# Vector space or inner product space - ambiguous

• spaghetti3451
In summary: R##. In other words, the space is defined by the triple (set, addition, scalar multiplication) and the inner product is given by the equation\begin{align}I&=\left( \begin{array}{ccccc}x_1&\quad &x_2\\y_1&\quad &y_2\\\end{array} \right)\end{align} where the dot product is defined on any two vectors in the space.The space ℝ2 is not an inner product space because the norm is not induced by the inner product. For example, consider the vectors ##(x_
spaghetti3451
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?

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.

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.

HallsofIvy said:
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.)

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).

morphism said:
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?

Office_Shredder said:
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.

And "almost all" such sums do not converge.

failexam said:
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.

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Office_Shredder said:
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.

Number Nine said:
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.

micromass said:
Notice that the sum is a finite sum. Even if there is an uncountable basis.

Not if you're working with a Schauder basis.

failexam said:
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.

Bacle2 said:
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.

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

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.

morphism said:
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.

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!

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.

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 $<u, v>= a_1b_1+ a_2b_2+ ...+ a_nb_n$.

Don't mean to over-nitpick, but we may want to choose a vector space together with a choice of inner-product _up to isomorphism_ , i.e., so that (V,<,>):=(V, <,>') iff. (def.)*

there is a linear bijection L:V-->V with <vi,vj>=<L(vi),L(vj)>

I don't know how many there are, but I suspect these may be unique in the finite-dimensional case, but I may be (even egregiously) wrong.* Explaining that iff. is meant here as a def. , to spare others the pain I went thru trying to prove (what I did not know were) definitions given with an iff.

Don't mean to over-nitpick, but we may want to choose a vector space together with a choice of inner-product _up to isomorphism_ , i.e., so that (V,<,>):=(V, <,>') iff. (def.)*

there is a linear bijection L:V-->V with <vi,vj>=<L(vi),L(vj)>

I don't know how many there are, but I suspect these may be unique in the finite-dimensional case, but I may be (even egregiously) wrong.

They are unique in the finite-dimensional case.

But the definition of an inner product space is not up to isomorphism. If you defined it up to isomorphism, then you're sort of modding out by isomorphisms, so you're losing information, which is bad.

Fredrik said:
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.

So does that mean that every non-zero finite-dimensional inner-product space has an orthonormal basis? would that basis then be unique?

Krovski said:
So does that mean that every non-zero finite-dimensional inner-product space has an orthonormal basis? would that basis then be unique?

No, not at all. An inner-product space has (in general) many orthonormal bases.
For example, in $\mathbb{R}^2$, we have (1,0) and (0,1) that constitute a good basis, but also $(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})$ and $(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}})$.

Krovski said:
So does that mean that every non-zero finite-dimensional inner-product space has an orthonormal basis? would that basis then be unique?

This in fact is true even for infinite-dimensional i-p spaces, but the construction there is inductive (if the dimension is countable, at least).

You may want to google "Gram-Schmidt Process", which shows how to build an orthonormal basis out of ANY given basis in

an i-p linear space.

DonAntonio

And of course, if the dimension is not countable, then you're going to need something like the axiom of choice to produce an orthonormal basis.

Krovski said:
So does that mean that every non-zero finite-dimensional inner-product space has an orthonormal basis? would that basis then be unique?

Try the scaling property of inner product:

<a,b>=0 , then <ca,db>=cd<a,b> to produce more orthonormal bases.

homeomorphic said:
They are unique in the finite-dimensional case.

But the definition of an inner product space is not up to isomorphism. If you defined it up to isomorphism, then you're sort of modding out by isomorphisms, so you're losing information, which is bad.

Well, yes, I guess this will depend on the goal of the specific situation, whether you want to generalize, or you want to isolate a specific type/category of spaces.

## 1. What is a vector space?

A vector space is a mathematical structure that consists of a set of vectors and operations, such as addition and scalar multiplication, that satisfy certain properties. These properties include closure, associativity, commutativity, and distributivity.

## 2. What is an inner product space?

An inner product space is a vector space that also has an inner product, which is a mathematical operation that takes two vectors as input and returns a scalar. This inner product is defined in such a way that it satisfies properties such as linearity, symmetry, and positive definiteness.

## 3. How are vector spaces and inner product spaces related?

Vector spaces and inner product spaces are related in that an inner product space is a special type of vector space that has an additional structure, namely an inner product. All inner product spaces are vector spaces, but not all vector spaces are inner product spaces.

## 4. What is the difference between a vector space and an inner product space?

The main difference between a vector space and an inner product space is that an inner product space has an additional structure, the inner product, which allows for the definition of notions such as length, angle, and orthogonality. In other words, an inner product space provides a notion of distance and direction, while a vector space does not.

## 5. Why is the term "vector space" sometimes used interchangeably with "inner product space"?

The terms "vector space" and "inner product space" are sometimes used interchangeably because all inner product spaces are vector spaces, and the additional structure of the inner product is often implied or understood in the context of the discussion. However, it is important to note that not all vector spaces are inner product spaces, and the two terms should not be used interchangeably in all cases.

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