A very important theorem in linear algebra that is rarely taught is:

A vector space has the same dimension as its dual if and only if it is finite dimensional.

I have seen a total of one proof of this claim, in Jacobson's "Lectures in Abstract Algebra II: Linear Algebra". The proof is fairly difficult and requires some really messy arguments about cardinality using, if I remember correctly, infinite sequences to represent $\mathbb{N}\times\mathbb{N}$ matrices. Has anyone come up with a better argument in the 57 years since Jacobson's book was published, or is the noted proof still the only way to prove this fact?

Edit: For reference, the proof is on pages 244-248 of Jacobson's

Lectures in Abstract Algebra: II. Linear Algebra.

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    $\begingroup$ "very important theorem" - I disagree. $\endgroup$ Jan 29, 2010 at 12:48
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    $\begingroup$ @darij: You're entitled to disagree, but saying so, without any justification whatsoever, is not a positive contribution. $\endgroup$ Jan 29, 2010 at 12:58
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    $\begingroup$ This theorem proves that we cannot ever under any circumstances extend the results of finite dimensional linear algebra without considering topology. I think that's pretty important. =p $\endgroup$ Jan 29, 2010 at 13:23
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    $\begingroup$ Your comments seem disrespectful to Harry and especially to Mariano and Andrea, who have taken time to write out very interesting and instructive answers. Please refrain from making purely negative remarks. $\endgroup$ Jan 29, 2010 at 21:18
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    $\begingroup$ +1 for the question and for the answers. Here is an example for an application that I will give in a class, which shows that the theorem in question is not purely a no-go-result: the Poincare duality in de Rham theory states that $H^k (M) \cong (H^{n-k}_{cpt} )^{\ast}$ for an oriented manifold. If $M$ is compact, then $H^k (M) \cong (H^k (M))^{\ast \ast}$, so $H^k (M)$ is finite-dimensional.. $\endgroup$ Dec 9, 2012 at 18:52

4 Answers 4


Here is a simple proof I thought, tell me if anything is wrong.

First claim. Let $k$ be a field, $V$ a vector space of dimension at least the cardinality of $k$ and infinite. Then $\operatorname{dim}V^{*} >\operatorname{dim}V$.

Indeed let $E$ be a basis for $V$. Elements of V* correspond bijectively to functions from $E$ to $k$, while elements of $V$ correspond to such functions with finite support. So the cardinality of $V^{*}$ is $k^E$, while that of $V$ is, if I'm not wrong, equal to that of $E$ (in this first step I am assuming $\operatorname{card} k \le \operatorname{card} E$).

Indeed $V$ is a union parametrized by $\mathbb{N}$ of sets of cardinality equal to $E$. In particular $\operatorname{card} V < \operatorname{card} V^{*}$, so the same inequality holds for the dimensions.

Second claim. Let $h \subset k$ two fields. If the thesis holds for vector spaces on $h$, then it holds for vector spaces on $k$.

Indeed let $V$ be a vector space over $k$, $E$ a basis. Functions with finite support from $E$ to $h$ form a vector space $W$ over $h$ such that $V$ is isomorphic to the extension of $W$, i.e. to $W\otimes_h k$. Every functional from $W$ to $h$ extends to a functional from $V$ to $k$, hence

$$\operatorname{dim}_k V = \operatorname{dim}_h W < \operatorname{dim}_h W^* \leq \operatorname{dim}_k V^*.$$

Putting the two claims together and using the fact that every field contains a field at most denumerable yields the thesis.

