Does a set and a field together always generate a vector space?

[kent davidge]

Does it make sense to say that a set together with a field generates a vector space? I came across this question after starting the thread https://www.physicsforums.com/threads/determine-vector-subspace.941424/

To be more specific, suppose we have a set consisting of two elements $A = \{x^2, x \}$ and let the reals $\mathbb{R}$ be the field. The set $A$ doesn't form a vector space, but it generates a vector space, the space of all polynomials of the form $ax^2 + bx$, which by the way is a subspace of $P^2(x)$ which is discussed on that thread.

Is my thought about generating instead of forming correct?

 

[Orodruin]

A set in itself does not generate a vector space. In order to do that you must have a notion about what it means to add elements of the set and what it means to multiply them by a scalar. However, in your specific example, you do. The set therefore spans a subset of polynomials of degree 2.

 

[kent davidge]

A set in itself does not generate a vector space. In order to do that you must have a notion about what it means to add elements of the set and what it means to multiply them by a scalar

That's why I said $A$ isn't a vector space over the reals --the vector space axioms couldn't be satisfied. However $A$ together with the reals is like a generator for the mentioned subspace of $P^2(x)$. Correct?

The set therefore spans a subset of polynomials of degree 2

Which is also a subspace of $P^2(x)$, correct?

 

[Orodruin]

In this case you had monomials so it is pretty clear what addition means (addition of functions is typically done pointwise). However, your set could contain elements for which this is much less clear.

 

[PeroK]

... for example, it's not clear that the set $\{apple, banana \}$ generates a Real vector space. Unless you come up with approriate rules for adding fruit.

Note, however, that any non-empty subset of a vector space spans a subspace. Typically that subset may not be a finite set of linearly independent vectors. Any subset will do.

 

[Orodruin]

it's not clear that the set $\{apple, banana \}$ generates a Real vector space.

I don't know why, but this is exactly the example I considered mentioning but decided not to. It was between that and $\{\mbox{red, blue}\}$ ...

 

[kent davidge]

... for example, it's not clear that the set $\{apple, banana \}$ generates a Real vector space. Unless you come up with approriate rules for adding fruit.

I understand this. My point is that unless you come up with an addition rule such that, say, $apple + banana = banana$ you wouldn't satisfy some of the vector space axioms.

In the exmple I gave of polynomials, if you add the two elements in the set $A$ with the usual addition of polynomials, you end up with an element not in the set $A$, thus $A$ couldn't be a vector space.

 

[Stephen Tashi]

Does it make sense to say that a set together with a field generates a vector space?

Are you asking whether that vague general concept makes sense? - or whether there is a standard formal definition that precisely formulates the idea?

I'd say the vague notion makes sense because we can have a similar notion for a "free group" on a set of abritrary things and that notion has a formal definition. The vague notion by itself fails to give a specific definition for "generated by".

I suspect we could write formal definitions for "free fields", "free vectors spaces" etc Whether such definitions are regarded as standard, I don't know.

A category theorist might be able to give a master definition that applies to any mathematical structure.

 

[mathwonk]

in algebra there is a standard construction that takes any set at all, and a field k, and constructs a vector space over k for which that given set is essentially a basis, hence generates it. you just treat the elements exactly as you treated the monomials x and x^2. I.e. the elements of the space are formal sums of formal multiples of the set elements by field scalars. in your example the elements of the vector space are symbols of form : c.apple + d.banana and you combine them in the obvious way.
e.g. b.(c.apple+d.banana) = bc.apple + bd.banana.

More formally, given any set S, consider all functions from S to the field k, which have value zero at all but a finite number of elements of S. Then like any space of k valued functions, this is a k vector space. The element x of S corresponds to the function with value 1 at x and value zero at all other elements of S. In the naive treatment above the symbol "c.apple" was the function with value c at the element apple and value zero at the element banana.

The symbol "c.apple + d.banana" is the function with value c at apple and value d at banana. This vector space is isomorphic to the usual coordinate space k^2 consisting of pairs of scalars (c,d) from k. In general a vector v in k^n has coordinates (v1,...,vn) and we can think of v as a function on the generating set {1,2,...,n}, the coordinate vj being just the value of the function v on the element j.In a categorical sense, given a set S, and a field k, there is an essentially unique vector space F(S) called the vector space (freely) generated by S, and determined by the following properties: 1) there is an injection of the set S into the space F(S); 2) for every vector space W, there is a one-one correspondence between linear transformations F(S)-->W and set functions S-->W.

I.e. a linear map F(S)-->W is determined by its restriction to S, and every function on S extends to a linear transformation on F(S).

in short, the answer to the question posed above: "Does it make sense to say that a set together with a field generates a vector space?", is yes. This is covered in every basic graduate algebra class. See for example, Dummit and Foote, Abstract Algebra, 3rd ed., p. 354. for the generalization to free modules over any commutative ring, rather than just a field.I have answered your question as asked. However having now read the linked thread, there is a given vector space there, and the elements are assumed to be elements of that space with the given operations, so in that setting there is no reason for the subspace generated by a given set within the given vector space, to be a basis. I understood from your question that we wanted to create the vector space just out of the elements of the given set, and the field. In short, in the construction I gave, if it so happens the given elements already belong to some vector space, that fact is just forgotten, and I have constructed a new vector space containing them and in which they become independent.

