Existence of linearly indépendants vectors reaching each minima of a lattice

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 $u_1,\dots,u_{m-1}$ 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
• 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 , $u_1,..u_{m-1}$ 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. Apr 15 '18 at 6:52
• $|u_m|\leqslant \lambda_m$ and not $\lambda_i$ I meant. Apr 15 '18 at 10:46
• @Swann: There exist $m$ linearly independent vectors in the ball $|u|\leqslant\lambda_m$. At least one of these is linearly independent of $u_1,\dots,u_{m-1}$, because the latter span an $(m-1)$-dimensional subspace. Hence the shortest vector that is linearly independent of $u_1,\dots,u_{m-1}$ has length at most $\lambda_m$, and this is what we called $u_m$. Apr 15 '18 at 13:43