每週問題 February 23, 2015

利用內積空間的基底表達兩個向量的內積。

If \{\mathbf{u}_1,\ldots,\mathbf{u}_n\} is an orthonormal basis for an inner-product space \mathcal{V}, show that

\displaystyle  \left\langle\mathbf{x},\mathbf{y}\right\rangle=\sum_{i=1}^n\left\langle\mathbf{x},\mathbf{u}_i\right\rangle\left\langle\mathbf{u}_i,\mathbf{y}\right\rangle

for every \mathbf{x},\mathbf{y}\in\mathcal{V}.

 
參考解答:

因為 \{\mathbf{u}_1,\ldots,\mathbf{u}_n\} 是內積空間 \mathcal{V} 的一組單範正交基底 (orthonormal basis),\mathbf{y}\in\mathcal{V} 可唯一表示成 \mathbf{y}=c_1\mathbf{u}_1+\cdots+c_n\mathbf{u}_n。利用內積性質,可得

\displaystyle\begin{aligned}  \left\langle\mathbf{u}_j,\mathbf{y}\right\rangle&=\left\langle\mathbf{u}_j,\sum_{i=1}^nc_i\mathbf{u}_i\right\rangle\\  &=\sum_{i=1}^n\left\langle\mathbf{u}_j,c_i\mathbf{u}_i\right\rangle\\  &=\sum_{i=1}^nc_i\left\langle\mathbf{u}_j,\mathbf{u}_i\right\rangle=c_j,  \end{aligned}

\mathbf{y}=\sum_{i=1}^n\left\langle\mathbf{u}_i,\mathbf{y}\right\rangle\mathbf{u}_i,稱為傅立葉展開式。類似前述推導步驟,

\displaystyle\begin{aligned}  \left\langle\mathbf{x},\mathbf{y}\right\rangle&=\left\langle\mathbf{x},\sum_{i=1}^n\left\langle\mathbf{u}_i,\mathbf{y}\right\rangle\mathbf{u}_i\right\rangle\\  &=\sum_{i=1}^n\left\langle\mathbf{x},\left\langle\mathbf{u}_i,\mathbf{y}\right\rangle \mathbf{u}_i\right\rangle\\  &=\sum_{i=1}^n\left\langle\mathbf{u}_i,\mathbf{y}\right\rangle\left\langle\mathbf{x},\mathbf{u}_i\right\rangle.  \end{aligned}

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