Assuming the axiom of choice, set $\mathbb F$ to be some field (we can assume it has characteristics $0$).

I was told, by more than one person, that if $\kappa$ is an infinite cardinal then the vector space $V=\mathbb F^{(\kappa)}$ (that is an infinitely dimensional space with basis of cardinality $\kappa$) is not isomorphic (as a vector space) to the algebraic dual, $V^*$.

I have asked several professors in my department, and this seems to be completely folklore. I was directed to some book, but could not have find it in there as well.

The Wikipedia entry tells me that this is indeed not a cardinality issue, for example $\mathbb R^{<\omega}$ (that is all the eventually zero sequences of real numbers) has the same cardinality as its dual $\mathbb R^\omega$ but they are not isomorphic.

Of course being of the same cardinality is necessary but far from sufficient for two vector spaces to be isomorphic.

What I am asking, really, is whether or not it is possible when given a basis and an embedding of a basis of $V$ into $V^*$, to say "This guy is not in the span of the embedding"?

Edit: I read the answers in the link given by Qiaochu. They did not satisfy me too much.

My main problem is this: suppose $\kappa$ is our basis then $V$ consists of $\{f\colon\kappa\to\mathbb F\Big| |f^{-1}[\mathbb F\setminus\{0\}]|<\infty\}$ (that is finite support), while $V^*=\{f\colon\kappa\to\mathbb F\}$ (that is all the functions).

In particular, the basis for $V$ is given by $f_\alpha(x) = \delta_{\alpha x}$ (i.e. $1$ on $\alpha$, and $0$ elsewhere), while $V^*$ needs a much larger basis. Why can't there by other linear functionals on $V$?

Edit II: After the discussions in the comments and the answers, I have a better understanding of my question to begin with. I have no qualms that under the axiom of choice given an infinite set $\kappa$ there are a lot more functions from $\kappa$ into $\mathbb F$, than functions with finite support from $\kappa$ into $\mathbb F$. It is also clear to me that the basis of a vector space is actually the set of $\delta$ functions, whereas the basis for the dual is a subset of characteristic functions.

My problem is, if so, why is the dual space composed from all functions from $A$ into $F$?

(And if possible, not to just show by cardinality games that the basis is much larger but actually show the algorithm for the diagonalization.)