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    $\begingroup$ You can extend functionals, but you must also show that if they were linearly independent over the base then they remain linearly independent after the extension. Once you fill in that detail, I think you have a full proof. $\endgroup$ Jan 29, 2010 at 18:40
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    $\begingroup$ Take a linear combination = 0 with coefficients in k. That involves finitely many functionals which are independent over h. Take elements of W which are dual to this functionals. Now evaluate the linear combination at each of these vectors to find that all coefficients are zero. $\endgroup$ Jan 29, 2010 at 19:20
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    $\begingroup$ I nominate this answer for a hypothetical future "Best of MO" collection. I find it almost magical, and with a moral -- don't just stick with the field you're given! -- that I find very appealing. $\endgroup$ Jan 29, 2010 at 21:45
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    $\begingroup$ @Andrew: Sorry to resurrect an old answer you probably haven't thought about since January; I was looking for an easy proof of precisely the statement in the title. This looks very nice, but I'm having trouble seeing how to justify the existence in W of elements that are dual to a given (finite) set of functionals. I assume you mean that given $f_1,\ldots,f_n$ distinct elements of $h^E$ with finite support, you will find elements $v_j$ in $W$ (or $V$) with $f_i(v_j)=\delta_{ij}$. But I don't see how to justify that such elements exist. $\endgroup$ Sep 24, 2010 at 20:41
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    $\begingroup$ @Andrea, @Arturo: functionals in $W^*$ need not have finite support. I would argue instead as follows. Given $h$-lin. indep. $f_1,...,f_n$ in $W^*$, consider $h$-linear map $W \rightarrow h^n$ where $w \mapsto (f_1(w),...,f_r(w))$. To show this is onto we show the only vector in $h^n$ orthogonal to image is $(0,...,0)$. If $(c_1,...,c_n)$ in $h^n$ is orthogonal to image then for all $w$ in $W$, $c_1f_1(w) + ... + c_nf_(w) = 0$. Thus $c_1f_1 + ... + c_nf_n$ is 0 in $W^*$, so each $c_i$ is 0 by $h$-lin. indep. of the $f_i$'s. Now use $w_j$ giving image $(0,...,1,...,0)$ with 1 in j-th slot. $\endgroup$
    – KConrad
    May 8, 2011 at 23:42

I know a fairly elementary proof in the case when the field is countable.

First, you prove that $Hom(\bigoplus_{i\in I}A_{i},B)\cong \prod_{i\in I}Hom(A_{i},B)$, where all terms are $R$-modules. (This should be fairly intuitive. A homomorphism from a direct sum is determined by its actions on each piece individually.)

Second, specialize $A_{i}$ and $B$ to equal your field. So the direct product is over a bunch of pieces (all isomorphic to your field).

Third, use the standard cardinality argument to show that a direct product of $I$ non-empty pieces has cardinality strictly greater than $I$.

This argument doesn't quite work when your field has large cardinality, but I still think it is nice. (Basically, this is thinking about the first part of Andrea's proof a little differently.)


It is clearly enough to show that an infinite dimensional vector space $V$ has smaller dimension that its dual $V^*$.

Let $B$ be a basis of $V$, let $\mathcal P(B)$ be the set of its subsets, and for each $A\in\mathcal P(B)$ let $\chi_A\in V^*$ be the unique functional on $V$ such that the restriction $\chi_A|_B$ is the characteristic function of $A$. This gives us a map $\chi:A\in\mathcal P(B)\mapsto\chi_A\in V^*$.

Now a complete infinite boolean algebra $\mathcal B$ contains an independent subset $X$ such that $|X|=|\mathcal B|$---here, that $X$ be independent means that whenever $n,m\geq0$ and $x_1,\dots,x_n,y_1,\dots,y_m\in X$ we have $x_1\cdots x_n\overline y_1\cdots\overline y_n\neq0$. (This is true in this generality according to [Balcar, B.; Franěk, F. Independent families in complete Boolean algebras. Trans. Amer. Math. Soc. 274 (1982), no. 2, 607--618. MR0675069], but when $\mathcal B=\mathcal P(Z)$ is the algebra of subsets of an infinite set $Z$, this is a classical theorem of [Fichtenholz, G. M; Kantorovich L. V. Sur les opérations linéaires dans l'espace des fonctions bornées. Studia Math. 5 (1934) 69--98.] and [Hausdorff, F. Über zwei Sätze von G. Fichtenholz und L. Kantorovich. Studia Math. 6 (1936) 18--19])

If $X$ is such an independent subset of $\mathcal P(B)$ (which is a complete infinite boolean algebra), then $\chi(X)$ is a linearly independent subset of $V^*$, as one can easily check. It follows that the dimension of $V^*$ is at least $|X|=|\mathcal P(B)|$, which is strictly larger than $|B|$.

Later: The proof of the existence of an independent subset is not hard; it is given, for example, in this notes by J. D. Monk as Theorem 8.9. In any case, I think this proof is pretty because it captures precisely the intuition (or, rather, my intuition) of why this is true. I have not seen the paper by Fichtenhold and Kantorovich (I'd love to get a copy!) but judging from its title one sees that they were doing similar things...

