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Is there an easy way to show that given a lattice $\Lambda \subset \mathbb{R}^n$ of full rank, exists a basis where each vector has norm $\lambda_i$ i.e the i-th successive minima ($\lambda_i(\Lambda)=inf \{\text{dim(span}( \Lambda \cap \overline{B}(0,r)) \geq i\})$?

The existence of such independent vectors is not a problem since the balls centered in $0$ of that radius are finite thanks to lattice definition.

My problem is that I don't how to show that those $\mathbb{Z}-$generate $\Lambda$. This claim could pass from the fact that something is a basis if and only if doesn't intersect the fundamental parallelepiped?

Any help or reference would be appreciated.

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  • $\begingroup$ what do you mean by each vector being the $i$-th successive minima? $\endgroup$
    – AnCar
    Commented Jun 16 at 12:26
  • $\begingroup$ @AnCar I edited the question, sorry if was unclear. $\endgroup$
    – jacopoburelli
    Commented Jun 16 at 12:41
  • $\begingroup$ It is not always possible to show that there exists a basis $B = [b_1, \dots, b_n]$ such that $\|b_i\| = \lambda_i(\Lambda)$. See the following post - math.stackexchange.com/questions/1681308/… $\endgroup$
    – Mahesh S R
    Commented Jun 16 at 14:53
  • $\begingroup$ The answer in the following post shows that there exists a basis $B = [b_1, \dots, b_n]$ such that $\|b_i\| \leq max\{1, \sqrt{i}/2\} \cdot \lambda_i(\Lambda)$. math.stackexchange.com/questions/4532410/… $\endgroup$
    – Mahesh S R
    Commented Jun 16 at 15:00
  • $\begingroup$ @MaheshSR Why this implies that exists a basis such that $||b_i|| = \lambda_i$? $\endgroup$
    – jacopoburelli
    Commented Jun 16 at 16:31

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