# Proving Parseval's Theorem

• May 26th 2010, 11:46 PM
kaelbu
Proving Parseval's Theorem
Problem: A) Let $\{v_1, v_2, ..., v_3\}$ be an orthonormal basis for V. For any $x,y \epsilon V$ prove that
$= \sum\limits_{i=1}^n \overline{}$
B) use (A) to prove that if $\beta$ is an orthonormal basis for V with inner product $<*, *>$ then for any $x,y\epsilon V$ $<\phi_\beta (x),\phi_\beta (y)>' = <[x]_\beta.[y]_\beta>'=$ where $<*, *>$ is the standard product on $F^n$

Thoughts:
I worked out (A) but I am almost certain I did it wrong, however, this is what I did:
$= \sum\limits_{i=1}^n \overline{} = \sum\limits_{i=1}^n $
$= \sum\limits_{i=1}^n\sum\limits_{j=1}^n x_j y_{ij} v_{ij} y_j = \sum\limits_{i=1}^n\sum\limits_{j=1}^n x_j y_j =$
And I haven't the slightest idea what (B) means, let alone how to prove it using A which I almost certainly did wrong.
• May 27th 2010, 04:59 AM
tonio
Quote:

Originally Posted by kaelbu
Problem: A) Let $\{v_1, v_2, ..., v_3\}$ be an orthonormal basis for V. For any $x,y \epsilon V$ prove that
$= \sum\limits_{i=1}^n \overline{}$
B) use (A) to prove that if $\beta$ is an orthonormal basis for V with inner product $<*, *>$ then for any $x,y\epsilon V$ $<\phi_\beta (x),\phi_\beta (y)>' = <[x]_\beta.[y]_\beta>'=$ where $<*, *>$ is the standard product on $F^n$

Thoughts:
I worked out (A) but I am almost certain I did it wrong, however, this is what I did:
$= \sum\limits_{i=1}^n \overline{} = \sum\limits_{i=1}^n $
$= \sum\limits_{i=1}^n\sum\limits_{j=1}^n x_j y_{ij} v_{ij} y_j = \sum\limits_{i=1}^n\sum\limits_{j=1}^n x_j y_j =$
And I haven't the slightest idea what (B) means, let alone how to prove it using A which I almost certainly did wrong.

For (A): write both vectors as lin. comb. of the given orthonormal basis: $x=\sum^n_{i=1}a_iv_i\,,\,\,y=\sum^n_{j=1}b_jv_j\,, \,\,a_i,\,b_j\in\mathbb{C}$ , so:

$\left=\left<\sum^n_{i=1}a_iv_i\,,\,\sum ^n_{j=1}b_jv_j\right>=\sum^n_{i,j=1}a_i\overline{b _j}\left$ $=\sum^n_{i=1}ai\overline{b_i}$ , as $\left=\delta_{i,j}$ .

OTOH, $\left\overline{\left}=\s um^n_{k=1}a_k\left\overline{\sum^n_ {k=1}b_k\left}$ $=a_i\overline{b_i}$ , and summing over i from 1 to n we get the same as above.

Wat you did is wrong because one step before the last you write there the product of $x_iy_{ij}v_{ij}$ and etc...what's this??

About (B) also I don't understand what's going on: what's $\phi_\beta$ , anyway? Though I suspect it is the canonical isomorphism from $V\,\,\,to\,\,\,\mathbb{F}^n$ assigning to every vectors its coordinates wrt the basis $\beta$ ...but then the exercise is pretty straightforward, after you fix the notation's mistakes.

Tonio