Graduate Vector space (no topology) basis

[mathman]

The standard definition of the basis for a vector space is that all the vectors can be defined as finite linear combinations of basis elements. Consider the vector space consisting of all sequences of field elements. Basis vectors could be defined as vectors which are zero except for one term in the sequence, where that term = 1. However if only finite combinations are considered then only vectors with a finite number of non-zero components would be considered as part of the space. How can a basis be defined so that all the sequences are in the space? Do we need a different definition for basis?

 

[member 587159]

The standard definition of the basis for a vector space is that all the vectors can be defined as finite linear combinations of basis elements. Consider the vector space consisting of all sequences of field elements. Basis vectors could be defined as vectors which are zero except for one term in the sequence, where that term = 1. However if only finite combinations are considered then only vectors with a finite number of non-zero components would be considered as part of the space. How can a basis be defined so that all the sequences are in the space? Do we need a different definition for basis?

First of all, this is not the definition of basis. We also need linear independency.

To make your question more formal. Consider a field $F$ and the space $F^\mathbb{N}:=\{(a_n)_{n=0}^\infty \mid \forall n \geq 0: a_n \in F\}$. Define sequences $(e^i_n)_{n=0}^\infty$ by $e^i_i = 1$ and $e^i_j = 0$ for $j \neq i$. Then you claim that these sequences form a (Hamel)basis for $F^\mathbb{N}$. This is not true, for the reason you cite. We cannot write every element as a finite linear combination of these elements. There is even more, a Hamel basis for this space cannot be countable. So the concept of basis in the linear algebra will be (mostly) useless in this setting.

Why are you confused? My guess is that you have read about the concept orthonormal basis of a Hilbert space (in functional analysis). The example you gave is in fact an orthonormal basis for the Hilbert space $l^2$. But an orthonormal basis in a Hilbert space is not a basis in the linear algebra sense!

Also relevant, and maybe what you are looking for: https://en.wikipedia.org/wiki/Schauder_basis

 

[fresh_42]

This is the difference between a Hamel basis (finite linear combinations) and a Schauder basis (infinitely many allowed).

 

[mathwonk]

you have identified a very interesting question. i noticed it when i was writing my notes on graduate algebra. it concerns the distinction between the "direct sum" of a countable number of copies of a field, and their "direct product". both are vector spaces, and it is very easy to write down a vector basis for the direct sum, wherein all the sequences allowed have only finitely many non zero entries. but i do not know at all how to write down a vector basis for the direct product ("hamel basis"), and presumably no one else does either. so this is a nice explicit example of something whose existence is guaranteed by zorn's lemma, but apparently no one has ever seen or explicitly described one.

 

[fresh_42]

Well, separable Hilbert spaces have an uncountable number of Hamel basis vectors (Wikipedia), and $\mathcal{l}_2$ is separable, even though the only one.

 

[nuuskur]

A basis is a minimal subset whose linear span is the entire space. A basis doesn't have to be unique (and it usually isn't). However, any two bases of the same (non-zero) vector space have the same cardinality.

What you envision is an orthonormal basis of, say, \mathbb C^n where the basis vectors are sequences e_j where the the j-th co-ordinate is the identity and the rest are zero. While this is convenient, a basis isn't required to be orthogonal (orthonormal).

 

[mathman]

I would like a simple answer to the question. Does this vector space (V) have a basis and what is it? The scalar field for V consists of two integers 0 and 1, using mod2 arithmetic. The vectors in V are all sequences, whose elements are either 0 or 1.

 

[member 587159]

I would like a simple answer to the question. Does this vector space (V) have a basis and what is it? The scalar field for V has two elements 0 and 1, using mod2 arithmetic. The vectors in V are all sequences, whose elements are either 0 or 1.

Yes, it has a basis (every vector space has one assuming the axiom of choice). The basis is uncountable. You can't write down the basis explicitely.

 

[fresh_42]

And it has the suggested countable basis $e_i = (0,\ldots,1,0,\ldots)$ if the basis is allowed to be a Schauder basis.

 

[WWGD]

The standard definition of the basis for a vector space is that all the vectors can be defined as finite linear combinations of basis elements. Consider the vector space consisting of all sequences of field elements. Basis vectors could be defined as vectors which are zero except for one term in the sequence, where that term = 1. However if only finite combinations are considered then only vectors with a finite number of non-zero components would be considered as part of the space. How can a basis be defined so that all the sequences are in the space? Do we need a different definition for basis?

