Question about orthogonal vectors and the cosine

In summary, the Orthogonality defect is the ratio of adjacent to the hypotenuse. This statement is easily explained with the help of the angle between b_i and span(b_1,...,b_{i-1})^{\perp}. The angle is measured between the Gram Schmidt vectors and is close to 0 if the vectors are close together.
  • #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?
 
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  • #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##.
 
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  • #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.
 
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Likes Peter_Newman and PeroK
  • #6
Hey @Office_Shredder , thanks again for your help! I see it absolutely the same!
 

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