Author Topic: Orthogonal systems: approximation  (Read 13100 times)

Shu Wang

  • Jr. Member
  • **
  • Posts: 11
  • Karma: 0
    • View Profile
Orthogonal systems: approximation
« on: October 20, 2012, 06:29:47 PM »
Hi all,
  First, I have a trivial question on the definition of the {u_n}(finite orthonormal system) that's mentioned at the introduction of this section in lecture note #14.
  Is this set/system defined in H, in which you did not mention whether it's infinite or finite? or does it span its own space(any) based on perhaps infinite/finite number of linear hulls including K, which in particular belongs to H?   :-\

Also  In the following Theorem 1
b) I think this implies w is linear independent of other elements in K, regardless of "unique" in a) or c)
c) do you mean "w in K"(summation of alpha times u_n) rather than equal to it?

Thanks in advance.
« Last Edit: October 20, 2012, 06:44:05 PM by Shu Wang »

Victor Ivrii

  • Administrator
  • Elder Member
  • *****
  • Posts: 2607
  • Karma: 0
    • View Profile
    • Personal website of Victor Ivrii
Re: Orthogonal systems: approximation
« Reply #1 on: October 21, 2012, 03:26:18 AM »
Hi all,
  First, I have a trivial question on the definition of the {u_n}(finite orthonormal system) that's mentioned at the introduction of this section in lecture note #14.
  Is this set/system defined in H, in which you did not mention whether it's infinite or finite? or does it span its own space(any) based on perhaps infinite/finite number of linear hulls including K, which in particular belongs to H?   :-\
Quote

At definition we consider any system (finite or infinite); in fact it could be even not enumerable.

Linear hull is defined as a set of finite sums, in the closed linear hull  converging infinite sums are also included.
Quote
Also  In the following Theorem 1
b) I think this implies w is linear independent of other elements in K, regardless of "unique" in a) or c)

How it can be? $w\in \mathsf{K}$. But $v-w$ is.

Quote
c) do you mean "w in K"(summation of alpha times u_n) rather than equal to it?

$\mathsf{K}$ is a set of elements represented as $\sum \alpha_n u_n$; so $w\in \mathsf{K} \iff w= \sum \alpha_n u_n$ with some coefficients $\alpha_n$.
« Last Edit: October 22, 2012, 05:28:36 AM by Victor Ivrii »

Shu Wang

  • Jr. Member
  • **
  • Posts: 11
  • Karma: 0
    • View Profile
Re: Orthogonal systems: approximation
« Reply #2 on: October 21, 2012, 06:40:55 PM »
Theorem 1
b) You are right I think; we need a) and c) to say w is linearly independent, if we let it to be an element in K. However, if we consider that every 'alpha*u_n' product spans a dimension for K then it implies every element(u_n) in K is independent of one another. Hence if we allow w to be one of them, then it's independent as well. (say w = a_1*u_1--> a_n*u_n = 0 for n != 1)
c) I'm still a little confused. Shouldn't we change one of the alpha_n notation? maybe to w = summation(alpha_m*u_n)? Otherwise the expression for w and K are the same...

Victor Ivrii

  • Administrator
  • Elder Member
  • *****
  • Posts: 2607
  • Karma: 0
    • View Profile
    • Personal website of Victor Ivrii
Re: Orthogonal systems: approximation
« Reply #3 on: October 21, 2012, 11:21:49 PM »
You are confused. Linearly independent are elements of the system not of its span. Look at "Linear Algebra" course what linear independence means.

Quote
Otherwise the expression for w and K are the same

Huh?
« Last Edit: October 22, 2012, 05:32:12 AM by Victor Ivrii »