# Vector space or inner product space - ambiguous!

1. ### failexam

378
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?

2. ### Number Nine

779
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.

3. ### HallsofIvy

40,241
Staff Emeritus
Given any (finite dimensional) vector space there are an infinite number of possible "inner products". For example, choose any basis, $\{e_1, e_2, ..., e_n\}$. We can now write two vectors, $u= a_1e_1+ a_2e_2+ ...+ a_ne_m$ and $v= b_1e_1+ b_2e_2+ ...+ b_ne_n$, written in terms of that basis.
We define the inner product $<u, v>= a_1b_1+ a_2b_2+ ...+ a_nb_n$.

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.

4. ### morphism

2,020
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.)

5. ### homeomorphic

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).

6. ### Office_Shredder

4,500
Staff Emeritus
If the dimension is, say 2c how the devil are you going to add all those numbers up?

7. ### Number Nine

779
We add an uncountable number of numbers all the time in calculus.

8. ### HallsofIvy

40,241
Staff Emeritus
And "almost all" such sums do not converge.

9. ### Fredrik

10,162
Staff Emeritus
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.

Last edited: Mar 28, 2012
10. ### micromass

18,448
Staff Emeritus
Notice that the sum is a finite sum. Even if there is an uncountable basis.

11. ### Bacle2

1,175
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.

12. ### Bacle2

1,175
Not if you're working with a Schauder basis.

13. ### SteveL27

803
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.

14. ### morphism

2,020
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.

15. ### Bacle2

1,175
Right, my bad, I lost focus and was making a general statement about sums in V.Spaces. The force is back with me now.

16. ### homeomorphic

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.

17. ### Fredrik

10,162
Staff Emeritus
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.

18. ### morphism

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!

19. ### Fredrik

10,162
Staff Emeritus
OK, I get it now. Thanks. It wouldn't work for an orthonormal basis (=maximal orthonormal set) for a Hilbert space, but it would work for a Hamel basis (=maximal linearly independent set). Of course, Hilbert spaces already have inner products, and there's usually no need to define another one.

20. ### Alesak

134
To OP interesting fact might be that at least in finite dimensional case, all inner products are kinda similar, since every inner product can be written as $<u, v>= k_1 a_1b_1+ k_2 a_2b_2+ ...+ k_n a_nb_n$ for suitable choice of basis. In real case, this can be reduced to $<u, v>= a_1b_1+ a_2b_2+ ...+ a_nb_n$.