Proving Parseval's Theorem

**kaelbu** · May 26th 2010, 10:46 PM

Problem: A) Let $\{v_1, v_2, ..., v_3\}$ be an orthonormal basis for V. For any $x,y \epsilon V$ prove that
$<x,y>= \sum\limits_{i=1}^n <x,v_i> \overline{<y,v_i>}$
B) use (A) to prove that if $\beta$ is an orthonormal basis for V with inner product $<*, *>$ then for any $x,y\epsilon V$ $<\phi_\beta (x),\phi_\beta (y)>' = <[x]_\beta.[y]_\beta>'=<x,y>$ 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:
$<x,y>= \sum\limits_{i=1}^n <x,v_i> \overline{<y,v_i>} = \sum\limits_{i=1}^n <x,v_i> <v_i,y>$
$= \sum\limits_{i=1}^n\sum\limits_{j=1}^n x_j y_{ij} v_{ij} y_j = \sum\limits_{i=1}^n\sum\limits_{j=1}^n x_j y_j =<x,y>$
And I haven't the slightest idea what (B) means, let alone how to prove it using A which I almost certainly did wrong.

**tonio** · May 27th 2010, 03:59 AM

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
$<x,y>= \sum\limits_{i=1}^n <x,v_i> \overline{<y,v_i>}$
B) use (A) to prove that if $\beta$ is an orthonormal basis for V with inner product $<*, *>$ then for any $x,y\epsilon V$ $<\phi_\beta (x),\phi_\beta (y)>' = <[x]_\beta.[y]_\beta>'=<x,y>$ 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:
$<x,y>= \sum\limits_{i=1}^n <x,v_i> \overline{<y,v_i>} = \sum\limits_{i=1}^n <x,v_i> <v_i,y>$
$= \sum\limits_{i=1}^n\sum\limits_{j=1}^n x_j y_{ij} v_{ij} y_j = \sum\limits_{i=1}^n\sum\limits_{j=1}^n x_j y_j =<x,y>$
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:

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

OTOH, $\left<x,v_i\right>\overline{\left<y,v_i\right>}=\s um^n_{k=1}a_k\left<v_k,v_i\right>\overline{\sum^n_ {k=1}b_k\left<v_k,v_i\right>}$ $=a_i\overline{b_i}$ , 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

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