# Mathematics Prelims

## July 18, 2008

### The Algebraic Dual Space

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