# Abstract Nonsense

## 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 $\dim_F\mathscr{V}<\infty$

8.

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 $\dim_F\mathcal{V}<\infty$

Proof:

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 ‘

$\ker(f+g)\subseteq\{x_1,\cdots,x_n,y_1,\cdots,y_m\}$

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

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

where

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

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

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

where

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

and

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

Thus,

\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.

Referenences:

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 -