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Suppose I want to show that two infinite-dimensional vector spaces are not isomorphic to each other. This is easy if my vector spaces are finite-dimensional as I just find a basis for each and show they are of different size, since finite-dimensional vector spaces are isomorphic if and only if they have the same dimension.

If the vector spaces have different cardinality then of course we would not be able to find a bijection between them, so they are not isomorphic, but what if the cardinality is the same? I know I need to show that every bijection is not linear, but that sounds like a lot of work. What is the best way to go about it generally?

user85798
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1 Answers1

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Two vector spaces (over the same field) are isomorphic iff they have the same dimension - even if that dimension is infinite.

Actually, in the high-dimensional case it's even simpler: if $V, W$ are infinite-dimensional vector spaces over a field $F$ with $\mathrm{dim}(V), \mathrm{dim}(W)\ge \vert F\vert$, then $V\cong W$ iff $\vert V\vert=\vert W\vert$. In particular, if $F=\mathbb{Q}$, two infinite-dimensional vector spaces over $F$ are isomorphic iff they have the same cardinality.

So, for example:

  • As vector spaces over $\mathbb{Q}$, $\mathbb{R}$ and $\mathbb{C}$ are isomorphic (this assumes the axiom of choice).

  • The algebraic numbers $\overline{\mathbb{Q}}$ are not isomorphic to the complex numbers $\mathbb{C}$ as vector spaces over $\mathbb{Q}$, since the former is countable while the latter is uncountable.

EDIT: All of this assumes the axiom of choice - without which, the idea of "dimension" doesn't really make sense. See the comments for a bit more about this.

Alex Ortiz
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Noah Schweber
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  • Ah, this wasn't covered in lectures. How do I in general find the dimension of an infinitely dimensional vector space? I heard it is not always possible to construct a basis, e,g for the set of continuous functions from [a,b] to R. – user85798 Oct 26 '15 at 19:19
  • Assuming the axiom of choice, it is always possible to construct a basis (or rather, a basis always exists; "construct" is a funny word where the axiom of choice is involved :D). Without the axiom of choice, infinite-dimensional vector spaces are indeed extremely strange. Surprisingly, the statement "Every vector space has a basis" is equivalent to the axiom of choice; this was proved by Andreas Blass. – Noah Schweber Oct 26 '15 at 19:20
  • The most straightforward way to show that bases exist is via Zorn's lemma: given a vector space $V$ over a field $F$, let $\mathbb{P}$ be the partial order of all linearly independent subsets of $V$, ordered by inclusion. Then by Zorn, $\mathbb{P}$ has a maximal element; and it's easy to see that a maximal linearly independent set must be a basis. Of course, you now really ought to prove Zorn's lemma from the axiom of choice, which is nontrivial, but usually in early algebra classes you are allowed to just assume Zorn. – Noah Schweber Oct 26 '15 at 19:22
  • Meanwhile, the proof that two vector spaces with bases $B_0$ and $B_1$ of the same cardinality are isomorphic is completely straightforward, and doesn't use choice at all. By the way, it is consistent with ZF (=set theory without choice) that $\mathbb{R}$ and $\mathbb{C}$ are not isomorphic as $\mathbb{Q}$-vector spaces, but this is quite hard to show. – Noah Schweber Oct 26 '15 at 19:24
  • By the way, note that without a basis, the word "dimension" makes little sense - at best, we can use the phrase "infinite-dimensional" as slang for "does not have a finite basis." But you should be careful about this - using the word "dimension" in the context of vector spaces may lead people to assume you're assuming choice. – Noah Schweber Oct 26 '15 at 19:25
  • Ok, thanks. Is the proof in the infinite dimensional case much harder and do you where I could find it? – user85798 Oct 26 '15 at 19:30
  • @LTS The proof of what in the infinite-dimensional case - that two vector spaces with bases of the same size are isomorphic? That proof is exactly the same: given a bijection $b$ between the bases, extend $b$ to a bijection $f$ between the vector spaces in the natural way, and show that $f$ is linear. – Noah Schweber Oct 26 '15 at 19:31
  • yes, and the converse? Would the first proposition and corollary on page 11 https://www.dpmms.cam.ac.uk/~sjw47/LinearAlgebraM15.pdf still be true for infinite-dimensional vector spaces? I cannot see why not but I assumed not as the lecturer takes great pain to point out that the vector spaces are finite dimensional. – user85798 Oct 26 '15 at 19:41
  • @LTS Yes, this is all still true, and it's a good exercise to work through the proofs in the infinite-dimensional case yourself (basically nothing changes). The reason the text is focusing on finite-dimensional vector spaces is to prevent any axiom-of-choice issues from arising - and this is fine, because most natural vector spaces are finite dimensional. – Noah Schweber Oct 26 '15 at 19:44
  • @NoahSchweber Do you have a reference for your claim about when two infinite-dimensional vector spaces are isomorphic? Thanks. – Alex Ortiz Oct 15 '18 at 22:24
  • @AOrtiz Not off the top of my head, but the proof is literally identical to the finite case: show that any bijection from a basis $A$ for $U$ to a basis $B$ for $V$ extends to an isomorphism from $U$ to $V$. – Noah Schweber Oct 16 '18 at 13:13
  • Of course, two comments are necessary: (1) by "basis" I mean "Hamel basis," and there is a separate notion of "Schauder basis" for infinite-dimensional vector spaces which is more useful but also more complicated in some respects; (2) we need the axiom of choice to show that all vector spaces have bases, and without it there is no similar characterization of isomorphism. – Noah Schweber Oct 16 '18 at 13:19
  • It's also worth noting that many basic facts about bases do fail to generalize to infinite-dimensional spaces. – Noah Schweber Oct 16 '18 at 13:24
  • "Actually, in the high-dimensional case it's even simpler". What followed was actually not so simple. – user5826 Jun 13 '20 at 23:28
  • @AlJebr Not sure what you mean. My point was that once the dimensions involved are at least as large as the field of scalars itself, then isomorphism just reduces to equicardinality. That's simpler than the general ("small-dimension") case, where we can have two vector spaces which have the same cardinality but are not isomorphic. – Noah Schweber Jun 13 '20 at 23:32
  • Why is $\mathrm{dim}(V), \mathrm{dim}(W) \geq |F|$ required? – Poscat Oct 21 '23 at 08:52
  • Ugh it’s because cardinal arithmetic requirement right? – Poscat Oct 21 '23 at 09:03