
#1
Nov1812, 02:31 AM

P: 68

Hey, here's a simple question.
I have been reading some materials and, for the nth time in my life, there was a definition of an inner product as a function [itex]V \times V \rightarrow F[/itex], where [itex]V[/itex] is an abstract vector space and [itex]F[/itex] 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 [itex]R^n[/itex] (if n is the dimension of the vector space), so we effectively consider only [itex]R^n[/itex] 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 [itex]R^n[/itex] 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? 



#2
Nov1812, 05:16 AM

HW Helper
P: 6,189

Hi Lajka!
First off, the definition of a vector space also includes two binary operations: addition and multiplication with a scalar. So you can add vectors. 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 ℝ^{n} (like in the example I just described). 



#3
Nov1812, 08:29 AM

Sci Advisor
P: 1,716

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
Nov1912, 10:37 AM

P: 302

A question about inner product
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. 



#5
Nov1912, 11:10 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,882

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




#6
Nov1912, 11:39 AM

P: 150





#7
Nov1912, 11:44 AM

P: 771

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
Nov1912, 08:25 PM

Sci Advisor
P: 1,716

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 ntuple 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. 



#9
Nov2012, 01:49 AM

P: 68

Thanks for your answers, everyone!
@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. 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. I like Serena's suggestion is to simply define it. I think that sounds reasonable. 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. Then, I can define inner product on this "underlying" R space as a standard multiplication. So, for example, inner product [itex]<2,4>[/itex] would be [itex]0.4 \cdot 0.8 = 0.32[/itex]. 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 



#10
Nov2012, 02:18 AM

Sci Advisor
P: 778





#11
Nov2012, 02:58 AM

P: 302

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
Nov2012, 06:45 AM

P: 150

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
Nov2012, 08:00 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,882





#14
Nov2012, 09:17 AM

P: 302

I think that you meant to talk about Banach algebras. Any complex Banach algebra which is a field is isomorphic to C. 



#15
Nov2012, 10:00 AM

Sci Advisor
P: 1,716





#16
Nov2012, 11:19 AM

Mentor
P: 16,591

Another useful generalization is that of a Hilbert [itex]C^*[/itex] module which is just a complex vector space E equipped a right Amodule structure and with an "inner product" [itex]E\times E\rightarrow A[/itex] which takes values in a [itex]C^*[/itex]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 [itex]A=\mathbb{C}[/itex]. But there are still many nice results that can be proven for things like this. 


Register to reply 
Related Discussions  
dot product question  Calculus  2  
Need help with combination of dot product and cross product question  Introductory Physics Homework  2  
Dot Product Question  Calculus & Beyond Homework  3  
DC Product Question.  Electrical Engineering  6  
inner product question.  Calculus & Beyond Homework  8 