Suppose is a vector space over a field . We say a function is a linear functional if for every and every , we have . We will always assume that is either or . Properties and theorems associated with “traditional” linear operators apply since and can be thought of as normed spaces with the “traditional” norms (absolute values). Note that the set of all linear functionals on , which is denoted , is itself a vector space if we allow scalar multiplication and addition of functions in the traditional way ( and ) .

The vector space is referred to as the algebraic dual of . Since is itself a vector space, we can define its algebraic dual, , which is called the second algebraic dual of . An important property of is that there exists an injective mapping called the canonical mapping of into . This mapping is given by taking an and considering the functional such that for each we map .

Since is a linear map from to (this follows from the fact that each is linear), we have is isomorphic to a subspace of of (recall that the range of a linear operator is a subspace of the operator’s codomain). For this reason, is sometimes called the canonical embedding of into . (A space is said to be embeddable in if is isomorphic to a subspace of .) In the event that is also surjective, so we have is isomorphic to all of , we say that is algebraically reflexive.

Every finite dimensional vector space is algebraically reflexive.

**Proof**: Suppose is an n-dimensional vector space with basis . Let and with . As is linear, we have . This implies that $f$ is uniquely determined by the values of , which means we can view as an n-tuple of scalars. That in turn means that and that the set of n-tuples which have a single one and then n-1 zeroes form a basis for . This is called the dual basis of . Applying the same procedure to , we see that . Now, since is an injective linear map from into we have that is isomorphic to an n-dimensional subspace of , and since the only n-dimensional subspace of is itself, we have that is isomorphic to , and so every finite dimensional vector space is algebraically reflexive.

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why is the canonical mapping C injective?

Comment by wilson — April 2, 2009 @ 9:20 pm |

It comes as a consequence of the Hahn-Banach theorem:

http://en.wikipedia.org/wiki/Reflexive_space#Normed_spaces

Comment by Felipo Bacani — May 24, 2010 @ 3:16 pm |