Mathematics Prelims

July 18, 2008

The Algebraic Dual Space

Filed under: Analysis,Functional Analysis — cjohnson @ 10:27 pm

Suppose X is a vector space over a field K.  We say a function f : X \to K is a linear functional if for every \alpha, \beta \in K and every x, y \in X, we have f(\alpha x + \beta y) = \alpha f(x) + \beta f(y).  We will always assume that K is either \mathbb{R} or \mathbb{C}.  Properties and theorems associated with “traditional” linear operators apply since \mathbb{R} and \mathbb{C} can be thought of as normed spaces with the “traditional” norms (absolute values).  Note that the set of all linear functionals on X, which is denoted X^*, is itself a vector space if we allow scalar multiplication and addition of functions in the traditional way ((\alpha f)(x) = \alpha f(x) and (f + g)(x) = f(x) + g(x)) .

The vector space X^* is referred to as the algebraic dual of X.  Since X^* is itself a vector space, we can define its algebraic dual, (X^*)^* = X^{**}, which is called the second algebraic dual of X.  An important property of X^{**} is that there exists an injective mapping C : X \to X^{**} called the canonical mapping of X into X^{**}.  This mapping is given by taking an x \in X and considering the functional g_x : X^* \to \mathbb{R} such that for each f \in X^* we map g_x(f) = f(x).

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

Every finite dimensional vector space is algebraically reflexive.

Proof: Suppose X is an n-dimensional vector space with basis \{ e_1, ..., e_n \}.  Let f \in X^* and x \in X with x = \alpha_1 e_1 + ... + \alpha_n e_n.  As f is linear, we have f(x) = f(\alpha_1 e_1 + ... + \alpha_n e_n) = \alpha_1 f(e_1) + ... + \alpha_n f(e_n).  This implies that $f$ is uniquely determined by the values of f(e_1), ..., f(e_n), which means we can view f as an n-tuple of scalars.  That in turn means that \dim X^* = n and that the set of n-tuples which have a single one and then n-1 zeroes form a basis for X^*.  This is called the dual basis of X.  Applying the same procedure to X^*, we see that \dim X^{**} = n.  Now, since C is an injective linear map from X into X^{**} we have that X is isomorphic to an n-dimensional subspace of X^{**}, and since the only n-dimensional subspace of X^{**} is X^{**} itself, we have that X is isomorphic to X^{**}, and so every finite dimensional vector space is algebraically reflexive.



  1. why is the canonical mapping C injective?

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

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