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After learning about finite dimensional vector spaces, the time came for learning about infinite dimensional ones. However, these seem much less intuitive to me, and proof of that is the question I wanted to ask here.

Let's consider a infinite dimensional vector space $(V, +, \cdot)$, with no topology or any other structure associated to it - not even a norm or inner product.

Let $V=L(S)$ ($L$ is the span operator) with $S=\{e_1, e_2,...\}$ where:

$$e_1=(1,0,0,0,...)\\ e_2=(0,1,0,0,...)\\ e_3=(0,0,1,0,...)$$

And so on. Also, let

$$v_a=(1,1,1,1,1,1,...)\\ v_b=(1,2,3,4,5,6,...)\\ v_c=(3,1,4,1,5,9,...)$$

Where in $v_c$ each coordinate is the n-th digit of $\pi$.

The questions

  1. Is $v_a$ in $V$? Intuitively, I would think it is not, since in order to represent $v_a$ as a linear combination of elements in our basis $S$, one would need to perform a sum of infinitely many summands

$$\sum_{i=1}^\infty e_i$$

which, since it's not a finite sum, it is not defined in $(V,+,\cdot)$.

  1. In case $v_a$ was not in $V$, let $W = L(S\cup v_a)$. Would $v_b$ or $v_c$ also be in $W$? If so, what would its coordinates be? Again intuitively, if $v_a$ were not part of $V$, it would also not be possible to create a finite linear combination of vectors from $S \cup v_a$ that would yield $v_b$. Is this correct? Then, how can we construct a basis that can generate $v_b$ and $v_c$?

  2. Does $V$ only contain vectors with a finite number of non-zero coordinates? If so, how can we create a basis for a vector space which contains all these other vectors with "infinitely long" coordinates?

Martin Sleziak
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user3141592
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  • See the difference between Hamel bases and Schauder bases. In the one case you only allow finite linear combinations of the basis elements. In the other you may allow infinite linear combinations of the basis elements. – JMoravitz Feb 20 '24 at 14:24
  • @JMoravitz but Schauder basis apply only to the case of topological vector spaces, isn't it? – user3141592 Feb 20 '24 at 14:27

2 Answers2

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  1. No, $v_a$ is not in $V$ because you can't write it as a finite linear combination of $e_i$.

  2. Even if you add $v_a$ to the set of generators, still $v_b$ and $v_c$ don't belong to $W$, because you still need an infinite number of elements of the basis.

  3. Yes, $V$ only contains vectors with a finite number of non-zero elements. In the same way, $W$ contains only vectors which are constant except that for a finite number of elements. You can't always "see" the basis of a vector space. The theorem of existence of a basis guarantees its existence, but it doesn't tell you how it is. Also, you use Zorn's Lemma in this proof (equivalent to Axiom of Choice).

The idea of the theorem of existence of the basis is that of gradually adding vectors to a starting set, ensuring that the new vectors are not in the span of the existing set, ensuring the set remains linearly independent. This process continues until the set spans the entire vector space, at which point it becomes a basis. Zorn's Lemma is used to guarantee the existence of a maximal linearly independent set. So, this theorem is not constructive, you don't know at the end how the basis is done.

Usually, on infinite-dimensional vector spaces you want to add more structure (a scalar product for instance) and use more flexible and usable notions of basis (Hilbert basis for instance, which means writing elements as infinite lineare combinations of basis elements Orthonormal basis).

SilvioM
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  • Thank you for your answer! Now everything is a bit more clear. Nevertheless, you mention the fact that we know that there exists a basis due to the theorem of existence. Do we know any basis for this case of countable number of dimensions i.e. $\mathbb{R}^{\mathbb{N}}$? – user3141592 Feb 20 '24 at 15:37
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    @user3141592 I think we don't explicitly know a basis of such space. It's a bit tricky because of the use of the Axiom of Choice. The same thing happens for instance to the existence of a well-order over the real numbers, which existence is equivalent to Axiom of Choice and Zorn's Lemma. I suggest you to read this https://math.stackexchange.com/questions/6501/is-there-a-known-well-ordering-of-the-reals if you want to have a better grasp on this topic – SilvioM Feb 20 '24 at 15:47
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    @user3141592 Clarification: the existence of a well order on $\mathbb{R}$ is not equivalent to the Axiom of Choice; it is the existence of a well ordering of any set that is equivalent to the Axoim of Choice. – Arturo Magidin Feb 20 '24 at 17:23
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Are your $v_i$ vectors in your vector space? It's up to you: you can impose the rule that only finitely many coordinates are non-zero or allow any number of non-zero coordinates.

Finitely many non-zero coordinates

Your $e_i$ vectors form a basis but your $v_i$ vectors are not in the space.

This space has a countable dimension.

Any number of non-zero coordinates

Your $v_i$ vectors are in the space but your $e_i$ vectors do not form a basis.

This space has an uncountable dimension so see below. There might not be any explicit basis.

Uncountably many dimensions

This is possible and even more challenging. One example is the set of functions from $\mathbb{R}$ to $\mathbb{R}$ which can be easily made into a vector space. Another is $\mathbb{R}$ considered as a vector space over $\mathbb{Q}$.

Whether or not any vector space has a basis is equivalent to the Axiom of Choice. So, there will be cases in which an explicit basis cannot be given.

Axiom of Choice (Wikipeida) Scan for "vector space".

Some more on the subject here: Linear space with (Hamel) basis and the axiom of choice (Math Overflow)

badjohn
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  • Thanks for the answer, this other approach also helps me understanding the issue better. However, do we know any basis for this case of countable number of dimensions i.e. $\mathbb{R}^{\mathbb{N}}$? It seems to me like something that should have been studied already – user3141592 Feb 20 '24 at 15:38
  • I need to do some research. I expect that it is possible but maybe not as neat as your example for the finitely many non-zero coordinate case. – badjohn Feb 20 '24 at 15:49
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    This follow up question probably deserves its own separate question. E.g. is there an explicit for $\mathbb{R}^\mathbb{N}$ as a vector space? – badjohn Feb 20 '24 at 15:53
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    Maybe not so easy. The case in which the coordinates are not limited has an uncountable dimension and hence there may be no explicit basis. See https://math.stackexchange.com/questions/575765/the-proof-of-the-infinity-base-of-mathbbr-infty/1577650#1577650 – badjohn Feb 20 '24 at 15:57
  • Yes, but maybe also $\mathbb{Q}^{\mathbb{N}}$ or $\mathbb{Z}^{\mathbb{N}}$ – user3141592 Feb 20 '24 at 20:46
  • Sorry, I am not sure what you mean. $\mathbb{Q}^\mathbb{N}$ is a vector space but $\mathbb{Z}^\mathbb{N}$ as $\mathbb{Z}$ is not a field. – badjohn Feb 20 '24 at 21:58
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    Sorry, I wanted to mean the infinite product $\prod^{\infty} \mathbb{Q}$; you're right about $\mathbb{Z}$ - my slip – user3141592 Feb 20 '24 at 22:00