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When I have two vector spaces $W, V$ over $k$ a field. And I know that the algebraic dual spaces of $V$ and $W$ are isomorphic. Can I conclude, (in the infinite dimensional case) that $V$ and $W$ are isomorphic?

I am saying algebraic dual, because i don't want to be confused with the dual of continuous linear functionals. But I am just using the definition of dual space, that everyone uses in linear algebra.

Say I know, that I have a linear map $\varphi: V \to W$ such that its dual map is an isomorphism, can I then conclude it was an isomorphism all along? I know this is true if $\varphi$ is assumed to be injective, because then the injectivity of the dual map, implies surjectivity of the original map. Can I say something, if my map $\varphi$ was assumed to be surjective and the dual map is an isomorphism?

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  • $\begingroup$ The answer is definitely yes if $W,V$ are reflexive. Why do you say algebraic dual? Are you defining the dual space slightly differently? $\endgroup$
    – FShrike
    Dec 22, 2022 at 20:55
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    $\begingroup$ @FShrike If $V$ is a $k$-vector space, its algebraic dual space $V^\vee$ is the $k$-vector space of all linear maps from $V$ to $k$. It is well-known that if $V$ is infinite dimensional, then $V^{\vee \vee}$ is not isomorphic to $V$. $\endgroup$
    – azif00
    Dec 22, 2022 at 21:00
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    $\begingroup$ I suspect it depends on your set theory. If $V$ is infinite dimensional, then $\dim(V^*)\gt\dim(V)$, and moreover, if $\dim(V)=\kappa$, then $|k|^{\kappa}=|V^*|=\max(\dim(V^*,|k|))$ (see, e.g., here). But then, if we can find two cardinal $\kappa\neq\kappa'$ such that $\kappa,\kappa'\gt|k|$ and $|k|^{\kappa}=|k|^{\kappa'}$, then you would have two vector spaces that are not isomorphic (one of dimension $\kappa$ and one of dimension $\kappa'$) but with isomorphic duals (both of dimension $|k|^{\kappa}=|k|^{\kappa'}$. $\endgroup$ Dec 22, 2022 at 21:25
  • $\begingroup$ Seconding @ArturoMagidin's comment: most of the time [sic] the cardinality of a(n) (algebraic) basis for a vector space is a pretty docile cardinal, and then probably the answer is affirmative, just for cardinality reasons. But, yeah, if we try to find funky cardinals (are there any of this particular ilk? I have no idea... dependent on Continuum Hypothesis? ...) possibly the "general" answer is negative. And, also, as in an answer, do we really mean natural isomorphism...? $\endgroup$ Dec 22, 2022 at 21:43

2 Answers 2

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Yes. If $\phi^\lor$ is an isomorphism, then $\phi$ is an isomorphism.

The most straightforward way of seeing this is through category theory. We must show that the dual functor is faithful. That is, if we have vector spaces $W, V$ and linear maps $f, g : W \to V$ such that $f^\lor = g^\lor$, then $f = g$.

Indeed, suppose we have such $W, V, f, g$, and consider some $x \in W$. Suppose $f(x) \neq g(x)$. Then take some basis $B$ of $V$ such that $f(x) - g(x) \in B$, and consider the unique linear map $h : V \to k$ such that for all $b \in B$,

$$h(b) = \begin{cases} 1 & b = f(x) - g(x) \\ 0 & otherwise \end{cases}$$

Then $h \circ f = f^\lor(h) = g^\lor(h) = h \circ g$, so in particular, $h(f(x)) = h(g(x))$. So $h(f(x) - g(x)) = 0$; contradiction.

Now that we’ve established the dual functor is faithful, we note that the category of vector spaces is balanced - a linear map which is both a monomorphism and an epimorphism is an isomorphism. It follows from faithfulness that since $\phi^\lor$ is an isomorphism, it must be an epimorphism, and hence $\phi$ is a monomorphism. Similarly, it follows that $\phi^\lor$ is a monomorphism, and hence $\phi$ is an epimorphism. Therefore, $\phi$ is an isomorphism.

