## 每週問題 October 14, 2013

Let $\mathcal{V}$ be an inner product vector space over $\mathbb{C}$. A linear functional on $\mathcal{V}$ is a linear transformation from $\mathcal{V}$ to $\mathbb{C}$ and the dual space of $\mathcal{V}$, denoted by $\mathcal{V}^\ast$, is the vector space of all linear functionals on $\mathcal{V}$.

(a) For $\mathbf{v}\in\mathcal{V}$, let $L_\mathbf{v}(\mathbf{x})=\left\langle\mathbf{v},\mathbf{x}\right\rangle$, for all $\mathbf{x}\in\mathcal{V}$. Show that $L_{\mathbf{v}}$ is a linear functional for every $\mathbf{v}$.
(b) Let $T$ be the transformation from $\mathcal{V}$ to $\mathcal{V}^\ast$ defined by $T(\mathbf{v})=L_{\mathbf{v}}$, for all $\mathbf{v}\in\mathcal{V}$. Show that $T$ is one-to-one and onto.

(a) 根據定義，對於 $\mathbf{x},\mathbf{y}\in\mathcal{V}$$c\in\mathbb{C}$

$\displaystyle L_{\mathbf{v}}(\mathbf{x}+\mathbf{y})=\left\langle\mathbf{v},\mathbf{x}+\mathbf{y}\right\rangle =\left\langle\mathbf{v},\mathbf{x}\right\rangle+\left\langle\mathbf{v},\mathbf{y}\right\rangle =L_{\mathbf{v}}(\mathbf{x})+L_{\mathbf{v}}(\mathbf{y})$

$\displaystyle L_{\mathbf{v}}(c\mathbf{x})=\left\langle\mathbf{v},c\mathbf{x}\right\rangle =c\left\langle\mathbf{v},\mathbf{x}\right\rangle =cL_{\mathbf{v}}(\mathbf{x})$

(b) 對於 $\mathbf{x},\mathbf{y}\in\mathcal{V}$$c\in\mathbb{C}$

\displaystyle\begin{aligned} T(\mathbf{x}+\mathbf{y})(\mathbf{u})&=L_{\mathbf{x}+\mathbf{y}}(\mathbf{u})=\left\langle\mathbf{x}+\mathbf{y},\mathbf{u}\right\rangle =\left\langle\mathbf{x},\mathbf{u}\right\rangle+\left\langle\mathbf{y},\mathbf{u}\right\rangle\\ &=L_{\mathbf{x}}(\mathbf{u})+L_{\mathbf{y}}(\mathbf{u})=(L_{\mathbf{x}}+L_{\mathbf{y}})(\mathbf{u})=(T(\mathbf{x})+T(\mathbf{y}))(\mathbf{u}),\end{aligned}

\displaystyle\begin{aligned} T(c\mathbf{x})(\mathbf{u})&=L_{c\mathbf{x}}(\mathbf{u})=\left\langle c\mathbf{x},\mathbf{u}\right\rangle =\overline{c}\left\langle\mathbf{x},\mathbf{u}\right\rangle\\ &=\overline{c}L_{\mathbf{x}}(\mathbf{u})=(\overline{c}L_{\mathbf{x}})(\mathbf{u})=(\overline{c}T(\mathbf{x}))(\mathbf{u}).\end{aligned}

$\displaystyle \left\langle\mathbf{x},\mathbf{u}\right\rangle=\left\langle\mathbf{y},\mathbf{u}\right\rangle$

$\mathbf{x}=\mathbf{y}$，證明 $T$ 是一對一。欲證明 $T$ 是滿射，假設 $\{\mathbf{e}_1,\ldots,\mathbf{e}_n\}$$\mathcal{V}$ 的一組單範正交基底 (orthonormal basis)，即 $\left\langle\mathbf{e}_i,\mathbf{e}_j\right\rangle=1$$i=j$$\left\langle\mathbf{e}_i,\mathbf{e}_j\right\rangle=0$$i\neq j$。給定任一 $f\in\mathcal{V}^\ast$，令

$\displaystyle \mathbf{x}=\overline{f(\mathbf{e}_1)}\mathbf{e}_1+\cdots+\overline{f(\mathbf{e}_n)}\mathbf{e}_n$

$\displaystyle \left\langle\mathbf{x},\mathbf{e}_i\right\rangle=\left\langle\sum_{j=1}^n\overline{f(\mathbf{e}_j)}\mathbf{e}_j,\mathbf{e}_i\right\rangle=\sum_{j=1}^n f(\mathbf{e}_j)\left\langle\mathbf{e}_j,\mathbf{e}_i\right\rangle=f(\mathbf{e}_i),~~~i=1,\ldots,n$

$f$ 是一線性泛函，對於 $\mathbf{u}=u_1\mathbf{e}_1+\cdots+u_n\mathbf{e}_n\in\mathcal{V}$，可得

\displaystyle\begin{aligned} (T(\mathbf{x}))(\mathbf{u})&=L_{\mathbf{x}}(\mathbf{u})=\left\langle\mathbf{x},\mathbf{u}\right\rangle=\left\langle\mathbf{x},\sum_{i=1}^nu_i\mathbf{e}_i\right\rangle\\ &=\sum_{i=1}^nu_i\left\langle\mathbf{x},\mathbf{e}_i\right\rangle=\sum_{i=1}^nu_if(\mathbf{e}_i)=f\left(\sum_{i=1}^nu_i\mathbf{e}_i\right)=f(\mathbf{u}),\end{aligned}

$T(\mathbf{x})=f$，證明 $T$ 是滿射。合併以上結果，$\mathbf{x}\rightleftharpoons L_{\mathbf{x}}$，我們稱 $T$ 為共軛同構。

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