Hence in the example in the linked thread, where the vector space was taken to be the polynomials of degree ≤ 2, the set of three elements {X, X^2, X+X^2} generate only a 2 dimensional vector subspace of the space of polynomials, but in my construction they would generate a new 3 dimensional vector space. We would also need to come up with a new symbol for vector addition, i.e. we could not use +, since in our new space the sum of X and X^2 would not be the polynomial already called X+X^2. I.e. if we use "+" for addition in the space of polynomials, we would need a new symbol in our new space.

 

[mathwonk]

after re - reading your question, I apologize for all this high flying baloney. you seem to be assuming you are in the setting wherein there is already a given vector space, and your set of elements belongs to that space. thus we are already given a vector space structure complete with addition and scalar multiplication, and then yes, your intuition is exactly correct: namely given any set of elements of that space, they do generate, exactly as you say, a subspace of that space in exactly the way you describe. Contrary to my construction above, there is no reason then for the elements to be a basis for the subspace they generate, i.e. they may not be independent. In this setting, given elements X, X^2, and X+X^2, we would assume the vector space to be the obvious one they belong to, namely polynomials, and then they would generate the subspace of all polynomials of form aX + bX^2 + c(X+X^2), which is 2 dimensiopnal, and is also generated (freely) by the two elements X and X^2.

 

[WWGD]

A set in itself does not generate a vector space. In order to do that you must have a notion about what it means to add elements of the set and what it means to multiply them by a scalar. However, in your specific example, you do. The set therefore spans a subset of polynomials of degree 2.

I believe that even when sums are not defined, just using formal sums is enough to generate a subspace/vector space.

 

[Mark44]

I believe that even when sums are not defined, just using formal sums is enough to generate a subspace/vector space.

I'm not sure what this means. A vector space consists of a set of objects, together with an operation defining addition of these objects, and another operation that defines multiplication of the objects by scalars belonging to some field. "Using formal sums" presumes that some sort of additions already defined, at least if I understand what you're saying.

 

[FactChecker]

I understand this. My point is that unless you come up with an addition rule such that, say, $apple + banana = banana$ you wouldn't satisfy some of the vector space axioms.

This is not exactly true. You can consider it a two dimensional vector space, where one dimension is apples and the other is bananas. Define apple + banana as (1 apple, 0 banana) + (0 apple, 1 banana) = (1 apple, 1 banana) to define addition in a two-dimensional vector space. Things that can not be added, can be kept as separate dimensions in the definition of a vector space addition. That is what is done in your example of polynomials ... different powers of x are kept separate. Just make them different dimensions in the $(x, x^2)$ vector system.

In the exmple I gave of polynomials, if you add the two elements in the set $A$ with the usual addition of polynomials, you end up with an element not in the set $A$, thus $A$ couldn't be a vector space.

Yes it can be considered a vector space. Addition like:
$r_1x+r_2x^2$ is defined as $(r_1x, 0) +(0, r_2x^2) = (r_1x, r_2x^2)$
When you add polynomials, you only add identical powers of x. You don't really add different powers of x except symbolically. So you can consider the symbology $r_1x+r_2x^2$ to be the same as $(r_1x, r_2x^2)$ for the purpose of considering it a vector space.

PS. In abstract algebra, this is called a direct product. (see https://en.wikipedia.org/wiki/Direct_product_of_groups )

 

[Orodruin]

I believe that even when sums are not defined, just using formal sums is enough to generate a subspace/vector space.

Yes, as already pointed out in detail in #9. However, I do not think this is what the OP was looking for. That you formally can do something does not necessarily mean that it has the meaning you were looking for.

When you add polynomials, you only add identical powers of x. You don't really add different powers of x except symbolically.

This depends on how you consider the construction. If you construct it formally as described in #9, then yes, but then there is an implicit assumption that the elements of the set are linearly independent (see #10). However, the normal construction when dealing with function spaces is to define addition and multiplication pointwise according to
$$
(f_1+f_2)(x) = f_1(x) + f_2(x), \quad (kf)(x) = k f(x),
$$
not referring to any special basis of the vector space.

 

[FactChecker]

However, the normal construction when dealing with function spaces is to define addition and multiplication pointwise according to
$$
(f_1+f_2)(x) = f_1(x) + f_2(x), \quad (kf)(x) = k f(x),
$$
not referring to any special basis of the vector space.

I agree. But to give the impression that polynomials can not be considered a vector space would leave one unprepared for a lot of powerful and beautiful math results.

 

[Orodruin]

But to give the impression that polynomials can not be considered a vector space would leave one unprepared for a lot of powerful and beautiful math results.

I think (hope?) that we have not given that impression anywhere in this thread. I was trying to explicitly point out in #2 that you indeed already have natural definitions of addition and multiplication with a scalar in the case of polynomials.

 

[FactChecker]

I think (hope?) that we have not given that impression anywhere in this thread. I was trying to explicitly point out in #2 that you indeed already have natural definitions of addition and multiplication with a scalar in the case of polynomials.

