이 포스트에서 $V, W$는 유한차원 $F$-벡터공간으로 취급한다.
Linear Functional
Definition 1. Let $T \in \mathcal{L}(V, F)$. Then we call $T$ a linear functional on $V$.
Dual Space
Definition 2. We define the dual space of $V$ to be the vector space $\mathcal{L}(V, F)$, denoted by $V^*$.
The double dual (or bidual) space $V^{**}$ is the dual space of $V^*$.
선형 변환 $T: V \rightarrow F$을 linear functional, 선형 범함수라고 부르고 이들을 모아놓은 공간을 dual space, 쌍대 공간이라고 부른다.
Theorem 1
Theorem 1. $V \cong V^*$.
Proof. dim($V^*$) = dim($\mathcal{L}(V, F)$) = dim($V$) $\cdot$ dim($F$) = dim($V$) $\Longleftrightarrow$ $V \cong V^*$. $\blacksquare$
Coordinate Function
Definition 3. Let $\beta$ be an ordered basis for $V$. Define $\mathsf{f}_i(x) = a_i (i = 1, ..., n),$ where $$[x]_{\beta} = \begin{pmatrix}
a_1 \\
\vdots \\
a_n \end{pmatrix}.$$ Then $\mathsf{f}_i$ is a linear functional on $V$ called the $i$th coordinate function with respect to the basis $\beta$.
$\beta = \{x_1, ..., x_n\}$라 하자. 이때 자명하게 모든 $i, j$에 대해서 $\mathsf{f}_i(x_j)$ = $\delta_{ij}$이 성립한다.
Theorem 2
Theorem 2. Let $\beta = \{x_1, ..., x_n\}$ be an ordered basis for $V$, and let $\beta^* = \{\mathsf{f}_1, ..., \mathsf{f}_n\}$, where $\mathsf{f}_i$ is the $i$th coordinate function with respect to $\beta$.
Then $\beta^*$ is an ordered basis for $V^*$, and, $\forall \mathsf{f} \in V^*$, we have $$\mathsf{f} = \sum_{i=1}^n \mathsf{f}(x_i)\mathsf{f}_i.$$
Proof. Since dim($V$) = dim($V^*$), we only need to show that $\beta^*$ is linearly independent by the replacement theorem.
Suppose that $\sum_{i=1}^n a_i\mathsf{f}_i = \mathsf{f}_0$ where $\mathsf{f}_0(x) = 0, \forall x \in V$, for $a_i \in F (i = 1, ..., n)$. Then $\forall x_j \in \beta$, $$(\sum_{i=1}^n a_i\mathsf{f}_i)(x_j) = \sum_{i=1}^n a_i\mathsf{f}_i(x_j) = \sum_{i=1}^n a_i\delta_{ij} = a_j = 0 (j = 1, ..., n).$$ Thus $\beta^*$ is linearly independent, so $\beta^*$ is a basis for $V^*$.
Fix $\mathsf{f} \in V^*$. Define $g = \sum_{i=1}^n \mathsf{f}(x_i)\mathsf{f}_i$. Then $\forall x_j \in \beta$, we have $$\mathsf{g}(x_j) = (\sum_{i=1}^n \mathsf{f}(x_i)\mathsf{f}_i)(x_j) = \sum_{i=1}^n \mathsf{f}(x_i)\mathsf{f}_i(x_j) = \sum_{i=1}^n \mathsf{f}(x_i)\delta_{ij} = \mathsf{f}(x_j).$$ Since $\mathsf{f}(x_j) = \mathsf{g}(x_j) (j = 1, ..., n)$, $\mathsf{f} = \mathsf{g}$. $\blacksquare$
Dual Basis
Definition 4. Let $\beta = \{x_1, ..., x_n\}$ be an ordered basis for $V$. We call the ordered basis $\beta^* = \{\mathsf{f}_1, ..., \mathsf{f}_n\}$ of $V^*$ that satisfies $\mathsf{f}_i(x_j) = \delta_{ij} (1 \leq i, j \leq n)$ the dual basis of $\beta$.
Theorem 3 (Transpose of a Linear Transformation)
Theorem 3. Let $\beta, \gamma$ be ordered bases for $V, W$, respectively. Then $\forall T \in \mathcal{L}(V, W)$, the mapping $T^t: W^* \longrightarrow V^*$ defined by $T^t(\mathsf{g}) = \mathsf{g}T, \forall \mathsf{g} \in W^*$ is a linear transformation with the property that $[T^t]_{\gamma^*}^{\beta^*} = ([T]_{\beta}^{\gamma})^t$.
