Vector space or inner product space - ambiguous!


by failexam
Tags: ambiguous, product, space, vector
failexam
failexam is offline
#1
Mar26-12, 11:52 PM
P: 338
Sometimes the way authors write a book makes you wonder if they are at their wits' end: 'A vector space with an inner product is an inner product space!'

I am not sure if I have gone crazy but to me it is obvious that if you have a vector space wrapped up in a nice gift basket and sent off to your address from Amazon, then the inner product comes with the vector space for free. After all, the inner product is obtained from any two vectors in a vector space, so I just can't understand why a vector space cannot be pre-equipped with an inner product, that is, why the hell isn't a vector space and an inner product space the same thing?
Phys.Org News Partner Science news on Phys.org
Review: With Galaxy S5, Samsung proves less can be more
Making graphene in your kitchen
Study casts doubt on climate benefit of biofuels from corn residue
Number Nine
Number Nine is offline
#2
Mar27-12, 01:08 AM
P: 771
Because the definition of vector space doesn't include an inner product.

An inner product is a function mapping a pair of vectors to an element of the underlying field; until you have defined such a function, you do not have an inner product. The definition of vector space does not include such a function, therefore, a vector space is not necessarily an inner product space.
HallsofIvy
HallsofIvy is online now
#3
Mar27-12, 08:12 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,890
Given any (finite dimensional) vector space there are an infinite number of possible "inner products". For example, choose any basis, [itex]\{e_1, e_2, ..., e_n\}[/itex]. We can now write two vectors, [itex]u= a_1e_1+ a_2e_2+ ...+ a_ne_m[/itex] and [itex]v= b_1e_1+ b_2e_2+ ...+ b_ne_n[/itex], written in terms of that basis.
We define the inner product [itex]<u, v>= a_1b_1+ a_2b_2+ ...+ a_nb_n[/itex].

Choosing a different basis will give a different inner product. (And the "theoretical meat" of the Gram-Schmidt orthogonalization process is that, given any abstractly defined inner product there exist a basis in which that inner product is as given above.

And note the "finite dimensional". There exist infinite dimensional vector spaces which do not have any inner product.

morphism
morphism is offline
#4
Mar27-12, 03:59 PM
Sci Advisor
HW Helper
P: 2,020

Vector space or inner product space - ambiguous!


Quote Quote by HallsofIvy View Post
And note the "finite dimensional". There exist infinite dimensional vector spaces which do not have any inner product.
That's not true - your method will produce an inner product on a real vector space of any dimension. (The trouble, of course, is "choosing" a basis. But once you have one, you're good to go.)
homeomorphic
homeomorphic is offline
#5
Mar27-12, 09:16 PM
P: 1,039
Originally Posted by HallsofIvy View Post

And note the "finite dimensional". There exist infinite dimensional vector spaces which do not have any inner product.

That's not true - your method will produce an inner product on a real vector space of any dimension. (The trouble, of course, is "choosing" a basis. But once you have one, you're good to go.)
Yes, they all have an inner product. The dimension, as in cardinality of a basis, is a complete invariant if all you are looking at is the vector space structure, so there's not much variety there. However, if you have a norm, not all norms are induced by inner products and that's probably what he was thinking. For example, L^p is not an inner product space for p not equal to 2 (check the parallelogram identity).
Office_Shredder
Office_Shredder is offline
#6
Mar28-12, 08:23 AM
Mentor
P: 4,499
Quote Quote by morphism View Post
That's not true - your method will produce an inner product on a real vector space of any dimension
If the dimension is, say 2c how the devil are you going to add all those numbers up?
Number Nine
Number Nine is offline
#7
Mar28-12, 10:08 AM
P: 771
Quote Quote by Office_Shredder View Post
If the dimension is, say 2c how the devil are you going to add all those numbers up?
We add an uncountable number of numbers all the time in calculus.
HallsofIvy
HallsofIvy is online now
#8
Mar28-12, 10:14 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,890
And "almost all" such sums do not converge.
Fredrik
Fredrik is online now
#9
Mar28-12, 10:58 AM
Emeritus
Sci Advisor
PF Gold
Fredrik's Avatar
P: 8,999
Quote Quote by failexam View Post
Sometimes the way authors write a book makes you wonder if they are at their wits' end: 'A vector space with an inner product is an inner product space!'

I am not sure if I have gone crazy but to me it is obvious that if you have a vector space wrapped up in a nice gift basket and sent off to your address from Amazon, then the inner product comes with the vector space for free. After all, the inner product is obtained from any two vectors in a vector space, so I just can't understand why a vector space cannot be pre-equipped with an inner product, that is, why the hell isn't a vector space and an inner product space the same thing?
A vector space over a field F is defined as a triple (set, addition operation, scalar multiplication operation) that satisfies a bunch of axioms. If F=ℝ or F=ℂ, then an inner product space over F is defined as a pair (vector space over F, inner product) that satisfies a bunch of axioms. For example, when we refer to ℝ2 as a vector space, we're actually being sloppy. ℝ2 is just a set. However, if we define three functions ##A:\mathbb R^2\times\mathbb R^2\to\mathbb R^2##, ##S:\mathbb R\times\mathbb R^2\to\mathbb R^2## and ##I:\mathbb R^2\times\mathbb R^2\to\mathbb R## by
$$
\begin{align}
A\big((x_1,x_2),(y_1,y_2)\big) &=(x_1+x_2,y_1+y_2)\\
S\big(a,(x_1,x_2)\big) &=(ax_1,ax_2)\\
I\big((x_1,x_2),(y_1,y_2)\big) &=x_1 x_2+y_1y_2
\end{align}
$$ for all ##(x_1,x_2), (y_1,y_2)\in\mathbb R^2## and all ##a\in\mathbb R##, then ##(\mathbb R^2,A,S)## is a vector space over ℝ, and if we denote that space by V, then ##(V,I)## is an inner product space over ℝ.