Martin Sleziak
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Asaf Karagila
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  • I am not really sure about the tags, if someone can approve/correct them for me I will be grateful! – Asaf Karagila Aug 19 '11 at 19:04
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    http://mathoverflow.net/questions/13322/slick-proof-a-vector-space-has-the-same-dimension-as-its-dual-if-and-only-if-it – Qiaochu Yuan Aug 19 '11 at 19:09
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    How about [this one](http://groups.google.com/group/sci.math/msg/82792c829f7fe139), courtesy of Bill Dubuque and sci.math? – Arturo Magidin Aug 19 '11 at 21:13
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    @Arturo: While no disrespect for Bill, his style is a little too concise for my taste (perhaps because this is an unfamiliar territory for me I need a somewhat more elaborated explanations). Also working without choice for the past few months really screwed up my cardinal arithmetic skills :) – Asaf Karagila Aug 19 '11 at 21:38
  • Andrea Ferretti's answer in Qiaochu's link is very nice. It reduces the question to a question where cardinality solves it. – Cheerful Parsnip Aug 19 '11 at 22:13
  • @Jim: I wrote in my edit why this argument, while true, bothers me. – Asaf Karagila Aug 19 '11 at 22:16
  • @Asaf: I don't know that your objection is well-formed. In your edit, you say $V$ consists of some set of functions, which seems to mean you are implictly fixing a potential isomorphism between $V$ and $V^*$. – Cheerful Parsnip Aug 19 '11 at 22:39
  • @Jim: Okay, so what about explicitly arguing for the other non-potential isomorphisms? – Asaf Karagila Aug 20 '11 at 06:56
  • Dear Asaf, sorry to bother you with such a detail. Is it on purpose that you wrote "Why vector spaces not isomorphic to their duals?", and not something like "Why *are* vector spaces not isomorphic to their duals?" – Pierre-Yves Gaillard Aug 20 '11 at 07:12
  • @Pierre-Yves: I have corrected the mistake, thank you :) – Asaf Karagila Aug 20 '11 at 07:14
  • Do we *really* know that «the basis for the dual is a subset of characteristic functions»? No function which takes infinitely many values on your basis for $V$ will be in the span of characteristic functions, no? – Mariano Suárez-Álvarez Aug 20 '11 at 22:12
  • @Asaf: I don't understand your edit. What is $A$? What vector space are you implicitly referring to? – Qiaochu Yuan Aug 21 '11 at 01:00
  • @Qiaochu: Is this better? – Asaf Karagila Aug 21 '11 at 05:52
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    @Asaf: I may be beating a dead horse, but a quick reply to your comment: "Why is the dual space composed of **all** functions from the set $A$ (presumably a basis for $V$) to $F$?" This is just an instance of the principle that to specify a linear mapping $f$ is equivalent to specifying its values on elements of a basis. Furthermore, we are free to do this any which way we want. This is what being a free module means! Also, the elements of $V$ are finite linear combinations of basis elements, so we have no problems `extending $f$ linearly'. – Jyrki Lahtonen Aug 21 '11 at 06:36
  • @Jyrki: I'm not sure I understand, actually. – Asaf Karagila Aug 21 '11 at 11:11
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    @Asaf: I still don't understand your edit. Is your question answered by Jyrki's comment? If not, which part is still unclear? – Qiaochu Yuan Aug 21 '11 at 12:43
  • @Qiaochu: It seems that Jyrki's comment *might* answer. I cannot say that I fully understand what he was trying to say. If he elaborates (in another comment, or an answer) it may help. I can say that it does seem to sort of hit the nerve of what I'm after. – Asaf Karagila Aug 21 '11 at 14:03
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    @Asaf: I hate to belabor this point, but I am still not sure what you don't understand. Do you agree that to say that a set $A$ is a basis of a vector space $V$ is equivalent to saying that $V$ is the coproduct of $A$ copies of the underlying field? Do you recall the universal property of coproducts? – Qiaochu Yuan Aug 21 '11 at 18:37
  • @Qiaochu: It becomes apparent to me that the content of this question is not what I am *really* trying to understand. I am currently discussing it with a friend who helps me understand exactly which points in the current answers I am missing. As for universal properties, I am not very knowledgeable with regarding to those. – Asaf Karagila Aug 21 '11 at 18:44
  • @Qiaochu: I have left a comment on Jyrki's answer. I am not sure whether or not this question has gone far enough, and I should accept one of the answers (which are good and correct) and just ask my question in a new thread. – Asaf Karagila Aug 21 '11 at 20:05
  • @Asaf: Have your issues been dealt with with my additions, or did I miss the point of your queries? – Arturo Magidin Aug 22 '11 at 17:24
  • @Arturo: I have printed a version of your answer and I'm currently reading this, I am very very grateful for the effort and will let you know soon! :) – Asaf Karagila Aug 22 '11 at 17:43
  • "Of course being of the same cardinality is necessary but far from sufficient for two vector spaces to be isomorphic." I don't agree with "very far"; for instance, over a finite field, it is sufficient. I also don't agree with "of course"--unless you had something else in mind, the proof uses the same lemma that answered the original question. – Joshua P. Swanson Oct 24 '14 at 03:40
  • @JSwanson: Seeing how there are only countably many finite fields; It's very far. – Asaf Karagila Oct 24 '14 at 03:43
  • @AsafKaragila: Well, it's sufficient whenever $\dim V \geq |F|$ in the notation below, or "half" the time. Just depends on your perspective, I suppose, and from mine it's not "very far". Ah well. – Joshua P. Swanson Oct 24 '14 at 18:03
  • @JSwanson: Yes, but now you're forcing an unintended interpretation to my words. Given two vector spaces of the same cardinality over an infinite field. Are they necessarily isomorphic? No. Not necessarily. The larger the field, the more options we have. Between that and being able to say "yes" there is a gap which is very wide. It's not that "it's consistent that the answer is always yes". It's provable that the answer is sometimes no. – Asaf Karagila Oct 24 '14 at 18:06
  • @AsafKaragila: Well, that's what I meant by it "depends on your perspective". Does one wish to fix a base field and ask the question, or not? You seemed to want to (though you didn't say so), whereas I'm more ambivalent. It feels like we're unproductively splitting semantic hairs at this point though. – Joshua P. Swanson Oct 24 '14 at 22:11

3 Answers3


This is just Bill Dubuque's sci.math proof (see Google Groups or MathForum) mentioned in the comments, expanded.

