idea of an inner product is not really universal for all vector spaces, only for Rn.

Yes, of course, I've been reckless, thank you. So, we have two dinstict objects, vectors and scalars. The question is how is it possible to map two vectors to a scalar.

I like Serena's suggestion is to simply define it. I think that sounds reasonable.

I think that, in an infinite dimensional case, you define a basis by saying that a closure of its span, well, spans the space. That way you deal with all those functions that represent the limits of your finite combinations.

Of course, it has to be possible that every function can be expressed as a limit of your finite combinations, that's what makes basis, well, a basis.
That is how I understand it, anyway.

This is a nice thought experiment! I tried by choosing a 5 to be my basis vector. Then, in this setting, number 1 is a represenation of a number 5, number 3 is a representation of a number 15, number 0.4 is a representation of a number 2 etc.

Then, I can define inner product on this "underlying" R space as a standard multiplication. So, for example, inner product $<2,4>$ would be $0.4 \cdot 0.8 = 0.32$.
But, it's clearly possible for a result of this inner product to be a real number. And that can't be, since the scalar field here are the rational numbers only. Now I, as well, am not sure how would I construct it!

The way you did it at the beginning of your post
was without using any kind of basis. I guess all I maunder here is to see if this is possible in general, when your abstract vectors can't just "multiply together" in order to produce back a number.

10. Nov 20, 2012

### pwsnafu

No. There is a difference between a Hamel basis (finite combinations) and a Schauder basis (countable combinations). The former is used in linear algebra the latter in linear analysis. Also the latter only makes sense in a topological vector space.

11. Nov 20, 2012

### Erland

Actually, I have never seen inner products for vector spaces over other fields than R and C. And it is not clear to me how an inner product can be defined for vector spaces over other fields. For one of the axioms for an inner product for a vector space over C is:

$\langle u,v\rangle=\overline{\langle v,u\rangle}$.

To generalize this, we must have a field F in which there is something corresponding to complex conjugation, and it is not obvious what that should be in a general case.

12. Nov 20, 2012

I don't think you'd have to generalize it. There is no general definition of inner product so definition for a given field is driven mainly by convenience.

13. Nov 20, 2012

### HallsofIvy

Staff Emeritus
Actually it is. All infinite fields are isomorphic to either the real numbers or the complex numbers and the isomorphism defines the "conjugate".

14. Nov 20, 2012

### Erland

If you with "infinite field" mean a field with infinitely many elements, this is certainly wrong. Q (the rational numbers) is an obvious counterexample.
I think that you meant to talk about Banach algebras. Any complex Banach algebra which is a field is isomorphic to C.

15. Nov 20, 2012

### lavinia

As I said there is no general method.

no. This is a different idea of basis entirely.

The basis is the number 1 for the reals over the reals so in this case the basis is implicit in the definition of the inner product. For the reals over the rationals no one will ever know what the basis is..

16. Nov 20, 2012

### micromass

Staff Emeritus
Well, it's not the general case, but I suppose that if you have a vector space over an ordered field F, then you can define an inner product on that. Furthermore, you can complexify F to $F+iF$ and make that into a field. There you got an obvious choice for complex conjugation, so I suppose that for ordered fields, you can define a good inner product.

Another useful generalization is that of a Hilbert $C^*$ module which is just a complex vector space E equipped a right A-module structure and with an "inner product" $E\times E\rightarrow A$ which takes values in a $C^*$-algebra. There is a good notion of "positive" and of "conjugations" in A, so we can mimic the construction of the inner product. Of course, A won't be a field unless $A=\mathbb{C}$. But there are still many nice results that can be proven for things like this.