Abstract Nonsense

Crushing one theorem at a time

Halmos Chapter one Sections 15, 16 and 17: Dual Bases, Reflexivity, and Annihilators (Part II)

Point of post: In this post I complete the second half of the problems in the 15th, 16th, and 17th sections of Halmos’s book. There are some pretty neat theorems in here, in particular:

\mathscr{V} is always embeddable into \text{Hom}\left(\mathscr{V},F\right),

-For any vector space \mathscr{V} with basis \mathscr{B} it’s true that \text{Hom}\left(\mathscr{V},F\right)\cong F^{\mathscr{B}}

– That if \mathscr{V} is infinite dimensional that \dim_F\mathscr{V}<\dim_F\text{Hom}\left(\mathscr{V},F\right) and thus the canonical identification between \mathscr{V} and \mathscr{V}^{**} is an isomorphism if and only if \dim_F\mathscr{V}<\infty



Problem: This exercise  is concerned with vector spaces that need not be finite dimensional.

a) Suppose that F is a field and \Omega a set, then if f,g:\Omega\to F and \alpha,\beta\in F write h=\alpha\varphi+\beta\psi for the function h(x)=\alpha f(x)+\beta g(x) for all x\in\Omega. The set of all such functions is a vector space with respect to this definition of linear operations, and the same is true of the set of all finitely non-zero functions (A function f:\Omega\to F is finitely non-zero if \Omega-\ker f is finite)

b) Every vector space is isomorphic to the set of all finitely non-zero functions on some set.

c) If \mathscr{V} is a vector space with basis \mathscr{B}, and if f:\mathscr{B}\to F then there exists a unique \varphi\in\text{Hom}\left(\mathscr{V},F\right)  such that [x,\varphi]=f(x) for all x\in\mathscr{B}

d) Use a)b), and c) to conclude that every vector space \mathscr{V} is isomorphic to a subspace of \text{Hom}\left(\mathscr{V},F\right).

e) Which vector spaces are isomorphic to their own duals?

f) If \mathcal{Y} is a l.i. subset of a vector space \mathscr{V}, then there exists a basis of \mathscr{V} containing \mathcal{Y}

g) If \Omega is a set and if y\in\Omega, write f_y:\Omega\to F by

f_y(x)=\begin{cases}1 & \mbox{if} \quad x=y\\ 0 &\mbox{if}\quad x\ne y\end{cases}

Let \mathcal{Y} be the set of all functions f_y together with the function g(x)=1. Prove that if \Omega is infinite, then \mathcal{Y} is l.i. subset of the vector space of all scalar valued functions on \Omega

h) The canonical identification between \mathcal{V} and \mathcal{V}^{**} is defined for all vector spaces. Prove that \mathcal{V} is reflexive iff \dim_F\mathcal{V}<\infty


a) Clearly the result is true for F^{\Omega} so we must only prove it’s true for finitely non-zero functions. But, if f,g:\Omega\to F are finitely non-zero then \Omega-\ker f=\{x_1,\cdots,x_n\} and \Omega-\ker g=\{y_1,\cdots,y_m\} and so if f(x)+g(x)\ne {0} then clearly f and g cannot be simultaneously zero. Thus, x\in\{x_1,\cdots,x_m\}\cup\{y_1,\cdots,y_m\} and so ‘


and is thus finitely non-zero. Also, notice that \alpha f(x)\ne 0 \implies f(x)\ne {0} so that \Omega-\ker \alpha f\subseteq\{x_1,\cdots,x_n\} and so \alpha f is finitely non-zero. It follows that the set of all finitely non-zero functions is a subspace of  F^{\Omega}.

b) Let \mathscr{V} be a vector space with basis \mathscr{B} and let

\mathscr{W}=\left\{f:\mathscr{B}\to F\mid \mathscr{B}-\ker f\text{ is finite}\right\}

We claim that \mathscr{V}\cong \mathscr{W}. To see this we first find a basis for \mathscr{W}. To do this consider the set

\overline{ \mathscr{B}}=\left\{f_x:x\in\mathscr{B}\right\}


f_x(y)=\begin{cases} 1 & \mbox{if} \quad x=y\\ 0 & \mbox{if}\quad x\ne y\end{cases}

we claim that \overline{\mathscr{B}} is a basis for W. To see this, first suppose that

\displaystyle \sum_{j=1}^{n}\alpha_j f_{x_j}(x)=\bold{0}(x)

Then, in particular we see that

\displaystyle \sum_{j=1}^{n}\alpha_j f_{x_j}(x_m)=\alpha_m=0,\quad{ }m=1,\cdots,n

from where it follows that any finite subset of \overline{\mathscr{B}} is l.i. and so by definition so is \overline{\mathscr{B}}. Furthermore, let f\in \mathscr{W}. If f=\bold{0}(x) the fact that f\in\text{Span }\overline{\mathscr{B}} is trivial, so assume not. Then \mathscr{B}-\ker f=\{x_1,\cdots,x_n\} for some x_1,\cdots,x_n\in\mathscr{B}. So then, we claim that

