# Parseval's Identity

• Nov 10th 2011, 07:08 PM
AlexP
Parseval's Identity
I'm trying to prove Parseval's Identity.

"Let $\{v_1, ..., v_n\}$ be an orthonormal basis for $V$. For any $x, y \in V$ prove that $\langle x,y \rangle = \displaystyle\sum^n_{i=1} \langle x,v_i \rangle \overline{\langle y,v_i \rangle}$."

I'm really not sure how to go about this. I've played around with the summand, including taking $x$ as a linear combination of the $v_i$ and breaking it down, but that actually just gets us right back to $\langle x,v_i \rangle$.

I'd like some hints, but only hints. Thanks.
• Nov 10th 2011, 10:11 PM
Drexel28
Re: Parseval's Identity
Quote:

Originally Posted by AlexP
I'm trying to prove Parseval's Identity.

"Let $\{v_1, ..., v_n\}$ be an orthonormal basis for $V$. For any $x, y \in V$ prove that $\langle x,y \rangle = \displaystyle\sum^n_{i=1} \langle x,v_i \rangle \overline{\langle y,v_i \rangle}$."

I'm really not sure how to go about this. I've played around with the summand, including taking $x$ as a linear combination of the $v_i$ and breaking it down, but that actually just gets us right back to $\langle x,v_i \rangle$.

I'd like some hints, but only hints. Thanks.

Just do what's natural. By definition you have that $x=\langle x,v_1\rangle v_1+\cdots+\langle x,v_n\rangle v_n$ and $y=\langle y,v_1\rangle y_1+\cdots+\langle y,v_n\rangle v_n$ and so

$\displaystyle \langle x,y\rangle=\left\langle \sum_{i=1}^{n}\langle x,v_i\rangle v_i,\sum_{j=1}^{n}\langle y,v_i\rangle v_i\right\rangle$

so what if we expand by sequilinearity?
• Nov 11th 2011, 08:12 PM
AlexP
Re: Parseval's Identity
ok, I got it. I was trying to get the RHS into a useful form, rather than the LHS...not entirely sure why. But now in the future I'll remember to try both sides. Lesson learned. Thanks.