Edit. I'm also reorganizing this so that it flows a bit better.

Let $F$ be a field, and let $V$ be the vector space of dimension $\kappa$.

Then $V$ is naturally isomorphic to $\mathop{\bigoplus}\limits_{i\in\kappa}F$, the set of all functions $f\colon \kappa\to F$ of finite support. Let $\epsilon_i$ be the element of $V$ that sends $i$ to $1$ and all $j\neq i$ to $0$ (that is, you can think of it as the $\kappa$-tuple with coefficients in $F$ that has $1$ in the $i$th coordinate, and $0$s elsewhere).

Lemma 1. If $\dim(V)=\kappa$, and either $\kappa$ or $|F|$ are infinite, then $|V|=\kappa|F|=\max\{\kappa,|F|\}$.

Proof. If $\kappa$ is finite, then $V=F^{\kappa}$, so $|V|=|F|^{\kappa}=|F|=|F|\kappa$, as $|F|$ is infinite here and the equality holds.

Assume then that $\kappa$ is infinite. Each element of $V$ can be represented uniquely as a linear combination of the $\epsilon_i$. There are $\kappa$ distinct finite subsets of $\kappa$; and for a subset with $n$ elements, we have $|F|^n$ distinct vectors in $V$.

If $\kappa\leq |F|$, then in particular $F$ is infinite, so $|F|^n=|F|$. Hence you have $|F|$ distinct vectors for each of the $\kappa$ distinct subsets (even throwing away the zero vector), so there is a total of $\kappa|F|$ vectors in $V$.

If $|F|\lt\kappa$, then $|F|^n\lt\kappa$ since $\kappa$ is infinite; so there are at most $\kappa$ vectors for each subset, so there are at most $\kappa^2 = \kappa$ vectors in $V$. Since the basis has $\kappa$ elements, $\kappa\leq|V|\leq\kappa$, so $|V|=\kappa=\max\{\kappa,|F|\}$. QED

Now let $V^*$ be the dual of $V$. Since $V^* = \mathcal{L}(V,F)$ (where $\mathcal{L}(V,W)$ is the vector space of all $F$-linear maps from $V$ to $W$), and $V=\mathop{\oplus}\limits_{i\in\kappa}F$, then again from abstract nonsense we know that $$V^*\cong \prod_{i\in\kappa}\mathcal{L}(F,F) \cong \prod_{i\in\kappa}F.$$ Therefore, $|V^*| = |F|^{\kappa}$.

Added. Why is it that if $A$ is the basis of a vector space $V$, then $V^*$ is equivalent to the set of all functions from $A$ to the ground field?

A functional $f\colon V\to F$ is completely determined by its value on a basis (just like any other linear transformation); thus, if two functionals agree on $A$, then they agree everywhere. Hence, there is a natural injection, via restriction, from the set of all linear transformations $V\to F$ (denoted $\mathcal{L}(V,F)$) to the set of all functions $A\to F$, $F^A\cong \prod\limits_{a\in A}F$. Moreover, given any function $g\colon A\to F$, we can extend $g$ linearly to all of $V$: given $\mathbf{x}\in V$, there exists a unique finite subset $\mathbf{a}_1,\ldots,\mathbf{a}_n$ (pairwise distinct) of $A$ and unique scalars $\alpha_1,\ldots,\alpha_n$, none equal to zero, such that $\mathbf{x}=\alpha_1\mathbf{a}_1+\cdots\alpha_n\mathbf{a}_n$ (that's from the definition of basis as a spanning set that is linearly independent; spanning ensures the existence of at least one such expression, linear independence guarantees that there is at most one such expression); we define $g(\mathbf{x})$ to be $$g(\mathbf{x})=\alpha_1g(\mathbf{a}_1)+\cdots \alpha_ng(\mathbf{a}_n).$$ (The image of $\mathbf{0}$ is the empty sum, hence equal to $0$). Now, let us show that this is linear.

