Undergrad Finding the orthogonal projection of a vector without an orthogonal basis

Click For Summary
The discussion focuses on proving the orthogonal projection of a vector onto a subspace without requiring an orthogonal basis. The main result states that for a subspace F spanned by vectors e_i, the projection p_F(x) satisfies the condition that the inner product (x-y, e_i) equals zero for all i. The original poster successfully proved this for orthogonal vectors but seeks clarification for the general case. A suggestion is made to use the Gram-Schmidt process to orthogonalize the basis vectors, which preserves the span of the subspace. Overall, the conversation emphasizes understanding the relationship between inner products and orthogonal projections in finite-dimensional spaces.
AimaneSN
Messages
5
Reaction score
1
Hi there,

I am currently reading a course on euclidian spaces and I came across this result that I am struggling to prove :

Let ##F## be a subspace of ##E## (of finite dimension) such that ##F=span(e_1, e_2, ..., e_p)## (not necessarily an orthogonal family of vectors), let ##x \in E##

Then we have:

##\forall y \in F## : ##y=p_F(x) \Leftrightarrow \forall i= 1,...,p : (x-y,e_i) = 0##

where ##(.,.)## denotes an inner product
and the linear map ##p_F## is the orthogonal projection onto ##F##.

I managed to prove the equivalence only when the family of the vectors ##(e_i)## is orthogonal but the result is more general.

Thank you and I would appreciate any hint or help.
 
Last edited:
Physics news on Phys.org
The inner product ##(x-y,e_i)## is zero for all ##i## if and only if ##x-y\in F^\perp## if and only if ##x## is of the form ##x=y+z## where ##z\in F^{\perp}## if and only if ##y## is the orthogonal projection of ##x## onto ##F##.

Your post was a bit of effort to read- it would be better to use latex.
 
Thank you for your reply, it's much clearer now. I just modified my post using Latex code.
 
OP: Notice you can always use Gram-Schmidt to orthogonalize your basis vectors, and this won't affect the span .
 
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 9 ·
Replies
9
Views
3K
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 9 ·
Replies
9
Views
4K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K