Graduate Proof of the inequality of a reduced basis

[Peter_Newman]

TL;DR

For $1 \leq i \leq n$ concider $H = span(b_1,...,b_{i-1},b_{i+1},...,b_n)$. Show that $2^{-n(n-1)/4}||b_i|| \leq dist(H,b_i) \leq || b_i ||$. Hint use $\prod ||b_i|| \leq 2^{n(n-1)/4} det(\Lambda)$. Own reasoning below.

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I would like to show that a LLL-reduced basis satisfies the following property (Reference):

For $1 \leq i \leq n$ concider $H = span(b_1,...,b_{i-1},b_{i+1},...,b_n)$. Show that $2^{-n(n-1)/4}||b_i|| \leq dist(H,b_i) \leq || b_i ||$. Hint use $\prod ||b_i|| \leq 2^{n(n-1)/4} det(\Lambda)$.

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My Idea:
I also have a first approach for the part $dist(H,b_i) \leq || b_i ||$ of the inequality, which I want to present here based on a picture, which is used to explain my thought:

So based on the picture we can see that the norm of the orthogonal part ($\tilde{b_i}$) of the vector $b_i$ gives the distance. Therefor I would argue that $dist(H,b_i) = ||\tilde{b_i}||$ and further $||\tilde{b_i}|| \leq ||b_i||$ since the norm of a Gram-Schmidt vector is less than the norm of the regular vector. If I put this together it makes for me a valid argument to say $dist(H,b_i) \leq || b_i ||$. This would prove one part of the inequality.

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Problem:
However I do not manage to show $2^{-n(n-1)/4}||b_i|| \leq dist(H,b_i)$ with the specific hint. There are other useful properties of an reduced basis (see e.g. the reference) with which it is easier to show, but with the hint there I have no idea. This brings me to the question, if someone here might have an idea.

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Remark:
This is not a homework exercise, it's just a question I'm thinking about for myself...

 

[Peter_Newman]

I wanted to ask again if anyone would consider my reasoning to be correct? Or can someone contradict that?

 

[Office_Shredder]

Do you know how $\det(\Lambda)$ is built up by computing $||\tilde{b_i}||$s?

 

[Peter_Newman]

Hey @Office_Shredder, If I remember correctly, it was this: $\det(\Lambda) = \prod_{i=1}^n ||\tilde{b_i}||$

 

[Office_Shredder]

That's really the only clever piece. I'll put the rest in spoilers but it's just doing the right algebra, so you should give it a shot yourself.

Spoiler

plugging this into
$\prod ||| b_j|| \leq 2^{n(n-1)/4} \det(\Lambda)$

$\prod ||b_j|| \leq 2^{n(n-1)/4} \prod ||\tilde{b_j}||$

For each $j\neq i$, $||\tilde{b_j}|| \leq ||b_j||$. Apply this to the right hand side to get

$\prod ||b_j|| \leq 2^{n(n-1)/4} ||\tilde{b_i}|||\prod_{j\neq i} ||b_j||$

Cancel all the norms left on both sides

 

[Peter_Newman]

Hey @Office_Shredder, then in the end remains: $2^{-n(n-1)/4}||b_i|| \leq ||\tilde{b_i}||$.

But with my thought above that $dist(H,b_i) = ||\tilde{b_i}||$, we can derive the required inequality (left side). But this assumes that $dist(H,b_i) = ||\tilde{b_i}||$ is right, and I'am not sure, can you confim that this is right?

 

[Office_Shredder]

You are correct.

 

[Peter_Newman]

Nice! Then thank you very much!

I must say, unconsciously I was able to show the inequality also over another property of the reduced basis (The hint has complicated this however rather :) ).