## 每週問題 February 23, 2015

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 $\mathbf{x},\mathbf{y}\in\mathcal{V}$.

\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}