First, note that $\mathbf{x}=\beta_1\mathbf{a}_{i_1}+\cdots\beta_m\mathbf{a}_{i_m}$ is any expression of $\mathbf{x}$ as a linear combination of pairwise distinct elements of the basis $A$, then it must be the case that this expression is equal to the one we already had, plus some terms with coefficient equal to $0$. This follows from the linear independence of $A$: take $$\mathbf{0}=\mathbf{x}-\mathbf{x} = (\alpha_1\mathbf{a}_1+\cdots\alpha_n\mathbf{a}_n) - (\beta_1\mathbf{a}_{i_1}+\cdots+\beta_m\mathbf{a}_{i_m}).$$ After any cancellation that can be done, you are left with a linear combination of elements in the linearly independent set $A$ equal to $\mathbf{0}$, so all coefficients must be equal to $0$. That means that we can likewise define $g$ as follows: given any expression of $\mathbf{x}$ as a linear combination of elements of $A$, $\mathbf{x}=\gamma_1\mathbf{a}_1+\cdots+\gamma_m\mathbf{a}_m$, with $\mathbf{a}_i\in A$, not necessarily distinct, $\gamma_i$ scalars not necessarily equal to $0$, we define $$g(\mathbf{x}) = \gamma_1g(\mathbf{a}_1)+\cdots+\gamma_mg(\mathbf{a}_m).$$ This will be well-defined by the linear independence of $A$. And now it is very easy to see that $g$ is linear on $V$: if $\mathbf{x}=\gamma_1\mathbf{a}_1+\cdots+\gamma_m\mathbf{a}_m$ and $\mathbf{y}=\delta_{1}\mathbf{a'}_1+\cdots+\delta_n\mathbf{a'}_n$ are expressions for $\mathbf{x}$ and $\mathbf{y}$ as linear combinations of elements of $A$, then $$\begin{align*} g(\mathbf{x}+\lambda\mathbf{y}) &= g\Bigl(\gamma_1\mathbf{a}_1+\cdots+\gamma_m\mathbf{a}_m+\lambda(\delta_{1}\mathbf{a'}_1+\cdots+\delta_n\mathbf{a'}_n)\Bigr)\\ &= g\Bigl(\gamma_1\mathbf{a}_1+\cdots+\gamma_m\mathbf{a}_m+ \lambda\delta_{1}\mathbf{a'}_1+\cdots+\lambda\delta_n\mathbf{a'}_n\\ &= \gamma_1g(\mathbf{a}_1) + \cdots \gamma_mg(\mathbf{a}_m) + \lambda\delta_1g(\mathbf{a'}_1) + \cdots + \lambda\delta_ng(\mathbf{a'}_n)\\ &= g(\mathbf{x})+\lambda g(\mathbf{y}). \end{align*}$$

Thus, the map $\mathcal{L}(V,F)\to F^A$ is in fact onto, giving a bijection.

This is the "linear-algebra" proof. The "abstract nonsense proof" relies on the fact that if $A$ is a basis for $V$, then $V$ is isomorphic to $\mathop{\bigoplus}\limits_{a\in A}F$, a direct sum of $|A|$ copies of $A$, and on the following universal property of the direct sum:

Definition. Let $\mathcal{C}$ be an category, let $\{X_i\}{i\in I}$ be a family of objects in $\mathcal{C}$. A coproduct of the $X_i$ is an object $C$ of $\mathcal{C}$ together with a family of morphisms $\iota_j\colon X_j\to C$ such that for every object $X$ and ever family of morphisms $g_j\colon X_j\to X$, there exists a unique morphism $\mathbf{f}\colon C\to X$ such that for all $j$, $g_j = \mathbf{g}\circ \iota_j$.

That is, a family of maps from each element of the family is equivalent to a single map from the coproduct (just like a family of maps into the members of a family is equivalent to a single map into the product of the family). In particular, we get that:

Theorem. Let $\mathcal{C}$ be a category in which the sets of morphisms are sets; let $\{X_i\}_{i\in I}$ be a family of objects of $\mathcal{C}$, and let $(C,\{\iota_j\}_{j\in I})$ be their coproduct. Then for every object $X$ of $\mathcal{C}$ there is a natural bijection $$\mathrm{Hom}_{\mathcal{C}}(C,X) \longleftrightarrow \prod_{j\in I}\mathrm{Hom}_{\mathcal{C}}(X_j,X).$$

The left hand side is the collection of morphisms from the coproduct to $X$; the right hand side is the collection of all families of morphisms from each element of $\{X_i\}_{i\in I}$ into $X$.

