# Abstract Nonsense

## Simple But Elegant

Point of post: This is a simple proof that the canonical identification between an $n$-dimensional vector space $\mathcal{V}$ and it’s dual space’s dual space $\mathcal{V}^{**}$ is actually a linear isomorphism.

Lemma: Let $\mathcal{W}$ be an $n$-dimensional subspace of an $n$-dimensional vector space $\mathcal{V}$ then $\mathcal{V}=\mathcal{W}$

Proof: Let $\{x_1,\cdots,x_n\}$ be a basis for $\mathcal{W}$. Suppose that $v\in\mathcal{V}-\mathcal{W}$, and suppose that

$\alpha_1+\cdots+\alpha_n x_n+\alpha v=\bold{0}\quad(1)$

We clearly must have that $\alpha=0$ otherwise

$\displaystyle v=\frac{-\alpha_1}{\alpha}x_1+\cdots+\frac{-\alpha_n}{\alpha}x_n$

contradicting $v\notin \mathcal{W}=\text{Span }\{x_1,\cdots,x_n\}$. But, this means that $(1)$ may be rewritten as

$\alpha_1x_1+\cdots+\alpha_n x_n=\bold{0}$

and since $\{x_1,\cdots,x_n\}$ are l.i. we may conclude that $\alpha_1=\cdots=\alpha_n=\alpha=0$. Thus, $\{x_1,\cdots,x_n,v\}$ is l.i. But, this contradicts that $\dim_F\left(\mathcal{V}\right)=n$. $\blacksquare$

Lemma: Let $x,y\in\mathcal{V}$ be distinct , then there exists some $\phi\in\mathcal{V}^{*}$ such that $[x,\phi]\ne[y,\phi]$ (where $[x,\phi]=\phi(x),[y,\phi]=\phi(y)$

Proof: Clearly since

$[x,\phi]\ne[y,\phi]\text{ }\Longleftrightarrow\text{ }[x,\phi]-[y,\phi]=[x-y,\phi]\ne 0$

we must merely show that given a nonzero vector $v$ there exists some $\phi\in\mathcal{V}^{*}$ such that $[v,\phi]\ne 0$. To do this suppose that $[v,\phi]=0$ for every $v\in\mathcal{V}$. Then, in particular if $\{x_1,\cdots,x_n\}$ is a basis for $\mathcal{V}$ we see that $[x_j,\phi]=0,\text{ }j=1,\cdots,n$. Thus, if $\{\theta_1,\cdots,\theta_n\}$ is the corresponding dual base and $v=\alpha_1x_1+\cdots+\alpha_n x_n$ then $\bold{0}=\theta(v)=\alpha_1,\cdots,0=\theta(v)=\alpha_n$ and so $v=\bold{0}$

Theorem: Let $\mathcal{V}$ be an $n$-dimensional vector space over the field $F$ then the natural identification

$\varphi:\mathcal{V}\to\mathcal{V}^{**}:x_0\mapsto[x_0,\psi]$

(where $x_0$ is a fixed number, and $[x_0,\psi]$ can be thought of as the function $f:\mathcal{V}^{*}\to F:\psi\mapsto \psi(x_0)$ where we just write $\psi(x_0)=[x_0,\psi]$ to emphasize that the $\psi$ is the independent variable)

Proof: Suppose that $x_0\ne y_0$ then by the second lemma we may find $\psi_0\in\mathcal{V}^{*}$ such that $[x_0,\psi_0]\ne[y_0,\psi_0]$ and thus the mappings

$[x_0,\psi],[y_0,\psi]:\mathcal{V}^{*}\to F$

are not equal, but $[x_0,\psi]=\varphi(x_0),[y_0,\psi]=\varphi(y_0)$ and so $\varphi$ is injective.

The fact that $\varphi$ is linear follows from the fact that

$\varphi\left(\alpha x_0+\beta y_0\right)=[\alpha x_0+\beta y_0,\psi]=\alpha[x_0,\psi]+\beta[y_0,\psi]=\alpha\varphi(x_0)+\beta\varphi(y_0)$

the penultimate step being true by virtue of the fact that each of the (and this is an informal way of putting it) $\psi$ you put into the function $[x_0,\psi]$ is linear.

Thus, the mapping $\varphi:\mathcal{V}\to\mathcal{V}^{**}$ is a linear embedding, in particular $\varphi\left(\mathcal{V}\right)$ is a linear subspace of $\mathcal{V}^{**}$. But,  since $\mathcal{V}^{**}$ is $n$-dimensional we may appeal to the first lemma to conclude that $\varphi\left(\mathcal{V}\right)=\mathcal{V}^{**}$ (in general given an $n$-dimensional vector space $\mathcal{W}$, it’s dual space $\mathcal{W}^{*}$ is $n$-dimensional and so the dual spaces dual space is $n$-dimensional). The conclusion follows. $\blacksquare$

References:

1. Halmos, Paul R. Finite-dimensional Vector Spaces,. New York: Springer-Verlag, 1974. Print.

October 1, 2010 -

## 1 Comment »

1. […] have seen in the past the proof that every finite dimensional vector space is isomorphic to its double dual. […]

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