We select \[\begin{align} v_1 & = & && \mathop{\mathrm{argmin}}_{v \in \Lambda \setminus \{ 0\}}\ \|v\|&& \\ v_2 &= & &&\mathop{\mathrm{argmin}}_{v \in \Lambda \setminus D v_1} \|v\|&&\\ v_3 &= & &&\mathop{\mathrm{argmin}}_{v \in \Lambda \setminus (D v_1 + D v_2)} \|v\|&&\\ v_4 & = & && \mathop{\mathrm{argmin}}_{v \in \Lambda \setminus (D v_1 + D v_2 + D v_3)} \|v\| &&\\ & \vdots \end{align}\] Since \(\Lambda\otimes_{\mathbb{Z}} \mathbb{Q}\) has \(\mathbb{Q}\)-dimensions \(k[D:\mathbb{Q}]\), so we know that \(\Lambda \setminus (D v_1 + D v_2 + \cdots D v_i)\) will always be non-empty for \(i \le k\). Now suppose that for some \(i\), \(a_1,a_2,\cdots,a_i \in D\) satisfy \[\begin{align} a_1 v_1 + a_2 v_2 + a_3 v_3 + \ldots a_i v_i= 0 , \ a_i \neq 0 \end{align}\] Then, this will imply that \(v_i\) lies in the \(D\)-span of \(v_1,v_2,\cdots,v_{i-1}\) which is impossible. Hence, the claim about linear independent follows.

Observe that \[\begin{align} \|v_1\| \le \|v_2\| \le \ldots \le \|v_k\|, \end{align}\] and furthermore for each \(i\) \[\begin{align} \|v_i\| \ge \lambda_i. \end{align}\] Now \(\|v_1\| = \lambda_{1}\) is clear. Suppose it is true upto \(i\) that \(\lambda_{i} = \|v_i\|\). Let \(\{ w_1,w_2,\cdots,w_{i+1}\} \subseteq \overline{B_{\lambda_{i+1}}(0)}\) such that they are \(D\)-linearly independent. Then, because the \(\mathbb{Q}\)-dimension of the space they span is larger, at least one of these vectors does not lie in \(Dv_1 + D v_2 + \cdots + D v_k\). Let’s say \(w_{i+1}\) is this vector. So \(\{ v_1,v_2,\cdots,v_i,w_{i+1}\}\) is a \(D\)-linearly independent set of vectors. Knowing that \(\|w_{i+1}\| \ge \lambda_1\), we try to find the smallest \(j \in \{2,\cdots, i+1\}\) so that \(\|w_{i+1}\| < \lambda_{j}\). Such a \(j\) is not possible by the definition of \(\lambda_j\) and hence \(\|w_{i+1}\| \ge \lambda_{j}\). In particular, \(\|w_{i+1}\|= \lambda_{i+1}\). So \(\|v_i\|\ge \|w_i\|\). On the other hand \(\|v_{i+1}\| > \|w_{i+1}\|\) is not possible by the definition of \(\|v_{i+1}\|\) so we have \(\|v_{i+1}\| = \|w_{i+1}\|= \lambda_{i+1}\).

Generate the vectors \(x_1,x_2,\cdots,x_k\) using the Gram-Schmidt process on \(v_1,v_2,\cdots,v_k\). Since \(\{ v_i\}_{i=1}^{k}\) is free with respect to left-\(D_{\mathbb{R}}\) action, we get that \(x_1,x_2,\cdots,x_k\) freely generate \(D_{\mathbb{R}}^{k}\) as a left-\(D_{\mathbb{R}}\) module. Caution is advised here, since this is a non-commutative variant of the usual Gram-Schmidt algorithm. See the page for details.

Now consider the \(\mathbb{R}\)-linear map given by \[\begin{align} T: y \mapsto \frac{y_1}{\lambda_1} x_1 + \frac{y_2}{ \lambda_2} x_2 + \cdots \frac{y_k}{ \lambda_k} x_k , \\ \text{ for }y = y_1x_1+y_2x_2 + \cdots +y_kx_k \in D_{\mathbb{R}}^{k}. \end{align}\]

Now create a lattice \(\Lambda'\) as \[\begin{align} \Lambda' = (\lambda_1 \lambda_2 \ldots \lambda_k)^{ {1}/{k}}T(\Lambda) \end{align}\] Observe that \(\det(\Lambda') = \det(\Lambda)\). For any \(y' \in \Lambda' \setminus \{ 0\}\), we can find a \(y = y_1 x_1 + \cdots + y_k x_k \in \Lambda\) such that \(y' = (\lambda_1\cdots \lambda_k)^{1/k} T ( y)\). Furthermore, there must be a smallest \(i_0 \ge 1\) that \(y_{i_0} \neq 0\) and \(y_{i_0+1} = y_{i_0+2} = \cdots = y_{k} = 0\). Then \(y \in \Lambda \cap ( D_{\mathbb{R}} v_1 + D_{\mathbb{R}} v_2 + \cdots + D_{\mathbb{R}} v_{i_0})\) and not in \(\Lambda \cap ( D_{\mathbb{R}} v_1 + \cdots + D_{\mathbb{R}} v_{i_0 - 1} )\). Therefore \(\|y\| \ge \lambda_{i_0}\). This implies that \[\begin{align} \|y'\|^{2} = \left( \lambda_1 \lambda_2 \ldots \lambda_k \right)^{2/k} \sum_{i=1}^{i_0} \left| \frac{y_i}{\lambda_i}\right|^{2} \ge \left( \lambda_1 \lambda_2 \ldots \lambda_k \right)^{2/k} \frac{1}{\lambda_{i_0}^{2}} \sum_{i=1}^{k}\left| {y_1}\right|^{2} \ge (\lambda_1 \ldots \lambda_k)^{2/k} \end{align}\] This lower bound is tight, since we can set \(y_1=1_{D_{\mathbb{R}}}\) and the rest \(y_i=0\) for \(i \ge 2\).