In the vector space case, the fact that a linear transformation is completely determined by its value on a basis is what establishes that a vector space $V$ with basis $A$ is the coproduct of $|A|$ copies of the one-dimensional vector space $F$. So we have that $$\mathcal{L}(V,W) \leftrightarrow \mathcal{L}\left(\mathop{\oplus}\limits_{a\in A}F,W\right) \leftrightarrow \prod_{a\in A}\mathcal{L}(F,W).$$ But a linear transformation from $F$ to $W$ is equivalent to a map from the basis $\{1\}$ of $F$ into $W$, so $\mathcal{L}(F,W) \cong W$. Thus, we get that if $V$ has a basis of cardinality $\kappa$ (finite or infinite), we have: $$\mathcal{L}(V,F) \leftrightarrow \mathcal{L}\left(\mathop{\oplus}_{i\in\kappa}F,F\right) \leftrightarrow \prod_{i\in\kappa}\mathcal{L}(F,F) \leftrightarrow \prod_{i\in\kappa}F = F^{\kappa}.$$

Lemma 2. If $\kappa$ is infinite, then $\dim(V^*)\geq |F|$.

Proof. If $F$ is finite, then the inequality is immediate. Assume then that $F$ is infinite. Let $c\in F$, $c\neq 0$. Define $\mathbf{f}_c\colon V\to F$ by $\mathbf{f}_c(\epsilon_n) = c^n$ if $n\in\omega$, and $\mathbf{f}_c(\epsilon_i)=0$ if $i\geq\omega$. These are linearly independent:

Suppose that $c_1,\ldots,c_m$ are pairwise distinct nonzero elements of $F$, and that $\alpha_1\mathbf{f}_{c_1} + \cdots + \alpha_m\mathbf{f}_{c_m} = \mathbf{0}$. Then for each $i\in\omega$ we have $$\alpha_1 c_1^i + \cdots + \alpha_m c_m^i = 0.$$ Viewing the first $m$ of these equations as linear equations in the $\alpha_j$, the corresponding coefficient matrix is the Vandermonde matrix, $$\left(\begin{array}{cccc} 1 & 1 & \cdots & 1\\ c_1 & c_2 & \cdots & c_m\\ c_1^2 & c_2^2 & \cdots & c_m^2\\ \vdots & \vdots & \ddots & \vdots\\ c_1^{m-1} & c_2^{m-1} & \cdots & c_m^{m-1} \end{array}\right),$$ whose determinant is $\prod\limits_{1\leq i\lt j\leq m}(c_j-c_i)\neq 0$. Thus, the system has a unique solution, to wit $\alpha_1=\cdots=\alpha_m = 0$.

Thus, the $|F|$ linear functionals $\mathbf{f}_c$ are linearly independent, so $\dim(V^*)\geq |F|$. QED

To recapitulate: Let $V$ be a vector space of dimension $\kappa$ over $F$, with $\kappa$ infinite. Let $V^*$ be the dual of $V$. Then $V\cong\mathop{\bigoplus}\limits_{i\in\kappa}F$ and $V^*\cong\prod\limits_{i\in\kappa}F$.

Let $\lambda$ be the dimension of $V^*$. Then by Lemma 1 we have $|V^*| = \lambda|F|$.

By Lemma 2, $\lambda=\dim(V^*)\geq |F|$, so $|V^*| = \lambda$. On the other hand, since $V^*\cong\prod\limits_{i\in\kappa}F$, then $|V^*|=|F|^{\kappa}$.

Therefore, $\lambda= |F|^{\kappa}\geq 2^{\kappa} \gt \kappa$. Thus, $\dim(V^*)\gt\dim(V)$, so $V$ is not isomorphic to $V^*$.

Added${}^{\mathbf{2}}$. Some results on vector spaces and bases.

Let $V$ be a vector space, and let $A$ be a maximal linearly independent set (that is, $A$ is linearly independent, and if $B$ is any subset of $V$ that properly contains $A$, then $B$ is linearly dependent).