\displaystyle f=\sum_{j=1}^{n}f(x_j)f_{x_j}

To see this, let x\in\mathscr{B} be arbitrary. If x\ne x_1,\cdots,x_n then

\displaystyle f(x)=0=\sum_{j=1}^{n}f(x_j)f_{x_j}(x)

since x\ne x_j\implies f_{x_j}(x)=0,\text{ }j=1,\cdots,n and x\notin\mathscr{B}-\ker f\implies x\in\ker f. Now, if x=x_m for some m\in\{1,\cdots,n\} then f_{x_j}(x_m)=\delta_{jm} so that

\displaystyle \sum_{j=1}^{n}f(x_j)f_{x_j}(x_m)=f(x_m)

from where the conclusion follows. Thus, \overline{\mathscr{B}} is a basis for \mathscr{W}. So, then define

\displaystyle T:\mathscr{V}\to \mathscr{W}:\sum_{j=1}^{n}\alpha_j x_j\mapsto \sum_{j=1}^{n}\alpha_j f_{x_j}(x)

we claim that T is an isomorphism. To show it’s linear we note that if x,y\in \mathscr{V} then \displaystyle x=\sum_{j=1}^{n}\alpha_j x_j and \displaystyle y=\sum_{k=1}^{m}\beta_j y_j for some x_1,\cdots,x_n,y_1,\cdots,y_m\in\mathscr{B} so then

\displaystyle \alpha x+\beta y=\sum_{j=1}^{n+m}\gamma_j z_j


z_j=\begin{cases} x_j & \mbox{if}\quad j=1,\cdots,n \\ y_{j-n} & \mbox{if}\quad j=n+1,\cdots,n+m\end{cases}


\gamma_j=\begin{cases}\alpha \alpha_j & \mbox{if}\quad j=1,\cdots,n \\ \beta \beta_{j-n} & \mbox{if}\quad j=n+1,\cdots,n+m\end{cases}


\displaystyle \begin{aligned} T(\alpha x+\beta y) &= T\left(\sum_{j=1}^{n+m}\gamma_j z_j\right)\\ &= \sum_{j=1}^{n+m}\gamma_j f_{z_j}(z)\\ &= \sum_{j=1}^{n}\gamma_j f_{z_j}(z)+\sum_{j=n+1}^{n+m}\gamma_j f_{z_j}(z)\\ &=\sum_{j=1}^{n}\alpha \alpha_j f_{x_j}(z)+\sum_{k=1}^{m}\beta \beta_k f_{y_k}(z)\\ &=\alpha T(x)+\beta T(y)\end{aligned}

from where T‘s linearity follows. Thus, it remains to show that T is bijective.  To see this, suppose that x and y may be written as the linear sum of elements of \mathscr{B} with the same notation as above. Suppose then that T(x)=T(y) we see that if x_k\notin\{y_1,\cdots,y_m\} that T(x)=\alpha_k and T(y)=0 and since we (implicitly) assumed that \alpha_k\ne 0 this is a contradiction. Conversely, if y_k\notin\{x_1,\cdots,x_n\} then T(y)=y_k and T(x)=0 which is a also a contradiction. It follows that \{x_1,\cdots,x_n\}=\{y_1,\cdots,y_m\}. We may assume though, (with the possibility of relabeling) that x_j=y_j,\text{ }j=1,\cdots, n (we may assume that m=n since the sets are equal, and so in particular have the same cardinality). Thus,

\displaystyle \sum_{j=1}^{n}\alpha_j f_{x_j}=\sum_{j=1}^{n}\beta_j f_{y_j}=\sum_{j=1}^{n}\beta_j f_{x_j}

from where the fact that \alpha_j=\beta_j,\text{ }j=1,\cdots,n follows from the “uniqueness of representation by basis” theorem. From there the fact that x=y clearly follows, and so T is injective.

