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.
Define the linearly independent vectors $u_1,\dots,u_n\in\Lambda$ recursively as follows. If $u_1,\dots,u_{m1}$ has already been defined, then let $u_m$ be the shortest lattice vector that is linearly independent of $u_1,\dots,u_{m1}$. By the definition of $\lambda_m$, it is clear that $u_m\leq\lambda_m$. By induction, it is also clear that $u_1\leq\dots\lequ_n$. Therefore, $u_m<\lambda_m$ would contradict the definition of $\lambda_m$, hence in fact $u_m=\lambda_m$. QED

$\begingroup$ Why $ u_m \leqslant \lambda_i $ ? As $u_1,..,u_m$ are independants and $u_1$ is the smallest vector we have $u_1=\min u_i\leqslant\lambda_m \leqslant \max u_i$ but i do not understand why $ u_m \leqslant \lambda_i $. Also , $u_1,..u_{m1}$ are specials and maybe by chosing them a little bigger u_m would have been smaller . For me, it is'nt clear, but thank you for this quick review. $\endgroup$– SwannApr 15 '18 at 6:52

$\begingroup$ $u_m\leqslant \lambda_m$ and not $\lambda_i$ I meant. $\endgroup$– SwannApr 15 '18 at 10:46

$\begingroup$ @Swann: There exist $m$ linearly independent vectors in the ball $u\leqslant\lambda_m$. At least one of these is linearly independent of $u_1,\dots,u_{m1}$, because the latter span an $(m1)$dimensional subspace. Hence the shortest vector that is linearly independent of $u_1,\dots,u_{m1}$ has length at most $\lambda_m$, and this is what we called $u_m$. $\endgroup$ Apr 15 '18 at 13:43