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How does one go about proving Parseval's identity?

Let ${v_1, v_2, ..., v_n}$ be an orthonormal basis for a a finite-dimensional inner product space $V$ over some field $F$. For any $x, y$ in $V$, prove: $\langle x, y \rangle$ $=$ $\sum\limits_{i = 1}^{n} \langle x, v_i \rangle \overline {\langle y, v_i \rangle}$.

Attempt: using $ x = \sum\limits_{i = 1}^{n} \langle x, v_i \rangle v_i$ and $ y = \sum\limits_{i = 1}^{n} \langle y, v_i \rangle v_i$

But I don't really know how to proceed from here.

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    $\begingroup$ If you don't know how to start you might want to calculate it for $n = 1, 2, 3$ and see where this leads you. $\endgroup$ – user79202 May 25 '13 at 20:32
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You were actually pretty close. With $$x=\sum_i\langle x,v_i\rangle v_i$$ and $$y=\sum_j\langle y,v_j\rangle v_j$$ you get

$$\langle x,y\rangle=\sum_i\langle x,v_i\rangle\sum_j \overline{\langle y,v_j\rangle} \langle v_i,v_j\rangle = \sum_i\langle x,v_i\rangle \overline{\langle y,v_i\rangle}$$

since $\langle v_i,v_j\rangle=\delta_{ij}$.

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