I was wondering : given a full rank lattice $\Lambda$ of $R^n$ (a discrete subgroup spanning $R^n$) the successive minima of $\Lambda$ are for $1\leqslant i \leqslant n$ $\lambda\_i= \min\{r>0 \mid \text{exists i linearly independants vectors of $\Lambda$ un the ball centered in 0 of radius r}\}$. Every text talking about theses say that there exists $u\_1,...,u\_n$ in $\Lambda$ such as for all $1\leqslant i \leqslant n$ $||u\_i||=\lambda\_i$, without giving a full proof. As I can't manage to proove it myself, maybe someone could help me.. Thank you.



Existence of linearly indépendants vectors reaching each minima of a lattice
Define the linearly independent vectors $u\_1,\dots,u\_n\in\Lambda$ recursively as follows. If $u\_1,\dots,u\_{m-1}$ has already been defined, then let $u\_m$ be the shortest lattice vector that is linearly independent of $u\_1,\dots,u\_{m-1}$. By the definition of $\lambda\_m$, it is clear that $|u\_m|\leq\lambda\_m$. By induction, it is also clear that $|u\_1|\leq\dots\leq|u\_n|$. Therefore, $|u\_m|<\lambda\_m$ would contradict the definition of $\lambda\_m$, hence in fact $|u\_m|=\lambda\_m$. QED


