The problem:

Given $\Lambda$ be a full-rank lattice in $\mathbb{R}^n$, which has $\lambda_1 < \lambda_2 < \; ... < \lambda_n$ as successive minima. There exist $\textbf{x}_1, \textbf{x}_2, ..., \textbf{x}_n \in \Lambda$ satisfying $\|\textbf{x}_i\| = \lambda_i \; \forall i$.

Decide whether the matrix $$\textbf{X} = [\textbf{x}_1, \textbf{x}_2, ..., \textbf{x}_n]$$ ($\textbf{x}_i$ are its columns) is always a basis of $\Lambda$.

$\text{ }$

Note: I believe the answer is Not always but I've failed to provide a counterexample.

  • $\begingroup$ If you relax $\lambda_1<\cdots<\lambda_n$ to $\lambda_1\leq\cdots\leq\lambda_n$, then I can give you an example for any $n\geq6$. $\endgroup$ Commented Apr 12, 2016 at 2:56
  • $\begingroup$ In that case, I had an example for every $n \geq 5$. So the question sticks. $\endgroup$ Commented Apr 21, 2016 at 8:50

1 Answer 1


This is a slight modification of the commonly known counterexample for $n=5$.

Let $a=1.00001$ and let $\Lambda$ be the lattice generated by the matrix $$\begin{pmatrix}a&0&0&0&a/2\\0&a^2&0&0&a^2/2\\0&0&a^3&0&a^3/2\\0&0&0&a^4&a^4/2\\0&0&0&0&1/2\end{pmatrix}.$$ Then the successive minima of $\Lambda$ are $1<a<a^2<a^3<a^4$ but the columns of the matrix $$X=\begin{pmatrix}a&0&0&0&0\\0&a^2&0&0&0\\0&0&a^3&0&0\\0&0&0&a^4&0\\0&0&0&0&1\end{pmatrix}$$ do not form a basis of $\Lambda$.

  • $\begingroup$ Is $n = 5$ the minimal dimension where there are counterexamples? $\endgroup$
    – Latrace
    Commented Sep 10, 2022 at 11:35
  • $\begingroup$ Yes, the answer in the following post shows that for $n \leq 4$, we can always find such a basis. math.stackexchange.com/questions/4532410/… $\endgroup$
    – Mahesh S R
    Commented Jun 16 at 15:03

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