In order to guarantee that there is a maximal linearly independent set in any vector space, one needs to invoke the Axiom of Choice in some manner, since the existence of such a set is, as we will see below, equivalent to a basis; however, here we are assuming that we already have such a set given. I believe that the Axiom of Choice is not involved in any of what follows.

Proposition. $\mathrm{span}(A) = V$.

Proof. Since $A\subseteq V$, then $\mathrm{span}(A)\subseteq V$. Let $\mathbf{v}\in V$. If $v\in A$, then $v\in\mathrm{span}(A)$. If $v\notin A$, then $B=V\cup\{v\}$ is linearly dependent by maximality. Therefore, there exists a finite subset $a_1,\ldots,a_m$ in $A$ and scalars $\alpha_1,\ldots,\alpha_n$, not all zero, such that $\alpha_1a_1+\cdots+\alpha_ma_m=\mathbf{0}$. Since $A$ is linearly independent, at least one of the $a_i$ must be equal to $v$; say $a_1$. Moreover, $v$ must occur with a nonzero coefficient, again by the linear independence of $A$. So $\alpha_1\neq 0$, and we can then write $$v = a_1 = \frac{1}{\alpha_1}(-\alpha_2a_2 -\cdots - \alpha_na_n)\in\mathrm{span}(A).$$ This proves that $V\subseteq \mathrm{span}(A)$. $\Box$

Proposition. Let $V$ be a vector space, and let $X$ be a linearly independent subset of $V$. If $v\in\mathrm{span}(X)$, then any two expressions of $v$ as linear combinations of elements of $X$ differ only in having extra summands of the form $0x$ with $x\in X$.

Proof. Let $v = a_1x_1+\cdots a_nx_n = b_1y_1+\cdots+b_my_m$ be two expressions of $v$ as linear combinations of $X$.

We may assume without loss of generality that $n\leq m$. Reordering the $x_i$ and the $y_j$ if necessary, we may assume that $x_1=y_1$, $x_2=y_2,\ldots,x_{k}=y_k$ for some $k$, $0\leq k\leq n$, and $x_1,\ldots,x_k,x_{k+1},\ldots,x_n,y_{k+1},\ldots,y_m$ are pairwise distinct. Then $$\begin{align*} \mathbf{0} &= v-v\\ &=(a_1x_1+\cdots+a_nx_n)-(b_1y_1+\cdots+b_my_m)\\ &= (a_1-b_1)x_1 + \cdots + (a_k-b_k)x_k + a_{k+1}x_{k+1}+\cdots + a_nx_n - b_{k+1}y_{k+1}-\cdots - b_my_m. \end{align*}$$ As this is a linear combination of pairwise distinct elements of $X$ equal to $\mathbf{0}$, it follows from the linear independence of $X$ that $a_{k+1}=\cdots=a_n=0$, $b_{k+1}=\cdots=b_m=0$, and $a_1=b_1$, $a_2=b_2,\ldots,a_k=b_k$. That is, the two expressions of $v$ as linear combinations of elements of $X$ differ only in that there are extra summands of the form $0x$ with $x\in X$ in them. QED

Corollary. Let $V$ be a vector space, and let $A$ be a maximal independent subset of $V$. If $W$ is a vector space, and $f\colon A\to W$ is any function, then there exists a unique linear transformation $T\colon V\to W$ such that $T(a)=f(a)$ for each $a\in A$.

Proof. Existence. Given $v\in V$, then $v\in\mathrm{span}{A}$. Therefore, we can express $v$ as a linear combination of elements of $A$, $v = \alpha_1a_1+\cdots+\alpha_na_n$. Define $$T(v) = \alpha_1f(a_1)+\cdots+\alpha_nf(a_n).$$ Note that $T$ is well-defined: if $v = \beta_1b_1+\cdots+\beta_mb_m$ is any other expression of $v$ as a linear combination of elements of $A$, then by the lemma above the two expressions differ only in summands of the form $0x$; but these summands do not affect the value of $T$.

Note also that $T$ is linear, arguing as above. Finally, since $a\in A$ can be expressed as $a=1a$, then $T(a) = 1f(a) = f(a)$, so the restriction of $T$ to $A$ is equal to $f$.

