Question about orthogonal vectors and the cosine

  • #1
Peter_Newman
155
11
Hi,

The orthogonality defect is ##\prod_i ||b_i|| / det(B)##. Now it is said: The relation between this quantity and almost orthogonal bases is easily explained. Let ##\theta_i## be the angle between ##b_i## and ##span(b_1,...,b_{i-1})##. Then ##||b_i^*|| = ||b_i|| cos(\theta_i)##. [...]

So the cosine is the ratio of the adjacent to the hypotenuse. That means between ##b_i^*## and ##b_i## there is always this ratio, I would accept that. But what irritates me a bit is the statement about the angle.

I have drawn this now for the case ##i=2## and I'm the opinion that what stands above is not completely correct, correctly would be, if it would be called ##\theta_i## is the angle between ##b_i## and ##span(b_1,...,b_{i-1})^{\perp}##.

For notation: ##b_i^*## are Gram Schmidt vectors.

What do you think?
 
Last edited:
Physics news on Phys.org
  • #2
Unfortunately no opinions about it yet? Are you perhaps missing information, or do you agree with me (and my comments)?
 
  • #3
Peter_Newman said:
Unfortunately no opinions about it yet? Are you perhaps missing information, or do you agree with me (and my comments)?
It's not a B level for one thing. And, the post is too far out of context to be easy to help. Also, at this level I'd imagine you could work out a specific example, doing the legwork yourself, rather than sitting back and letting us start from square one. There's also the risk that the "Gram-Schmidt" vectors form your source are non-standard.
 
  • #4
Ok, from your answer I can see that there might be some missing information here. To your suggestion with the sit back and wait for answers: (It sounds a bit accusatory for me) So I have a relatively concrete example tried to provide in the first post, by drawing once the facts and based on it, I would not agree with what was said. However, I have made a guess as to how it could be correct in the penultimate paragraph.

Unfortunately, I also do not have any background information other than what I had previously (quoted):

A quantity that has been used to measure how close a basis is to orthogonal is the orthogonality defect ##\prod_i ||b_i|| / det(B)##. The relation between this quantity and almost orthogonal bases is easily explained. Let ##\theta_i## be the angle between ##b_i## and ##span(b_1,...,b_{i-1})##. Then ##||b_i^*|| = ||b_i|| cos(\theta_i)##.

Perhaps it could and should be noted that the ##b_i^*##'s are not normalized Gram-Schmidt vectors. We are in a lattice context here, but it is not relevant for the quote for now. Otherwise I have no further information.



If we now consider only what the actual statement is, which makes me a bit wonder, then we could reduce this to the following statement:

Let ##\theta_i## be the angle between ##b_i## and ##span(b_1,...,b_{i-1})##. Then ##||b_i^*|| = ||b_i|| cos(\theta_i)##.

Here one needs then actually only, the information that the asterisks are Gram Schmidt vectors, which are not normalized and the ##b_i##'s are vectors of ##R^n##.
 
Last edited:
  • #5
I think you are right that you need to add a little orthogonal symbol. If the angle between ##b_i## and the previous vectors is close to 0 , then ##||b_i^*||## should be close to 0.

You can also fix this by replacing cosine with sine.
 
  • Like
Likes Peter_Newman and PeroK
  • #6
Hey @Office_Shredder , thanks again for your help! I see it absolutely the same!
 

Similar threads

Replies
7
Views
1K
Replies
20
Views
2K
Replies
19
Views
2K
Replies
25
Views
2K
Replies
33
Views
3K
Replies
2
Views
1K
Replies
2
Views
3K
Replies
5
Views
2K
Back
Top