# explain below part of the proof of Parseval's relation

explain $\langle \sum_{j=1}^n \langle v,u_j \rangle u_j, \sum_{k=1}^n \langle w,u_k \rangle u_k \rangle =\sum_{j=1}^n \langle v,u_j \rangle \sum_{k=1}^n \langle u_j, \langle w, u_k \rangle u_k \rangle$, V is complex inner product space with orthonormal basis $R={u_1, u_2,..., u_n}$

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Explain what? How you get from the left hand side to the right hand side? –  Arturo Magidin Nov 2 '10 at 14:28
As in your question math.stackexchange.com/questions/8632/… I suggest you enter the proper tag. –  AD. Nov 2 '10 at 20:05
-1: This is a bit of a one-way communication is it not? Also, you don't read our comments or accept the answers, do you? –  AD. Nov 2 '10 at 20:14

The inner product is sesquilinear, and each $\langle v,u_j\rangle$ and $\langle w,u_k\rangle$ are scalars. Rewrite $\alpha_j = \langle v,u_j\rangle$ and $\beta_k = \langle w,u_k\rangle$. Then the left hand side is $$\left\langle \sum_{j=1}^n\langle v,u_j\rangle u_j, \sum_{k=1}^{n}\langle w,u_k\rangle u_k\right\rangle = \left\langle \sum_{j=1}^n \alpha_ju_j,\sum_{k=1}^n\beta_k u_k\right\rangle.$$ So by additivity and homogeneity in the first component you have: \begin{eqnarray*} \left\langle \sum_{j=1}^n\langle v,u_j\rangle u_j, \sum_{k=1}^n\langle w,u_k\rangle u_k\right\rangle & = & \left\langle \sum_{j=1}^n \alpha_ju_j,\sum_{k=1}^n\beta_k u_k\right\rangle\\ & = & \sum_{j=1}^{n}\left\langle \alpha_j u_j, \sum_{k=1}^n\beta_k u_k\right\rangle\\ & = & \sum_{j=1}^n\left( \alpha_j\left\langle u_j, \sum_{k=1}^n\beta_k u_k\right\rangle\right). \end{eqnarray*} Then applying linearity on the second component you get the given expression.
Maybe it is even easier to see what is going on if you set $v_n= \langle v,u_j\rangle$ and $w_j=\langle w,u_j\rangle$? –  AD. Nov 2 '10 at 20:09