This post may be useful.
micromass
micromass is offline
#10
Mar28-12, 11:03 AM
Mentor
micromass's Avatar
P: 16,652
Quote Quote by Office_Shredder View Post
If the dimension is, say 2c how the devil are you going to add all those numbers up?
Notice that the sum is a finite sum. Even if there is an uncountable basis.
Bacle2
Bacle2 is offline
#11
Mar28-12, 02:04 PM
Sci Advisor
P: 1,168
Quote Quote by Number Nine View Post
We add an uncountable number of numbers all the time in calculus.
Not quite; if you're thinking about integration, you are selecting countably-many points, and countably-many partitions.

Any uncountable sum with more than countably-many non-zero terms, necessarily diverges. Just partition your uncountable support-set into sets An:={x:x>1/n}; at least one of the sets will have infinitely-many terms.
Bacle2
Bacle2 is offline
#12
Mar28-12, 02:06 PM
Sci Advisor
P: 1,168
Quote Quote by micromass View Post
Notice that the sum is a finite sum. Even if there is an uncountable basis.
Not if you're working with a Schauder basis.
SteveL27
SteveL27 is offline
#13
Mar28-12, 04:25 PM
P: 799
Quote Quote by failexam View Post
Sometimes the way authors write a book makes you wonder if they are at their wits' end: 'A vector space with an inner product is an inner product space!'

I am not sure if I have gone crazy but to me it is obvious that if you have a vector space wrapped up in a nice gift basket and sent off to your address from Amazon, then the inner product comes with the vector space for free. After all, the inner product is obtained from any two vectors in a vector space, so I just can't understand why a vector space cannot be pre-equipped with an inner product, that is, why the hell isn't a vector space and an inner product space the same thing?
This thread's drifted a bit, but the answer to your question is that when you care about the vector space properties, there's no point adding in extraneous details.

If you want to intersect two sets, there's no point mentioning that the sets are also vector spaces or groups or manifolds or anything else. If you loaded up every definition with all of its derived types, it would be incredibly confusing.

A set is such and so. A group is a set with such and so. A Lie group is a group with such and so. You build up complex definitions in terms of simpler ones.

If all you care about is the properties of a vector space, why add on additional properties that you don't care about?
morphism
morphism is offline
#14
Mar28-12, 08:31 PM
Sci Advisor
HW Helper
P: 2,020
Quote Quote by Bacle2 View Post
Not if you're working with a Schauder basis.
Yes but Schauder bases are irrelevant here.

The claim was that we can construct an inner product on any vector space. The argument was: (1) Choose a (Hamel) basis. (2) Every vector is a finite linear combination of vectors from this basis (even if the basis itself is infinite). (3) HallsofIvy's recipe for an inner product still works, regardless of whether the vector space is finite- or infinite-dimensional.
Bacle2
Bacle2 is offline
#15
Mar28-12, 08:40 PM
Sci Advisor
P: 1,168
Right, my bad, I lost focus and was making a general statement about sums in V.Spaces. The force is back with me now.
homeomorphic
homeomorphic is offline
#16
Mar28-12, 09:26 PM
P: 1,039
This thread's drifted a bit, but the answer to your question is that when you care about the vector space properties, there's no point adding in extraneous details.
That's true, but it's not the only issue. The other issue is that there is no canonical choice of inner product. They all have an inner product. Requiring the existence of an inner product doesn't add anything because it follows from the definition of a real or complex vector space. But which inner product?

The definition isn't a vector space on which there exists an inner product. It's a vector space with some chosen inner product that you have singled out.
Fredrik
Fredrik is online now
#17
Mar28-12, 09:29 PM
Emeritus
Sci Advisor
PF Gold
Fredrik's Avatar
P: 8,999
Quote Quote by morphism View Post
HallsofIvy's recipe for an inner product still works, regardless of whether the vector space is finite- or infinite-dimensional.
Am I missing something obvious here? Clearly it works for those vectors that belong to the smallest vector subspace that contains all the basis vectors (because the members of that set are linear combinations of basis vectors), but if the vector space we're talking about is the closure of that subspace, then it seems to me that it should fail.
morphism
morphism is offline
#18
Mar28-12, 09:38 PM
Sci Advisor
HW Helper
P: 2,020
But we're working with a (Hamel) basis for the entire vector space, so (by definition!) every vector is a finite linear combination of elements in the basis!


Register to reply

Related Discussions
Tensor product and infinite dimensional vector space Linear & Abstract Algebra 1
Basis - Complex Vector Space and Real Vector Space Calculus & Beyond Homework 5
normed linear space and inner product space Calculus & Beyond Homework 8
Vector space of the product of two matrices Linear & Abstract Algebra 2
cross product and vector space Linear & Abstract Algebra 3