For surjectivity, we merely notice that \displaystyle f(x)=\sum_{j=1}^{n}f(x_j)f_{x_j}(x) for some x_1,\cdots,x_n\in\mathscr{B}. It clearly follows then that \displaystyle f=T\left(\sum_{j=1}^{n}f(x_j)x_j\right) . Thus, T is an isomorphism and so \mathscr{V}\cong \mathscr{W} as was to be proven.

c) So, let f be the function described above and define [x,\varphi]=f(x) for all x \in\mathscr{B}. We may then extend \varphi:\mathscr{B}\to F to a linear functional by the following observation. Let us index \mathscr{B} by \mathscr{B}=\left\{x_\beta\right\}_{\beta\in\mathcal{B}}. We then know that for each v\in\mathscr{V} it’s true that

\displaystyle v=\sum_{\beta\in\mathcal{B}}\alpha_\beta x_\beta

where it’s understood that \alpha_\beta=0 for all but finitely many \beta. So, then define \psi by

\displaystyle \varphi\left(v\right)=\varphi\left(\sum_{\beta\in\mathcal{B}}\alpha_\beta x_\beta\right)=\sum_{\beta\in\mathcal{B}}\alpha_\beta f(x_\beta)

clearly then, for each x\in\mathscr{B} we have that [x,\varphi]=f(x), and thus we have found one linear functional. To prove that it’s unique we merely note that if \psi\in\text{Hom}\left(\mathscr{V},F\right) were such that [x,\psi]=f(x) for all x\in\mathscr{B} then

\displaystyle \psi(v)=\psi\left(\sum_{\beta\in\mathcal{B}}\alpha_{\beta} x_\beta\right)=\sum_{\beta\in\mathcal{B}}\alpha_\beta\psi\left(x_\beta\right)=\sum_{\beta\in\mathcal{B}}\alpha_\beta f(x_\beta)=\varphi(v)

for all v\in\mathscr{V}, thus proving uniqueness.

d) Let \mathscr{V} have the basis \mathscr{B} and let \mathscr{W} be defined as in b) Then, define

\displaystyle T:\mathscr{W}\to\text{Hom}\left(\mathscr{V},F\right):f(x) \mapsto \sum_{\beta\in\mathcal{B}}\alpha_\beta\varphi\left(x_\beta\right)

where \displaystyle \sum_{\beta\in\mathcal{B}}\alpha_\beta\ f\left(x_\beta\right) is as in c) (we see by c) that T is  well-defined). Noticing then that

\displaystyle \begin{aligned} T\left(\alpha \varphi_1+\beta \varphi_2\right) &=\sum_{\beta\in\mathcal{B}}\alpha_\beta\left(\alpha \varphi_1(x_{\beta})+\varphi_2(x_{\beta})\right)\\ &=\alpha\sum_{\beta\in\mathcal{B}}\alpha_{\beta}\varphi(x_{\beta})+\beta\sum_{\beta\in\mathcal{B}}\alpha_{\beta}\varphi(x_{\beta}) \\ &=\alpha T\left(\varphi_1\right)+\beta T(\varphi_2)\end{aligned}

we may conclude that T is linear. To see that T is injective suppose that f,g\in\mathcal{W} are distinct. Then, there exists some x\in\mathcal{B} such that f(x)\ne g(x), then (T(f))(x)=f(x)\ne g(x)=(T(g))(x) so that T(f)\ne T(g). It follows that \mathscr{W} is isomorphic with it’s image T\left(\mathscr{W}\right), and thus \mathscr{V}\cong\mathscr{W}\cong T\left(\mathscr{W}\right)\subseteq \text{Hom}\left(\mathscr{W},F\right)

e) It was proven in the book that if \dim_F\mathscr{V}<\infty then \mathscr{V}\cong\text{Hom}\left(\mathscr{V},F\right), and we will show that this is precisely when this isomorphism is true. To do this, we first show that regardless of the cardinality of the basis \mathscr{B} for \mathscr{V} that F^{\mathscr{B}}\cong\text{Hom}\left(\mathscr{V},F\right). To do this define

\displaystyle T:F^{\mathscr{B}}\to\text{Hom}\left(\mathscr{V},F\right):f(x)\mapsto \sum_{\beta\in\mathcal{B}}\alpha_\beta f(x_{\beta})

where \displaystyle \sum_{\beta\in\mathcal{B}}\alpha_\beta f(x_{\beta}) is the function described in c). Using the exact same argument as in d) one may show that T is both injective and linear, and so to prove that it’s a linear isomorphism it suffices to show that T is surjective. So, we note as was proven many times before, that if \varphi\in\text{Hom}\left(\mathscr{V},F\right) that \varphi is the unique \psi\in\text{Hom}\left(\mathscr{V},F\right) such that \varphi_{\mid \mathscr{B}}=\psi_{\mid\mathscr{B}}. So, in particular \varphi_{\mid \mathscr{B}}\in F^{\mathscr{B}} and as can easily be seen that T\left(\varphi_{\mid \mathscr{B}}\right)=\varphi. Thus, surjectivity, and consequently T‘s status as an isomorphism follow.

