MHB Proof of Inner Product in E with Orthonormal Sequence (n=positive integer)

Poirot1
Messages
243
Reaction score
0
let E be an inner product space and (e_n) an orthonormal sequence in E. For x in E and any positive integer n, prove that

Re(<x,(<x,e_1>+...<x,e_k>)e_n>)= |<x,e_1>|^2+...+|<x,e_n>|^2

I got <x,(<x,e_1>+...<x,e_k>)e_n>= <<x,e_1>e_1,x>+...<<x,e_n>e_n,x>

but haven't a clue how to find the real part of this. Sorry for the ugly subscript notation.
 
Physics news on Phys.org
Re: inner product proof

Poirot said:
let E be an inner product space and (e_n) an orthonormal sequence in E. For x in E and any positive integer n, prove that

Re(<x,(<x,e_1>+...<x,e_k>)e_n>)= |<x,e_1>|^2+...+|<x,e_n>|^2

I got <x,(<x,e_1>+...<x,e_k>)e_n>= <<x,e_1>e_1,x>+...<<x,e_n>e_n,x>

but haven't a clue how to find the real part of this. Sorry for the ugly subscript notation.
This looks wrong to me. Why is there a $k$ on the left side but not on the right? I think that the result should be $$\bigl\langle x,\langle x,x_1\rangle e_1 + \ldots + \langle x,x_n\rangle e_n\bigr\rangle = |\langle x,e_1\rangle|^2 + \ldots + |\langle x,e_n\rangle|^2.$$
 
Thread 'How to define a vector field?'
Hello! In one book I saw that function ##V## of 3 variables ##V_x, V_y, V_z## (vector field in 3D) can be decomposed in a Taylor series without higher-order terms (partial derivative of second power and higher) at point ##(0,0,0)## such way: I think so: higher-order terms can be neglected because partial derivative of second power and higher are equal to 0. Is this true? And how to define vector field correctly for this case? (In the book I found nothing and my attempt was wrong...

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
Replies
2
Views
2K
  • · Replies 23 ·
Replies
23
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 20 ·
Replies
20
Views
4K