I assume a field of chracteristic 0 , so no torsion or other weird issues and the elements are " Infinite tuples" ( f_i1, f_i2,,..,...) ; f_ij in F ? Edit: And no associated norm that gives rise to a topology?

 

[WWGD]

This is the difference between a Hamel basis (finite linear combinations) and a Schauder basis (infinitely many allowed).

But no topology here, so no Schauder Basis,

 

[nuuskur]

I would like a simple answer to the question. Does this vector space (V) have a basis and what is it? The scalar field for V consists of two integers 0 and 1, using mod2 arithmetic. The vectors in V are all sequences, whose elements are either 0 or 1.

It has a basis due to the axiom of choice. We can't give it explicitly, because there is no constructive proof of said fact. You would have to explicitly be able to tell what a maximal element looks like, but Zorn's (Kuratowski's) lemma merely states that under certain assumptions, a maximal element exists.

The basis can't be countable, because that would only yield a countable span. The vector space in question is uncountable.

I should specify, that we are talking in the ordinary linear algebra language: span consists of all Finite linear combinations.

A way to justify this is by the axioms of a vector space: it is, among other things, closed with respect to finite addition.In general, there's no guarantee an arbitrary infinite combination is well-defined. Indeed, what is \sum _{n\in\mathbb N} 1 modulo $2$?

 

[Stephen Tashi]

The standard definition of the basis for a vector space is that all the vectors can be defined as finite linear combinations of basis elements.

If you add words to say the the linear combination is unique, then you have a "standard" definition in the sense of an often-seen definition.

Do we need a different definition for basis?

What creates a need? Are you asking if we need a different definition of basis in order to prove that every vector space has a basis?

This question is addressed by http://www.math.lsa.umich.edu/~kesmith/infinite.pdf Example 3 of those notes is similar to your example.

 

[FactChecker]

The basis can't be countable, because that would only yield a countable span.

I don't follow this. The vector space, V, is uncountable. But consider the set of vectors with only one non-zero component. That is a countable set. Isn't it a basis?

 

[nuuskur]

I don't follow this. The vector space, V, is uncountable. But consider the set of vectors with only one non-zero component. That is a countable set. Isn't it a basis?

A linearly independent subset, yes, but does not span the entire space. As I mentioned, infinite combinations are not permitted.

 

[mathman]

It looks to me that for the example I described, the "basis" could consists of all vectors with one component = 1 and all others = 0, if we allow countable combinations. It seems that all is needed is a different name. If you want a topology, use the sup norm. If the scalar field is discrete, there is no way to get convergence.

 

[fresh_42]

It looks to me that for the example I described, the "basis" could consists of all vectors with one component = 1 and all others = 0, if we allow countable combinations. It seems that all is needed is a different name.

Yep, Schauder basis. See posts #3 and #9.

If you want a topology, use the sup norm. If the scalar field is discrete, there is no way to get convergence.

 

[FactChecker]

It looks to me that for the example I described, the "basis" could consists of all vectors with one component = 1 and all others = 0, if we allow countable combinations. It seems that all is needed is a different name. If you want a topology, use the sup norm. If the scalar field is discrete, there is no way to get convergence.

Yes. Actually, you are the one who specified no topology. With a topology, you can talk about convergence. Without a topology, you must talk about algebraic equivalence. Your vectors have been specified in a way other than an infinite summation. You have to consider when an infinite summation of the basis vectors is equal to your vector.

 

[WWGD]

It looks to me that for the example I described, the "basis" could consists of all vectors with one component = 1 and all others = 0, if we allow countable combinations. It seems that all is needed is a different name. If you want a topology, use the sup norm. If the scalar field is discrete, there is no way to get convergence.

So the cardinality of the field is countable and you are forming the product $F^{|F|}$?

 

[nuuskur]

It looks to me that for the example I described, the "basis" could consists of all vectors with one component = 1 and all others = 0, if we allow countable combinations. It seems that all is needed is a different name.

That would work for finite dimensional vector spaces, because an $n$-dimensional $\mathbb K$-vector space is isomorphic to $\mathbb K^n$.

That's not true for infinite dimensions, so what you propose is not guaranteed to work. Even Schauder basis doesn't quite do what you are looking for. In case of a Schauder basis, for any $x\in V$ there exists a (necessarily unique) sequence of scalars such that $\sum s_je_j = x$ which is an infinite sum, but notice that this is not an arbitrary combination. On the other hand, this requires quite a bit of structure to make sense and afaik it's defined on Banach spaces (perhaps on general topological vector spaces with enough nice properties), which have topologies.