To replicate the previous paragraph without category theory, we first show $\phi : V \to W$ is injective. Consider some $x \in V$ such that $\phi(x) = 0$. Then consider the unique linear map $f : k \to V$ such that $f(1) = x$. We see that $\phi \circ f = 0$, so $f^\lor \circ \phi^\lor = (\phi \circ f)^\lor = 0^\lor = 0$. Take the equation $f^\lor \circ \phi^\lor = 0$ and compose $(\phi^\lor)^{-1}$ on the right to get $f^\lor = 0 = 0^\lor$. It follows from faithfulness that $f = 0$, so $x = f(1) = 0$. Thus, $\phi$ is injective. You said in your question that you can take it from there.

Finally, this argument is not sufficient to show that if $V^\lor$ is isomorphic to $W^\lor$, then $V \cong W$. In general, dual spaces are far larger than their underlying spaces, and there are many maps between them that don’t arise as a dual map. I don’t know whether this implication holds, but I have a strong sense it doesn’t (or may even be independent from ZFC). An analogous question arises in set theory: if there is a bijection between $P(S)$ and $P(R)$, is there a bijection between $S$ and $R$? It turns out the question cannot be answered in ZFC.

But I think it’s likely there’s a well-known answer somewhere. You just have to look harder.

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  • $\begingroup$ @GrishaTaroyan That was the question. We were asked to consider a $\phi$ such that $\phi^\lor$ is an isomorphism, and show $\phi$ is an isomorphism. $\endgroup$ Dec 22, 2022 at 21:26
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    $\begingroup$ That's the question in the third paragraph; but it need not address the question in the first paragraph, since not every linear function between the duals need be the dual f a linear transformation between the original spaces, does it? $\endgroup$ Dec 22, 2022 at 21:27
  • $\begingroup$ Very clear answer, I like it very much, and it solves the case that I need for some exercise, namely, when the map is the dual of a map of the original spaces. Thank you very much! I think I will accept it, though technically as Grisha said, for the first question i didn't assume the dual map to come from a linear map $\endgroup$ Dec 22, 2022 at 21:28
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    $\begingroup$ Yes, this answers "the question as it was secretly intended to be asked", I guess, where it's not really about existence of isomorphisms, but about isomorphisms "naturally induced" in various senses. Still, the question @ArturoMagidin answered is perhaps more entertaining than the intended (saner?) question. :) :) $\endgroup$ Dec 22, 2022 at 22:08
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For the question in the first paragraph, the answer is "no" in the general case, at least in ZFC. However, as Mark Saving notes in the comments, depending on your field $k$ the answer may be independent of ZFC, and indeed depend on your Set Theory.

If $V$ is a vector space of infinite dimension $\kappa$, then we know that $\dim(V^*)\gt\dim(V)$. Moreover, we know that $\dim(V^*)=|k|^{\dim(V)}$, by a theorem of Kaplansky and Erdős. See for example this answer and this answer, the latter with citation for this theorem.

Now take a field $k$ of cardinality $2^{\mu}$, with $\mu\gt 2^{\aleph_0}$, and take a $k$-vector space $V$ of dimension $\aleph_0$, and a $k$-vector space $W$ of dimension $2^{\aleph_0}$.

Then $$\dim(V^*) =|k|^{\aleph_0} = (2^{\mu})^{\aleph_0} = 2^{\mu\aleph_0} = 2^{\mu},$$ and $$\dim(W^*) = |k|^{2^{\aleph_0}} = (2^{\mu})^{2^{\aleph_0}}= 2^{\mu 2^{\aleph_0}} = 2^{\mu}.$$ Thus, $V^*$ and $W^*$ are isomorphic (they have the same dimension), but $V$ and $W$ are not isomorphic, since one is countably-dimensional and the other is uncountably-dimensional.

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    $\begingroup$ Interestingly, for $k$ any finite field, it seems the answer is independent of ZFC. Assuming GHC, the answer is “yes”; assuming Martin’s Axiom and the failure of CH, the answer is “no”. $\endgroup$ Dec 22, 2022 at 22:13
  • $\begingroup$ @MarkSaving I suspected there would be cardinalities for $k$ for which the answer would be independent of ZFC, but wasn't certain it would always be independent of ZFC when I made my first comment... Thanks. I wonder in particular about real vector spaces, and whether the answer might depend on CH. $\endgroup$ Dec 22, 2022 at 22:14

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