Failing at projection question

Click For Summary
SUMMARY

The discussion centers on proving that the function pf(x) = sum(from i=1 to k) ui is a projection in a vector space V with an orthogonal basis {u1, ..., un}. The key to demonstrating this is to verify that pf satisfies the definition of a projection, which requires showing that pf(pf(x)) = pf(x) for all x in V. Participants emphasize the importance of understanding the properties of orthogonal projections in vector spaces.

PREREQUISITES
  • Understanding of vector spaces and their properties
  • Familiarity with orthogonal bases in linear algebra
  • Knowledge of projection operators in mathematics
  • Basic concepts of inner product spaces
NEXT STEPS
  • Study the definition and properties of projection operators in linear algebra
  • Learn about orthogonal projections and their applications in vector spaces
  • Explore the concept of inner products and their role in defining projections
  • Review examples of proving functions as projections using specific vector spaces
USEFUL FOR

Students and professionals in mathematics, particularly those studying linear algebra, as well as educators teaching concepts related to vector spaces and projections.

reha
Messages
7
Reaction score
0
let pf(x)= sum( from i=1 to k) <x, ui>ui, show pf is a projection.

Ive tried to show this fact myself but i failed. Please some one help me out. thanks

Note ui = u1...un an orthogonal basis of V where V is a vector space.
 
Physics news on Phys.org


Welcome to PF, reha.

The usual procedure in showing that a given function is a projection, is to verify that is satisfies the definition of a projection. (By the way, that's true for many other things as well.)

So, what's the definition of a projection in a vector space?
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 23 ·
Replies
23
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 3 ·
Replies
3
Views
4K
Replies
12
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 7 ·
Replies
7
Views
3K