##### 对偶基 - 对偶基 - **对偶基**是[[对偶空间]] $V'$ 中的[[向量空间的基|基]], 与[[向量空间]] $V$ 的基成对存在, 同样也有[[对偶基的变换]], 若 $\mathcal{B} = \{\mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_n\}$, 则有 $\mathcal{B}' = \{\varphi_1, \varphi_2, \dots, \varphi_n\}$. 对偶基是与原基对应的[[线性泛函]]集, 每个对偶基向量满足[[克罗内克函数]] $\varphi_i(\mathbf{v}_j) = \delta_{ij}$, 它们分别作用在原基向量上, 使得每个线性泛函只选择性检测与它对应的原基向量, 对偶基能提取向量的坐标. 把对偶基和原基表示为矩阵, 则乘积为单位矩阵 - $V$ , $\mathcal{B} = \{\mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_n\}$ - $V'$ , $\mathcal{B}' = \{\varphi_1, \varphi_2, \dots, \varphi_n\}$ , $\varphi_i(\mathbf{v}_j) = \left\{\begin{matrix} 1,i=j \\ 0,i\neq j \end{matrix}\right.$ - 对每个 $\mathbf{v} \in V$ 有坐标向量 - $\displaystyle \mathbf{v} = \sum_{j=1}^{n} a_j \mathbf{v}_j$ - $\displaystyle \varphi_i(\mathbf{v}) =\varphi_i(\sum_{j=1}^{n} a_j \mathbf{v}_j)=\sum_{j=1}^{n} a_j \varphi_i(\mathbf{v}_j)=a_i$ - $\displaystyle \mathbf{v} = \sum_{i=1}^{n} \varphi_i(\mathbf{v}) \mathbf{v}_i$ - 把对偶基和原基表示为矩阵 - $B'^TB=I$, $B'=(B^{-1})^T$ - $\begin{bmatrix} \varphi_1^T \\ \varphi_2^T \\ \vdots \\ \varphi_n^T \end{bmatrix} \cdot \begin{bmatrix} \mathbf{v}_1 & \mathbf{v}_2 & \cdots & \mathbf{v}_n \end{bmatrix} = \begin{bmatrix} \langle \varphi_1 , \mathbf{v}_1 \rangle & \langle \varphi_1 , \mathbf{v}_2 \rangle & \cdots & \langle \varphi_1 , \mathbf{v}_n \rangle \\ \langle \varphi_2, \mathbf{v}_1 \rangle & \langle \varphi_2, \mathbf{v}_2 \rangle & \cdots & \langle \varphi_2, \mathbf{v}_n \rangle \\ \vdots & \vdots & \ddots & \vdots \\ \langle \varphi_n, \mathbf{v}_1 \rangle & \langle \varphi_n, \mathbf{v}_2 \rangle & \cdots & \langle \varphi_n, \mathbf{v}_n \rangle \end{bmatrix} = \begin{bmatrix} 1 & 0 & \cdots & 0 \\ 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 1 \end{bmatrix}$ >[!example]- 对偶基 >- $V = \mathbb{R}^2$, $B=\{v_1,v_2\}$, $V'$ 中有对偶基 $B' = \{\varphi_1, \varphi_2\}$ , $\varphi_i(\mathbf{v}_j) = \left\{\begin{matrix} 1,i=j \\ 0,i\neq j \end{matrix}\right.$ > - $v_1=(1,1)$, $v_2=(1,-1)$ > - 设 $\varphi_1((x,y))=a_1x+b_1y$ > - $\varphi_1(v_1)=a_1+b_1=1$ > - $\varphi_1(v_2)=a_1-b_1=0$ > - $\varphi_1((x,y))=0.5x+0.5y$ > - 设 $\varphi_2((x,y))=a_2x+b_2y$ > - $\varphi_2(v_1)=a_2+b_2=0$ > - $\varphi_2(v_2)=a_2-b_2=1$ > - $\varphi_2((x,y))=0.5x-0.5y$ > - 对偶基 $B' = \{\varphi_1, \varphi_2\}$, $\varphi_1=(0.5,0.5)$, $\varphi_2=(0.5,-0.5)$ > - $\varphi_1((x,y))=0.5x+0.5y$ > - $\varphi_2((x,y))=0.5x-0.5y$ > - $B'^TB=\begin{bmatrix} 0.5 & 0.5 \\ 0.5 & -0.5 \end{bmatrix}\begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}=I$ > - $v_1=(1,0)$, $v_2=(0,1)$ > - 对偶基 $B' = \{\varphi_1, \varphi_2\}$, $\varphi_1=(1,0)$, $\varphi_2=(0,1)$ > - $B'^TB=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}=I$