B Question about orthogonal vectors and the cosine

[Peter_Newman]

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?

 

[Peter_Newman]

Unfortunately no opinions about it yet? Are you perhaps missing information, or do you agree with me (and my comments)?

 

[PeroK]

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.

 

[Peter_Newman]

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.

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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 $R^n$.

 

[Office_Shredder]

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.

 

[Peter_Newman]

Hey @Office_Shredder , thanks again for your help! I see it absolutely the same!