## Simple But Elegant

**Point of post: **This is a simple proof that the canonical identification between an -dimensional vector space and it’s dual space’s dual space is actually a linear isomorphism.

**Lemma:** Let be an -dimensional subspace of an -dimensional vector space then

**Proof:** Let be a basis for . Suppose that , and suppose that

We clearly must have that otherwise

contradicting . But, this means that may be rewritten as

and since are l.i. we may conclude that . Thus, is l.i. But, this contradicts that .

**Lemma:** Let be distinct , then there exists some such that (where

**Proof:** Clearly since

we must merely show that given a nonzero vector there exists some such that . To do this suppose that for every . Then, in particular if is a basis for we see that . Thus, if is the corresponding dual base and then and so

**Theorem:** Let be an -dimensional vector space over the field then the natural identification

(where is a fixed number, and can be thought of as the function where we just write to emphasize that the is the independent variable)

**Proof:** Suppose that then by the second lemma we may find such that and thus the mappings

are not equal, but and so is injective.

The fact that is linear follows from the fact that

the penultimate step being true by virtue of the fact that each of the (and this is an informal way of putting it) you put into the function is linear.

Thus, the mapping is a linear embedding, in particular is a linear subspace of . But, since is -dimensional we may appeal to the first lemma to conclude that (in general given an -dimensional vector space , it’s dual space is -dimensional and so the dual spaces dual space is -dimensional). The conclusion follows.

**References:**

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

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

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