So, considering the above paragraph’s result that F^{\mathscr{B}}\cong\text{Hom}\left(\mathscr{B},F\right) it suffices to show that \mathscr{W}\not\cong F^{\mathscr{B}} if \mathscr{B} is not finite. To do this, we use a cardinality argument. We first note that if K is the prime subfield of F that \dim_F\mathscr{W}=\dim_K\overline{\mathscr{W}} where \overline{\mathscr{W}}=\left\{f:\mathscr{B}\to K:\mathscr{B}-\ker f\text{ is finite}\right\}. This is clear though since \left\{f_x:x\in\mathscr{B}\right\} is a basis for both.  We next claim that \dim_K K^{\mathscr{B}}\leqslant \dim_F F^{\mathscr{B}}. To see this it suffices to prove that any l.i. subset of K^{\mathscr{B}} over K is l.i. as a subset of F^{\mathscr{B}} over F. To do this suppose that \{f_1,\cdots,f_n\}\subseteq K^{\mathscr{B}} is l.i. and \displaystyle \sum_{j=1}^{n}\alpha_j f_j=\bold{0}. Then, as K is a subfield of F we may consider the vector space F over K, and by an earlier theorem we must have that this admits a basis B. Thus, for each \alpha_j we may express it in the form \displaystyle \alpha_j=\sum_{i}\beta_{i,j}y_i. Thus, by earlier assumption

\displaystyle \bold{0}=\sum_{j=1}^{n}\alpha_j f_j=\sum_{j}\sum_{i}\beta_{i,j}y_i f_j=\sum_{i}\left(\sum_{j}\beta_{i,j}f_j(v)\right)y_i(v).

and since the inner sum is an element of K and \{y_i\} is l.i. over K we see that the inner sum is zero, and since this was true for all v\in\mathscr{V} we may deduce that \beta_{i,j}=0,\text{ for all }i,j and so \{f_1,\cdots,f_n\} is l.i. over F.

Lastly, we claim that \dim_K \overline{\mathscr{W}}<\dim_K K^{\mathscr{B}}, but we can do this equally well by showing that \text{card }\overline{\mathscr{W}}<\text{card }K^{\mathscr{B}}. So, let  \text{card }\mathscr{B}=\kappa. We clearly have that

\displaystyle \text{card }\overline{\mathscr{W}}=\text{card }\bigcup_{S\in\mathcal{P}_0\left(\mathscr{W}\right)}F^S

where \mathcal{P}_0\left(\mathscr{B}\right) is the set of all finite subsets of \mathscr{B}, since we may associate with function f\in \overline{\mathscr{W}} it’s values on \mathscr{B}-\ker f. Next, we appeal to a common cardinal arithmetic fact that  \text{card }\mathcal{P}_0\left(\mathscr{B}\right)=\kappa. Thus,

\displaystyle \text{card }\overline{\mathscr{W}}=\text{card }\bigcup_{S\in\mathcal{P}_0\left(\mathscr{B}\right)}K^S\leqslant \aleph_0\cdot \kappa=\kappa<2^{\kappa}\leqslant \lambda^\kappa=\text{card }K^{\mathscr{B}}

where we’ve used the fact that \text{card }K\geqslant 2.

Combining these inequalities gives

\dim_F \mathscr{W}=\dim_K\overline{\mathscr{W}}<\dim_K K^{\mathscr{B}}\leqslant \dim_F F^{\mathscr{B}}

and so in particular

\dim_F \mathscr{W}<\dim_F F^{\mathscr{B}}

and so \mathscr{W}\not\cong F^{\mathscr{B}} as desired.

Remark: The operative part of the proof was transferring from F to the prime subfield K. For, imagine if we’d not done that and we’d just considered F. Then, if \dim_F \mathscr{W}=\lambda we’d get into a whole bunch of problems (whether trying to prove that the dimensionality or actual cardinality of the first space is less than the second) like trying to assert that \lambda\cdot\kappa<\lambda^\kappa which clearly isn’t true in general (take \lambda=\mathfrak{c}=\text{card }\mathbb{R} and \kappa=\aleph_0

f) This was already done here

g) This was already done in general in part b)

h) You can use the other exercises as hints and literally construct an element of \mathscr{V}^{**} which isn’t in the image of the canonical identification. Or, more quickly we can notice that by the proof in e) we have that \dim_F\mathscr{V}<\dim_F\mathscr{V}^{*}<\dim_F\mathscr{V}^{**} so that \mathscr{V}\not\cong\mathscr{V}^{**} and since the canonical identification is a linear injection, it can’t be a surjection.




1. Halmos, Paul R. “Dual Bases, Reflexivity, Annihilators.” Finite-dimensional Vector Spaces,. New York: Springer-Verlag, 1974. 23-28. Print.

2.Jech, Thomas J. Set Theory. Berlin: Springer, 2003. Print.


October 12, 2010 - Posted by | Fun Problems, Halmos, Linear Algebra | , , , ,


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