Uniqueness. If $U$ is any linear transformation $V\to W$ such that $U(a)=f(a)$ for all $a\in A$, then for every $v\in V$, write $v=\alpha_1a_1+\cdots+\alpha_na_n$ with $a_i\in A$. Then. $$\begin{align*} U(v) &= U(\alpha_1a_1+\cdots + \alpha_na_n)\\ &= \alpha_1U(a_1) + \cdots + \alpha_n U(a_n)\\ &= \alpha_1f(a_1)+\cdots + \alpha_n f(a_n)\\ &= \alpha_1T(a_1) + \cdots + \alpha_n T(a_n)\\ &= T(\alpha_1a_1+\cdots+\alpha_na_n)\\ &= T(v).\end{align*}$$ Thus, $U=T$. QED

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Arturo Magidin
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The "this guy" you're looking for is just the function that takes each of your basis vectors and sends them to 1.

Note that this is not in the span of the set of functions that each take a single basis vector to 1, and all others to 0, because the span is defined to be the set of finite linear combinations of basis vectors. And a finite linear combination of things that have finite-dimensional support will still have finite-dimensional support, and thus can't send infinitely many independent vectors all to 1.

You may want to say, "But look! If I add up these infinitely many functions, I clearly get a function that sends all my basis vectors to 1!" But this is actually a very tricky process. What you need is a notion of convergence if you want to add infinitely many things, which isn't always obvious how to define.

In the end, it boils down to a cardinality issue - not of the vector spaces themselves, but of the dimensions. In the example you give, $\mathbb{R}^{<\omega}$ has countably infinite dimension, but the dimension of its dual is uncountable.

(Added, in response to comment below): Think of all the possible ways you can have a function which is 1 on some set of your basis vectors and 0 on the rest. The only ways you can do these and stay in the span of your basis vectors is if you take the value 1 on only finitely many of those vectors. Since your starting space was infinite-dimensional, there's an uncountable number of such functions, and so uncountably many of them lie outside the span of your basis. You can only ever incorporate finitely many of them by "adding" them in one at a time (or even countably many at a time), so you'll never establish the vector isomorphism you're looking for.

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  • Oh, I will not want to say that I am adding infinitely many functions. However given *one* more guy which you defined by a canonical embedding is no good... assuming the axiom of choice $\kappa+1=\kappa$ for every infinite cardinal. This means I can also define a linear embedding from $V$ which catches this new guy. Then who's going to be the vector not in my basis? – Asaf Karagila Aug 19 '11 at 21:48
  • I don't understand why "this guy" can not be in the image of any embedding from $V$ to $V^\ast$. Let's say $B$ is a basis for $V$ and $f:V\to\mathbb{F}$ is "this guy" (i.e. the linear map which sends all the elements of $B$ to $1$). Can't we embed $V$ into $V^\ast$ by choosing a linearly independent subset of $V^\ast$, say $B'$, which includes $f$, map one of elements of $B$ to $f$ and all the other elements injectively into $B'\setminus\{f\}$? Certainly $f$ will be in the image of this embedding. – LostInMath Aug 19 '11 at 23:12
  • @MartianInvader: Just to nitpick, countable sequences on $\mathbb Q$ which have finitely many nonzero coordinates is a countable set. Either way you have reduced it to a cardinality game which I am not comfortable with here. See the second edit of my question. Thanks a lot! – Asaf Karagila Aug 20 '11 at 06:26

Only attempting to address that one point Asaf raised in comments/edits. I refer to the CW answer by Arturo & Bill for the cardinality argument and an actual answer to the original question.

Assume that $A=\{e_i\mid i \in I\}$ is a basis for $V$. Let $f:A\rightarrow F$ be any function. This function can be extended linearly to an element of $V^*$ as follows. An arbitrary element $x\in V$ can be written as a finite linear combination of the basis elements in a unique way $$ x=\sum_{j=i}^nc_j e_{i_j}, $$ where $e_{i_1},e_{i_2},\ldots,e_{i_n}$ are the basis vectors needed to write $x$. This finite subset of $A$ (as well as the natural number $n$) obviously depends on $x$. Anyway, we can define $$ f(x)=\sum_{j=i}^n c_j f(e_{i_j}). $$ As the sum is finite, we do only vector space operations on the r.h.s. (no convergence question or some such). As the presentation of $x$ as a linear combinations of element of $A$ is unique (up to addition of terms with coefficients equal to zero), $f(x)$ is well defined. It is straightforward to check that the mapping $f$ defined in this way is linear, i.e. an element of $V^*$.

