## 線性泛函與伴隨

$f(x_1,x_2,x_3)=2x_1-3x_2+x_3$

$\left\langle f,g\right\rangle\overset{\underset{\mathrm{def}}{}}{=}\displaystyle\int_0^1f(t)g(t)dt$

$L(p)=\displaystyle\int_0^{1}p(t)(\cos\pi t)dt$

$L(p)=\displaystyle\int_0^1a(t)p(t)dt$

$f$ 是一定義於有限維內積空間 $\mathcal{V}$ 的線性泛函，則存在唯一向量 $\mathbf{a}\in\mathcal{V}$ 使得

$f(\mathbf{x})=\left\langle\mathbf{a},\mathbf{x}\right\rangle$

\begin{aligned} \left\langle\mathbf{x},\mathbf{y}+\mathbf{z}\right\rangle&=\left\langle\mathbf{x},\mathbf{y}\right\rangle+\left\langle\mathbf{x},\mathbf{z}\right\rangle\\ \left\langle\mathbf{x}+\mathbf{z},\mathbf{y}\right\rangle&=\left\langle\mathbf{x},\mathbf{y}\right\rangle+\left\langle\mathbf{z},\mathbf{y}\right\rangle\\ \left\langle\mathbf{x},c\mathbf{y}\right\rangle&=c\left\langle\mathbf{x},\mathbf{y}\right\rangle\\ \left\langle c\mathbf{x},\mathbf{y}\right\rangle&=\overline{c}\left\langle\mathbf{x},\mathbf{y}\right\rangle\end{aligned}

$\mathbf{x}=c_1\mathbf{q}_1+\cdots+c_n\mathbf{q}_n$

$\mathbf{x}=\left\langle\mathbf{q}_1,\mathbf{x}\right\rangle\mathbf{q}_1+\cdots+\left\langle\mathbf{q}_n,\mathbf{x}\right\rangle\mathbf{q}_n$

\begin{aligned} f(\mathbf{x})&=f\left(\left\langle\mathbf{q}_1,\mathbf{x}\right\rangle\mathbf{q}_1+\cdots+\left\langle\mathbf{q}_n,\mathbf{x}\right\rangle\mathbf{q}_n \right)\\ &=\left\langle\mathbf{q}_1,\mathbf{x}\right\rangle f(\mathbf{q}_1)+\cdots+\left\langle\mathbf{q}_n,\mathbf{x}\right\rangle f(\mathbf{q}_n)\\ &=\left\langle\overline{f(\mathbf{q}_1)}\mathbf{q}_1,\mathbf{x}\right\rangle +\cdots+\left\langle\overline{f(\mathbf{q}_n)}\mathbf{q}_n,\mathbf{x}\right\rangle\\ &=\left\langle\overline{f(\mathbf{q}_1)}\mathbf{q}_1+\cdots+\overline{f(\mathbf{q}_n)}\mathbf{q}_n,\mathbf{x}\right\rangle\end{aligned}

$\mathbf{a}=\overline{f(\mathbf{q}_1)}\mathbf{q}_1+\cdots+\overline{f(\mathbf{q}_n)}\mathbf{q}_n$

(1)

\begin{aligned} u_1(t)&=p_1(t)=1\end{aligned}

(2)

\begin{aligned} u_2(t)&=p_2(t)-\displaystyle\frac{\left\langle u_1,p_2\right\rangle}{\left\langle u_1,u_1\right\rangle}u_1(t)\\ &=t-\frac{1/2}{1}1=t-\frac{1}{2}\end{aligned}

(3)

\begin{aligned} u_3(t)&=p_3(t)-\displaystyle\frac{\left\langle u_1,p_3\right\rangle}{\left\langle u_1,u_1\right\rangle}u_1(t) -\frac{\left\langle u_2,p_3\right\rangle}{\left\langle u_2,u_2\right\rangle}u_2(t)\\ &=t^2-\frac{1/3}{1}1-\frac{1/12}{1/12}\left(t-\frac{1}{2}\right)\\ &=t^2-t+\frac{1}{6}\end{aligned}

\begin{aligned} q_1(t)&=\displaystyle\frac{u_1(t)}{\Vert u_1\Vert}=1\\ q_2(t)&=\frac{u_2(t)}{\Vert u_2\Vert}=\sqrt{12}\left(t-\frac{1}{2}\right)\\ q_3(t)&=\frac{u_3(t)}{\Vert u_3\Vert}=\sqrt{180}\left(t^2-t+\frac{1}{6}\right)\end{aligned}