 

[mathman]

A schauder basis almost works, but using a sup norm, I don't see how it can work. Assume the space is simply infinite sequences of elements of the scalar field. How do you get (0,1,0,1,0,1,...) as a limit of any sequence of vectors with a finite number of non-zero elements?

 

[nuuskur]

See, among others, example 2 in here

 

[WWGD]

A schauder basis almost works, but using a sup norm, I don't see how it can work. Assume the space is simply infinite sequences of elements of the scalar field. How do you get (0,1,0,1,0,1,...) as a limit of any sequence of vectors with a finite number of non-zero elements?

Any metric can ultimately be completed, i.e., a metric space that is not complete can be isometrically embedded in one that is complete.

 

[mathman]

Any metric can ultimately be completed, i.e., a metric space that is not complete can be isometrically embedded in one that is complete.

I don't understand your comment. The metric space is complete - any Cauchy sequence has a limit in the space. The question I am raising is the reverse - getting a Cauchy sequence using Schauder base?

 

[nuuskur]

For a topic that is specified to be not topological, there is sure a lot of topology involved To be quite frank, we don't seem to understand what your objective is, mathman.

The definition of Cauchy sequence (assuming at least some kind of metric present) states roughly that $d(a_n,a_m)$ eventually becomes arbitrarily small. Therefore, whether a sequence is Cauchy depends on given topology (is a given $\varepsilon$-ball contained in a certain open set or not). A basis is just a way to express the sequence elements in terms of basis elements.

Furthermore, we have no control how the Schauder basis operates. The definition states that every element can be expressed as a uniquely determined infinite combination of basis elements, that's it. (For instance, you could have almost all scalars equal to zero). We're not allowed to make arbitrary combinations and declare them elements in the space.

On the other hand the problem seems kind of trivial. There is a representation $\sum _{n\in\mathbb N} s_ne_n = (0,1,0,1,\ldots)$. By definition the series converges so the partial sums give us a Cauchy sequence.

 

[mathman]

For a topic that is specified to be not topological, there is sure a lot of topology involved To be quite frank, we don't seem to understand what your objective is, mathman.

The definition of Cauchy sequence (assuming at least some kind of metric present) states roughly that $d(a_n,a_m)$ eventually becomes arbitrarily small. Therefore, whether a sequence is Cauchy depends on given topology (is a given $\varepsilon$-ball contained in a certain open set or not). A basis is just a way to express the sequence elements in terms of basis elements.

Furthermore, we have no control how the Schauder basis operates. The definition states that every element can be expressed as a uniquely determined infinite combination of basis elements, that's it. (For instance, you could have almost all scalars equal to zero). We're not allowed to make arbitrary combinations and declare them elements in the space.

On the other hand the problem seems kind of trivial. There is a representation $\sum _{n\in\mathbb N} s_ne_n = (0,1,0,1,\ldots)$. By definition the series converges so the partial sums give us a Cauchy sequence.

I am not sure what the terms in the sequence are. If they are (0,1,0,1,...) for n terms, the sup norm for the difference is always 1, so they don't converge. If you drop a requirement that only finite combinations of basis elements are allowed, then I have no problem with the Schauder basis.

 

[nuuskur]

Certain infinite combinations are allowed (the ones provided by definition). That's the best you can do and the whole point of a Schauder basis. If you want to make arbitrary combinations, then you are restricted to finite combinations, because of potential well-definedness problems.

The last example I gave is sloppy for I didn't specify where - if we are talking about $\ell_\infty$ (sup-norm), it doesn't have a Schauder basis, because it isn't separable. Indeed, if $E$ is a Schauder basis (wo.l.o.g normalised), then $X := \{\sum _{j=1}^m x_je_j \mid e_j\in E, x_j\in \mathbb Q, m\in\mathbb N\}$ is a countable dense subset (a good exercise for FA1 students).

To be completely fair, the example is rotten to the core. The sequence $(0,1,0,1,\ldots )$ is not in any $\ell _p$ for $1\leq p<\infty$. Replace with appropriate sequence and my argument about a Cauchy sequence applies.

 

[mathman]

When all s said and done, does $l_\infty$ have a basis by any definition?

 

[nuuskur]

Yes, it is a nonzero space, thus it has a basis. It is known that the axiom of choice holds if and only if every spanning subset contains a minimal spanning subset.

It is also known that the axiom of choice is independent of ZF. Thus, if you don't accept AC, then you will have to relinquish the existence of bases for arbitrary nonzero vectorspaces.

 

[mathman]

The term basis is still confusing to me. Is the set of vectors of the form (0,0,0,...0,0,1,0,..) where the 1 is at any position a basis?