  • $\begingroup$ Is it easier to prove the relevant results about boolean algebras, or does that also rely on a proof like Jacobson's? $\endgroup$ Jan 29, 2010 at 4:26
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    $\begingroup$ The paper by G. Fichtenholz and L. Kantorovitch can be found here : matwbn.icm.edu.pl/ksiazki/sm/sm5/sm519.pdf . $\endgroup$
    – Ady
    Jan 29, 2010 at 11:58

Here is a proof for $\dim(V^*)>\dim(V)$ for every infinite dimensional vector space $V$ over a field $k$. More precisely, we prove that $$\mathrm{dim}(V^*)=|V^*|=|k|^{\dim(V)}\geq 2^{\dim(V)}>\dim(V).$$

We first show that $\dim(V^*)\geq |k|$. Fix a basis $B$ for $V$ and choose a countable infinite subset $v_0,v_1,v_2,\ldots$ of vectors of $B$. Given $\alpha\in k$, we define $f_\alpha\in V^*$ to be the unique functional defined on the basis elements by setting $f_\alpha(v_n):=\alpha^n$ for all $n=0,1,2,\ldots$ and $f_\alpha(v)=0$ for all other elements $v\in B$. Notice that $(f_\alpha)_{\alpha\in k}$ is a family of pairwise distinct functionals of $V$.

We claim that the family $(f_\alpha)_{\alpha\in k}$ is linearly independent. Indeed, if $\alpha_1,\ldots,\alpha_n\in k$ are distinct elements and $$x_1f_{\alpha_1}+\ldots+x_nf_{\alpha_n}=0$$ for certain scalars $x_i\in k$, then evaluating at $v_0, v_1,\ldots, v_{n-1}$ we get a system of $n$ linear equations of the form $$x_1\alpha_1^i+\ldots+x_n\alpha_n^i=0,\quad i=0,1,\ldots,n-1$$ in the variables $x_1,\ldots,x_n$. The $n\times n$ matrix of this system is $$\left(\begin{array}{ccc} 1& 1 &\ldots & 1\\ \alpha_1& \alpha_2 &\ldots & \alpha_n\\ \alpha_1^2& \alpha_2^2 &\ldots & \alpha_n^2\\ \vdots & \ldots & \ldots &\vdots \\ \alpha_1^{n-1} &\alpha_2^{n-1} &\ldots &\alpha_n^{n-1} \end{array}\right)$$ This is a (transpose of a) Vandermonde matrix, which is therefore invertible, so that $x_1=x_2=\ldots=x_n=0$, as desired.

It follows that $$|V^*|=\max\{|k|,\dim(V^*)\}=\dim(V^*).$$ Since there is an isomorphism of vector spaces $V^*\cong k^B$ (where the right hand side denotes the space of all functions $B\to k$), it also follows $$\dim(V^*)=|k^B|=|k|^{\dim(V)}.$$

  • $\begingroup$ This is a restatement of Pace Nielsen's argument. $\endgroup$ Apr 16, 2022 at 19:09
  • $\begingroup$ @Ryan Budney, I didn't pay too much attention to Pace Nielsen's argument in the first place. They are based partially on the same ideas. My point was too give a possibly elementary proof of the fact that $|V^*|\geq |k|$. Pace Nielsen adds at the end a comment that the argument does not work for fields of large cardinality. And my argument actually works for every field, of arbitrary cardinality. $\endgroup$ Apr 17, 2022 at 23:11
  • $\begingroup$ I meant $\dim(V^*)\geq |k|$ above, proving this inequality was the main point of my argument. And then it follows $\dim(V^*)=|V^*|$. $\endgroup$ Apr 18, 2022 at 14:16
  • $\begingroup$ Nice proof when the field has characteristic $0$. But may need adjustment in characteristic $p$, as $\alpha^p$ could be the same as $\alpha$ and the Vandermonde matrix may not be invertible. $\endgroup$
    – PatrickR
    Jan 21, 2023 at 6:38
  • $\begingroup$ @PatrickR, I didn't understand your comment. The Vandermonde matrix is always invertible, over any field, despite of its characteristic. $\endgroup$ Jan 31, 2023 at 12:57

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