$a(t)=\overline{L(q_1)}q_1(t)+\overline{L(q_2)}q_2(t)+\overline{L(q_3)}q_3(t)$

\begin{aligned} L(q_1)&=\displaystyle\int_0^11(\cos\pi t)dt=0\\ L(q_2)&=\int_0^1\sqrt{12}\left(t-\frac{1}{2}\right)(\cos\pi t)dt=-\frac{2\sqrt{12}}{\pi}\\ L(q_3)&=\int_0^1\sqrt{180}\left(t^2-t+\frac{1}{6}\right)(\cos\pi t)dt=2\sqrt{180}\left(\frac{1}{\pi}-\frac{1}{\pi^2}\right)\end{aligned}

\begin{aligned} a(t)&=\displaystyle-\frac{24}{\pi}\left(t-\frac{1}{2}\right)+360\left(\frac{1}{\pi}-\frac{1}{\pi^2}\right)\left(t^2-t+\frac{1}{6}\right)\\ &=360\left(\frac{1}{\pi}-\frac{1}{\pi^2}\right)t^2+\left(\frac{360}{\pi^2}-\frac{372}{\pi}\right)t+\frac{66}{\pi}-\frac{60}{\pi^2}\end{aligned}

$L(\mathcal{V},\mathcal{W})$ 代表所有從向量空間 $\mathcal{V}$ 映至 $\mathcal{W}$ 的線性變換組成的集合 (見“線性變換集合構成向量空間”)。若 $T\in L(\mathcal{V},\mathcal{W})$，線性泛函的唯一內積表達定理最主要的用處在於架構線性變換 $T$ 的自然且唯一的反向變換 $T^{\ast}:\mathcal{W}\to\mathcal{V}$，稱為 $T$ 的伴隨 (adjoint)。選定 $\mathbf{w}\in\mathcal{W}$，考慮下面這個定義於 $\mathcal{V}$ 的線性泛函：對於 $\mathbf{v}\in\mathcal{V}$

$f(\mathbf{v})=\left\langle \mathbf{w},T\mathbf{v}\right\rangle$

$\left\langle\mathbf{w},T\mathbf{v}\right\rangle=\left\langle T^{\ast}\mathbf{w},\mathbf{v}\right\rangle$

$T(x_1,x_2,x_3)=(x_1-x_2+3x_3,2x_1)$

\begin{aligned} \left\langle T^{\ast}(y_1,y_2),(x_1,x_2,x_3)\right\rangle&=\left\langle(y_1,y_2),T(x_1,x_2,x_3)\right\rangle\\ &=\left\langle(y_1,y_2),(x_1-x_2+3x_3,2x_1)\right\rangle\\ &=y_1x_1-y_1x_2+3y_1x_3+2y_2x_1\\ &=\left\langle(y_1+2y_2,-y_1,3y_1),(x_1,x_2,x_3)\right\rangle\end{aligned}

$T^{\ast}(y_1,y_2)=(y_1+2y_2,-y_1,3y_1)$

$A=\left[\!\!\begin{array}{crc} 1&-1&3\\ 2&0&0 \end{array}\!\!\right],~A^T=\left[\!\!\begin{array}{rc} 1&2\\ -1&0\\ 3&0 \end{array}\!\!\right]$

\begin{aligned} \left\langle T^{\ast}(\mathbf{w}_1+\mathbf{w}_2),\mathbf{v}\right\rangle&=\left\langle\mathbf{w}_1+\mathbf{w}_2,T\mathbf{v}\right\rangle\\ &=\left\langle\mathbf{w}_1,T\mathbf{v}\right\rangle+\left\langle\mathbf{w}_2,T\mathbf{v}\right\rangle\\ &=\left\langle T^{\ast}\mathbf{w}_1,\mathbf{v}\right\rangle+\left\langle T^{\ast}\mathbf{w}_2,\mathbf{v}\right\rangle\\ &=\left\langle T^{\ast}\mathbf{w}_1+T^{\ast}\mathbf{w}_2,\mathbf{v}\right\rangle\end{aligned}