Yes. Your answer (post #9) was very good, complete and correct. I was just responding to the earlier post by @kent davidge that seemed to indicate his belief that it was not a vector space. I wanted to correct that misconception as clearly and concisely as I could.

 

[Orodruin]

Your answer (post #9) was very good, complete and correct.

Just for the record, that post is by @mathwonk, but I agree.

 

[kent davidge]

I realized (after reading about) that what I was thinking of "generator" is called span. The span of the considered set $A$ is equal to $P^2(x)$.

 

[WWGD]

I realized (after reading about) that what I was thinking of "generator" is called span. The span of the considered set $A$ is equal to $P^2(x)$.

It may be more precise to say that the span of the _subset_ A , in this case, generates $P^2(x)$. The span is the smallest subspace containing the set.

 

[mathwonk]

I just realized after reading the several previous posts that polynopmials provide a perfect illustration of the formal construction described in post #9. I.e. one formal definition of a real polynomial is as a formal linear combination of the symbols 1,x,x^2,x^3,...,x^n,... i.e. any symbol of form a + bx + cx^2+...+ex^n. This is exzctly the formal vector space having the set {1,x,x^2,x^3,...,x^n,... } as a basis. So all of us have seen this formal construction in that one case.

It is true of course that many people prefer to consider them as certain functions, defined by these symbols, but that intyerpetation fails when the field of scalars is finite. I.e. a polynomial with coefficients from the field k, can always be viewed as a k valued function,. but if k is finite, that function does not determine the polynomial back again. I.e. there are many polynomials all of whose values are the same. E.g. if the elements of the field are say {0,1}, the field of 2 elements, then the polynomial X(X-1) has value zero at both elements of the field, and similarly if f(X) is any polynomial, then f(X)+X(X-1) have the same values at both elements of the field. This is easy to generalize for any finite field.

So in the general case, a polynomial over a field k, must be defined as an elememnt of the abstract vecor space over k having the set {1,X,X^2,X^3,...,X^n,...} as basis, or something more sophisticated that does not prefer these variables in terms of X. E.g. one could also define the space of polynomials as above in #9, initially as the unique formal vector space with basis {0,1,2,3,4...} but then we would also have to say how to define the multiplication. In this approach a polynomial over k is a sequence (a0,a1,a2,...) of elements of k which are eventually all zero. Addition is clear, and multiplication follows the rule you could probably guess.
I.e. (a0,a1,a2,...).(b0,b1,b2,b3,...) = (a0b0, a0b1+a1b0, a0b2+a1b1+a2b0, a0b3+a1b2+a2b1+a3b0,...). This is the approach taken in Zariski-Samuel, vol. 1, p.24-25.

 

[FactChecker]

I realized (after reading about) that what I was thinking of "generator" is called span. The span of the considered set $A$ is equal to $P^2(x)$.

You need to include a constant also. $\{1, x, x^2\}$

 

[kent davidge]

You need to include a constant also. $\{1, x, x^2\}$

It was shown that the set $A$ spans $P^2(x)$ without need for a constant. It became clear from the thread https://www.physicsforums.com/threads/determine-vector-subspace.941424/ that any vector in $P^2(x)$ can be written as a linear combination of elements of $A$.

 

[Orodruin]

It was shown that the set $A$ spans $P^2(x)$ without need for a constant. It became clear from the thread https://www.physicsforums.com/threads/determine-vector-subspace.941424/ that any vector in $P^2(x)$ can be written as a linear combination of elements of $A$.

He is not talking about the other thread. He is talking about the set you gave in the OP here. You obviously cannot span $P^2(x)$ with only two monomials.

 

[WWGD]

Davidge: Specifically in this case, how would you go about obtaining a 1 with just the set $\{ x, x^2\}$? You would need to have constants $c_1, c_2$(Field elements) with $c_1x+c_2x^2 =1$, with '=' meaning equal as polynomials. This is not possible ( except possibly for finite fields), e.g., by the fundamental theorem of algebra.

 

[kent davidge]

Davidge: Specifically in this case, how would you go about obtaining a 1 with just the set $\{ x, x^2\}$?

I thought he was referring to the set $A$ discussed in the thread https://www.physicsforums.com/threads/determine-vector-subspace.941424/

 

[WWGD]

I thought he was referring to the set $A$ discussed in the thread https://www.physicsforums.com/threads/determine-vector-subspace.941424/

Ah, OK, sorry.

 

[Orodruin]

I thought he was referring to the set $A$ discussed in the thread https://www.physicsforums.com/threads/determine-vector-subspace.941424/

Ah, OK, sorry.

I don't think there is a need to be sorry. @kent davidge, the typical thing to do when you talk about a set $A$ in a thread where you have defined a set $A$ is to assume that you are talking about the set you defined in the thread, not a set you defined in another thread. Even if you also linked to that thread.

 

[FactChecker]

I was not referring to another thread. It seems like everyone agrees that $\{x, x^2\}$ does not generate $P^2(x)$. That is all I was trying to say.

I see that the OP correctly states that it generates a subspace of $P^2(x)$.