Proof. For $c \in F, \mathsf{f}, \mathsf{g} \in W^*$, $T^t(c\mathsf{f} + \mathsf{g}) = (c\mathsf{f} + \mathsf{g})T = c\mathsf{f}T + \mathsf{g}T = cT^t(\mathsf{f}) + T^t(\mathsf{g})$. Thus $T^t$ is linear.
Denote $\beta = \{x_1, ..., x_n\}, \gamma = \{y_1, ..., y_m\}$ with dual bases $\beta^* = \{\mathsf{f}_1, ..., \mathsf{f}_n\}, \gamma^* = \{\mathsf{g}_1, ..., \mathsf{g}_m\}$, respectively.
Let $A := [T]_{\beta}^{\gamma}$. Then for each $j = 1, ..., m$, we have $$T^t(\mathsf{g}_j) = \mathsf{g}_jT = \sum_{i=1}^n \mathsf{g}_jT(x_i)\mathsf{f}_i = \sum_{i=1}^n \mathsf{g}_j(\sum_{k=1}^m A_{ki}y_k)\mathsf{f}_i \\ = \sum_{i=1}^n\sum_{k=1}^m A_{ki}\mathsf{g}_j(y_k)\mathsf{f}_i = \sum_{i=1}^n\sum_{k=1}^m A_{ki}\delta_{jk}\mathsf{f}_i = \sum_{i=1}^n A_{ji}\mathsf{f}_i.$$ Thus $[T^t]_{\gamma^*}^{\beta^*} = A^t = ([T]_{\beta}^{\gamma})^t$. $\blacksquare$
위에서 정의한 $T^t$를 transpose of $T$라고 부르고, $T^t$가 unique함은 자명하다.
Definition 5
Definition 5. For $x \in V$, we define $\widehat{x}: V^* \longrightarrow F$ by $\widehat{x}(\mathsf{f}) = \mathsf{f}(x), \forall \mathsf{f} \in V^*$.
Theorem 4
Theorem 4. Define $\psi: V \longrightarrow V^{**}$ by $\psi(x) = \widehat{x}$. Then $\psi$ is an isomorpshim.
Proof.
(1) $\psi$ is linear: Let $x, y \in V, c \in F$. $\forall \mathsf{f} \in V^*$, we have $$\psi(cx + y)(\mathsf{f}) = \mathsf{f}(cx + y) = c\mathsf{f}(x) + \mathsf{f}(y) = c\widehat{x}(\mathsf{f}) + \widehat{y}(\mathsf{f}) = (c\widehat{x} + \widehat{y})(\mathsf{f}) \\ \Longrightarrow \psi(cx+y) = c\widehat{x} + \widehat{y} = c\psi(x) + \psi(y).$$
(2) $\psi$ is an isomorphism: Suppose that $\psi(x) = \psi(y)$ for $x, y \in V$. We denote $[x]_{\beta} = \begin{pmatrix} a_1 \\ \vdots \\ a_n \end{pmatrix}$ and $[y]_{\beta} = \begin{pmatrix} b_1 \\ \vdots \\ b_n \end{pmatrix}$. If $\beta^* = \{\mathsf{f}_1, ..., \mathsf{f}_n\}$, then we have $$\widehat{x} = \widehat{y} \\ \Longrightarrow \widehat{x}(\mathsf{f}_i) = \widehat{y}(\mathsf{f}_i) (i = 1, ..., n) \\ = \mathsf{f}_i(x) = \mathsf{f}_i(y) \\ = a_i = b_i (i = 1, ..., n) \\ \Longrightarrow x = y.$$ Thus $\psi$ is injective. Since dim($V$) = dim($V^{**}$), $\psi$ is an isomorphism. $\blacksquare$
Corollary. Every ordered basis for $V^*$ is the dual basis for some basis for $V$.
Proof. Let $\beta^* = \{\mathsf{f}_1, ..., \mathsf{f}_n\}$ be an ordered basis for $V^*$. If $\{x_1, ..., x_n\}$ is an ordered basis for $V$, then $\{\widehat{x_1}, ..., \widehat{x_n}\}$ is an ordered basis for $V^{**}$ by Theorem 4. Note that $\widehat{x_i}(\mathsf{f}_j) = \mathsf{f}_j(x_i) = \delta_{ij}$ for all $i, j$. Thus $\{\mathsf{f}_1, ..., \mathsf{f}_n\}$ is the dual basis of $\{x_1, ..., x_n\}$. $\blacksquare$