What may have been confusing is that we do not require $f$ to have finite support for the above `linear extension' to work as described. The upshot is that we only need to use a finite number of vectors from the basis to write a given vector $x$. IOW the finiteness of the sum in the definition of the linear extension of $f$ comes from $A$ being a basis - not from the support of $f$ (that does not need to be finite).

We can similarly extend a function with singleton support. If $\chi_i:A\rightarrow F$ is the function defined by $e_i\mapsto 1, e_j\mapsto 0$, for all $j\in I, j\neq i$, let's call its linear extension to an element of $V^*$ also $\chi_i$. What's the span of the mappings $\chi_i$? Only those linear functions $f$ with $|A\setminus\ker f|<\infty$ can be written as linear combinations of $\chi_i, i\in I$. Therefore the span of the linear mappings $\chi_i,i\in I$ is not a all of $V^*$ unless $\dim V$ is finite.

Jyrki Lahtonen
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  • Many thanks, Jyrki. This seem to address the part which I know. My question is actually why are the elements of $V^*$ do not contain anything except the functions from $A$ into $F$. I am given to understand that this is a standard "universal property" proof, however as I am uncomfortable with those I am still trying to figure this out in my head. – Asaf Karagila Aug 21 '11 at 20:05
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    @Asaf: To see that the map from $\mathbb F^A$ to $V^*$ is surjective, note that each element $T$ of $V^*$ determines an element $T\vert_A$ of $\mathbb F^A$. By linearity, $T$ must be the function constructed from $T\vert_A$ as in Jyrki's answer. – Jonas Meyer Aug 21 '11 at 20:16
  • @Asaf: If your question is on why we go from $\mathcal{L}(\oplus A,B)$ to $\prod\mathcal{L}(A,B)$, I can certainly replace my invocation of abstract nonsense with an actual proof. – Arturo Magidin Aug 21 '11 at 20:22
  • @Arturo: That would be delightful! Assuming that the axiom of choice is involved, could you try and make a note where? Many, many, many thanks. – Asaf Karagila Aug 21 '11 at 20:27
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    @Asaf: As far as I know, it's not, in the sense that we begin with the assumption that we have a vector space with a basis; so it would come in if you simply assume "vector space". (Or if your definition of "infinite dimensional vector space" is "vector space that does not have a finite basis"). – Arturo Magidin Aug 21 '11 at 20:51
  • @Asaf: what is your definition of a basis? From your definition do you see why a linear operator out of a vector space is determined by what it does to a basis? – Qiaochu Yuan Aug 21 '11 at 20:55
  • @Asaf: Ok, sorry about misunderstanding your comment. I thought that the problematic part was: ``My problem is, if so, *why* is the dual space composed from **all** functions from $A$ into $F$?'' For the remaining part I would argue the same as way as Jonas: $T\vert_A$ determines a linear mapping $T$ uniquely. – Jyrki Lahtonen Aug 21 '11 at 20:59
  • @Asaf: Are we clear that since there is no analytic structure on $V$, by a basis the rest of us mean a *Hamel basis*? The existence of one for any vector space follows from a relatively straightforward application involving Zorn's lemma. – Jyrki Lahtonen Aug 21 '11 at 21:10
  • @Qiaochu, Arturo, Jyrki: The definition of a basis is the same as before. Linearly independent and maximal. In my application while the axiom of choice is absent there still exists a basis for the vector space. I am not sure how much of Zorn's lemma you can apply in this case, however. I do see why an operator is determined by the action on the basis. As per usual, I am always getting very suspicious on things which are "just" extended. I may be overly suspicious, but this is just a few intensive months without the axiom of choice... :-) – Asaf Karagila Aug 21 '11 at 22:50
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    @Asaf: once you declare that a basis exists, there is nothing suspicious going on. Really. Given a basis $A$ of a vector space $V$ over a field $k$ and a function $f : A \to k$, the function $f$ extends uniquely by linearity to any finite linear combination of elements of $A$, which by hypothesis extends $f$ to all of $V$. What part of that is suspicious? (If the extension to some $v \in V$ does not exist, $A$ is not maximal. If it is not unique, $A$ is not linearly independent.) – Qiaochu Yuan Aug 21 '11 at 23:38