📂Linear Algebra

Dual Pair Spaces

Definition¹

Let us call $X$ a vector space. Let $X^{\ast\ast}$ be the dual space of $X$’s dual space, $X^{\ast}$.

$$X^{\ast\ast} = (X^{\ast})^{\ast}$$

This is called the bidual space of $X$.

Theorem

If $X$ is a finite-dimensional vector space, then $X$ and $X^{\ast\ast}$ are isomorphic.

$$X \approx X^{\ast\ast}$$

Explanation

Bidual, double dual, and second dual all mean the same thing.

The above theorem holds only when $X$ is finite-dimensional. If it is not finite-dimensional, it generally does not hold, and if it does, $X$ is called a reflexive space.

An important point in the theorem is that the isomorphism $\psi : X \to X^{\ast\ast}$ can be derived very naturally. For an arbitrary $x \in X$, let’s define $\hat{x} : X^{\ast} \to \mathbb{R}$ as follows.

$$\hat{x} (f) = f(x) \quad \forall f \in X^{\ast}$$

Then, $\hat{x}$ becomes a linear functional on $X^{\ast}$. Now let’s define $\psi : X \to X^{\ast\ast}$ as follows.

$$\psi (x) = \hat{x}\quad \forall x \in X$$

Then the theorem is described as follows.

Let us consider $X$ as a finite-dimensional vector space. Define $\psi : X \to X^{\ast\ast}$ as $\psi (x) = \hat{x}$. Then, $\psi$ is an isomorphism.

Proof

• Linearity

  Let $x, y \in X$ and $k \in \mathbb{R}$. Then, for any $f \in X^{\ast}$,

  $$\begin{align*} \psi (kx + y)(f) = \widehat{kx + y}(f) &= f(kx + y) \\ &= kf(x) + f(y) \\ &= k\hat{x}(f) + \hat{y}(f) \\ &= k\psi (x)(f) + \psi (y)(f) \\ \end{align*}$$

  Therefore,

  $$\psi (kx + y) = k\hat{x} + \hat{y} = k\psi (x) + \psi (y)$$

• Injectivity

  Auxiliary Theorem

  Let $X$ be a finite-dimensional vector space, and $x \in X$. If for all $f\in X^{\ast}$, $\hat{x}(f) = 0$ holds, then $x=0$ holds.

  Proof

  The proof is by contraposition. Assume $x \ne 0$. Then it is sufficient to show that there exists $f$ such that $\hat{x}(f) \ne 0$. Choose any ordered basis $\beta = \left\{ x_{1}=x, x_{2}, \dots, x_{n} \right\}$. Then, for the dual basis $\beta^{\ast} = \left\{ f_{1}, \dots, f_{n} \right\}$, $f_{1}(x_{1}) = 1 \ne 0$ holds.

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  Let’s say for some $x \in X$ the functional $\psi (x)$ on $X^{\ast}$ is $\psi (x) = 0$. Then for all $f \in X^{\ast}$, $\psi (x)(f) = \hat{x}(f) = 0$ holds. Therefore, by the auxiliary theorem, $x=0$ holds. Since $N(\psi) = \left\{ 0 \right\}$, $\psi$ is injective.

• Isomorphism

  Since $X$ is finite-dimensional, so is $X^{\ast}$, and they have the same dimension. Similarly, the dimensions of $X^{\ast}$ and $X^{\ast\ast}$ are also the same,

  $$\dim (X) = \dim (X^{\ast}) = \dim (X^{\ast\ast})$$

  Therefore, if $\psi : X \to X^{\ast\ast}$ is injective, it is surjective, and $\psi$ is an isomorphism.

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