Abstract Nonsense

Crushing one theorem at a time

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.

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October 1, 2010 - Posted by | Algebra, Linear Algebra | , , ,

1 Comment »

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

    Pingback by Canonical Isomorphism Between a Finite Dimensional Inner Product Space and its Dual « Abstract Nonsense | November 12, 2011 | Reply


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