Orthonormal Bases: Determining Coefficients for Arbitrary Vector

  • Thread starter mrxtothaz
  • Start date
  • #1
14
0
If we have a vector that can be expressed in terms of some finite list of basis elements. If we have an orthonormal basis for a vector space V, then a vector v can be expressed as <v,e1>e1 +...+ <v,en>en. This appears to be widely used for many results (such as Gram-Schmidt), but the motivation for this is not clear to me. Not only that, I don't understand why this is the case (geometrically).

Obtaining the coefficients for a given vector (in terms of an orthonormal basis) using the inner product of the vector v with each basis element... why does this work?
 

Answers and Replies

  • #2
95
0
If you consider the version of the dot product involving cosine, [itex]v\cdot e_{1}=|v||e_{1}|cos\theta=|v|cos\theta[/itex] because [itex]e_{1}[/itex] is a unit vector. This is just the "[itex]e_1[/itex] component" of v (rather than the x or y component of v).
I included a poorly done illustration in Paint.
 

Attachments

  • #3
534
1
It works because it works on basis vectors. Define the linear operator A by
[tex]Av = \sum_{i=1}^n \langle v, e_i \rangle e_i.[/tex]
Then for any k,
[tex]Ae_k = \sum_{i=1}^n \langle e_k, e_i \rangle e_i
= \sum_{i=1}^n \delta_{ki} e_i = e_k.[/tex]
It follows that A = I, so Av = v for any v.
 

Related Threads on Orthonormal Bases: Determining Coefficients for Arbitrary Vector

  • Last Post
Replies
1
Views
2K
Replies
2
Views
2K
Replies
4
Views
771
Replies
6
Views
1K
Replies
1
Views
1K
  • Last Post
Replies
10
Views
2K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
2
Views
2K
Replies
1
Views
2K
Top