$T^{\ast}(\mathbf{w}_1+\mathbf{w}_2)=T^{\ast}\mathbf{w}_1+T^{\ast}\mathbf{w}_2$

\begin{aligned} \left\langle T^{\ast}(c\mathbf{w}),\mathbf{v}\right\rangle&=\left\langle c\mathbf{w},T\mathbf{v}\right\rangle=\overline{c}\left\langle\mathbf{w},T\mathbf{v}\right\rangle=\overline{c}\left\langle T^{\ast}\mathbf{w},\mathbf{v}\right\rangle=\left\langle cT^{\ast}\mathbf{w},\mathbf{v}\right\rangle\end{aligned}

$T^{\ast}(c\mathbf{w})=cT^{\ast}\mathbf{w}$

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

\begin{aligned} \left\langle\mathbf{y},T\mathbf{x}\right\rangle&=\left\langle T^{\ast}\mathbf{y},\mathbf{x}\right\rangle=\overline{\left\langle\mathbf{x},T^{\ast}\mathbf{y}\right\rangle}=\overline{\left\langle (T^{\ast})^{\ast}\mathbf{x},\mathbf{y}\right\rangle}=\left\langle\mathbf{y},(T^{\ast})^{\ast}\mathbf{x}\right\rangle\end{aligned}

$T\in L(\mathcal{V},\mathcal{W})$$S\in L(\mathcal{W},\mathcal{U})$，複合變換 $ST$ 的伴隨為

$(ST)^{\ast}=T^{\ast}S^{\ast}$

$ST$ 連續使用兩次定義，即得

\begin{aligned} \left\langle (ST)^{\ast}\mathbf{y},\mathbf{x}\right\rangle&=\left\langle\mathbf{y},ST\mathbf{x}\right\rangle=\left\langle S^{\ast}\mathbf{y},T\mathbf{x}\right\rangle=\left\langle T^{\ast}S^{\ast}\mathbf{y},\mathbf{x}\right\rangle\end{aligned}

\begin{aligned} (S+T)^{\ast}&=S^{\ast}+T^{\ast}\\ (cT)^{\ast}&=\overline{c}T^{\ast}\end{aligned}

$T:\mathcal{V}\to\mathcal{W}$ 是一線性變換，則 $T^{\ast}:\mathcal{W}\to\mathcal{V}$ 滿足 $\left\langle\mathbf{w},T\mathbf{v}\right\rangle=\left\langle T^{\ast}\mathbf{w},\mathbf{v}\right\rangle$，此式的意思是向量空間 $\mathcal{W}$$\mathbf{w}$$\mathbf{v}$$T$ 映射而得的像（$T\mathbf{v}$）的內積等於向量空間 $\mathcal{V}$$\mathbf{w}$$T^{\ast}$ 映射而得的像 ($T^{\ast}\mathbf{w}$) 和 $\mathbf{v}$ 的內積，見下圖：

\begin{aligned} N(T^{\ast})&=R(T)^{\perp}\\ N(T)&=R(T^{\ast})^{\perp}\end{aligned}

\begin{aligned} A&=[T(\boldsymbol{\beta})]_{\boldsymbol{\gamma}}\\ A^{\ast}&=[T^{\ast}(\boldsymbol{\gamma})]_{\boldsymbol{\beta}}\end{aligned}

$T\mathbf{v}_j=\left\langle\mathbf{w}_1,T\mathbf{v}_j\right\rangle\mathbf{w}_1+\cdots+\left\langle\mathbf{w}_m,T\mathbf{v}_j\right\rangle\mathbf{w}_m$

$a_{ij}=\left\langle \mathbf{w}_i,T\mathbf{v}_j\right\rangle$

$B=[b_{ij}]$ 代表 $T^{\ast}$ 參考基底 $\boldsymbol{\gamma}$$\boldsymbol{\beta}$ 的表示矩陣，將 $T$ 替換為 $T^{\ast}$，再交換基底 $\boldsymbol{\beta}$$\boldsymbol{\gamma}$ 的角色，可得

\begin{aligned} b_{ij}&=\left\langle\mathbf{v}_i, T^{\ast}\mathbf{w}_j\right\rangle=\left\langle T\mathbf{v}_i,\mathbf{w}_j\right\rangle=\overline{\left\langle\mathbf{w}_j,T\mathbf{v}_i\right\rangle}=\overline{a_{ji}}\end{aligned}

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