An full $n$-dimensional lattice $\Lambda$ is a discrete subgroup of $\mathbb{R}^n$ (equipped with some norm $\lVert \cdot \rVert$) containing $n$ linearly independent points. If $\Lambda = \{ A z, z\in \mathbb{Z}^n\}$ for $A \in GL(n,\mathbb{R})$, we call the $n$ columns of $A$ a basis of $\Lambda$ (every full lattice has a basis). We call $n$ points $l_1,\dots,l_n \in \Lambda$ a minimal system if for $k\in\{1,\dots,n\},$ $$\lVert l_k \rVert = \min \{ \lVert l \rVert: l \in \Lambda \setminus \left< l_1,\dots,l_{k-1}\right>_\mathbb{R}\}.$$ Let's just consider the standard $2$-norm $\lVert x \rVert = \left< x,x\right>^\frac{1}{2}$. In two dimensions, every minimal system is also a basis. In five dimensions, this is no longer true, as is stated in the answer to this post. The lattice generated by $$ A = \begin{pmatrix} 1 & 0 & 0 & 0 & \frac{1}{2} \\ 0 & 1 & 0 & 0 & \frac{1}{2} \\ 0 & 0 & 1 & 0 & \frac{1}{2} \\ 0 & 0 & 0 & 1 & \frac{1}{2} \\ 0 & 0 & 0 & 0 & \frac{1}{2} \\ \end{pmatrix} $$ has the five standard unit vectors as a minimal system, but they do not form a basis of the lattice.
In four dimensions, the lattice generated by the basis $$ A = \begin{pmatrix} 1 & 0 & 0 & \frac{1}{2} \\ 0 & 1 & 0 & \frac{1}{2} \\ 0 & 0 & 1 & \frac{1}{2} \\ 0 & 0 & 0 & \frac{1}{2} \\ \end{pmatrix} $$ has both $A$ and the standard basis as minimal systems. This is not a stable property: If the last vector is perturbed a little, the minimal system is unique (up to reflections) and a basis again. This has led to my feeling that this is an extremal case, and the situation is as follows:
Every four-dimensional lattice has a minimal system that's also a lattice basis. Furthermore, only a zero set of matrices generate lattices that have minimal systems that are not bases.
Is that true?
Josef E. Greilhuber's conjecture is correct, and the "extremal case" he found is the only one up to scaling.
This follows from the following estimate:
Proposition. Let $l_1,\ldots,l_n$ be a $\bf Z$-basis for a lattice $\Lambda_0 \subset {\bf R}^n$; and let $x = \sum_{i=1}^n c_i l_i$ for some real $c_i$ with $0 \leq c_i \leq 1$ for each $i$. Then there exists $x' \in x + \Lambda_0$ such that $$ \| x' \|^2 \leq M := \sum_{i=1}^n c_i (1-c_i) \|l_i\|^2. $$ Moreover, if $\min_{x' \in x + \Lambda_0} \| x' \|^2 = M$ then each $c_i \in \{0, 1/2, 1\}$, and the $l_i$ with $c_i = 1/2$ are pairwise orthogonal.
Assume this for now. If $l_1,\ldots,l_n$ form a minimal system in some lattice $\Lambda$ but generate some proper sublattice $\Lambda_0$ then we may find some $x \in \Lambda$ that is not in $\Lambda_0$. Write $x = \sum_{i=1}^n c_i l_i$ for some $c_i \in \bf R$, not all integers; translating $x$ by a $\Lambda_0$ vector, we may assume that $0 \leq c_i \leq 1$ for each $i$. Then $\min_{x' \in x + \Lambda_0} \| x' \|^2 \geq \max_i \|l_i\|^2$. But by the Proposition, if $n \leq 4$ then $$ \min_{x' \in x + \Lambda_0} \| x' \|^2 \leq \frac14 \sum_{i=1}^n \|l_i\|^2 \leq \max_i \|l_i\|^2, $$ with equality if and only if $n=4$, each $c_i = 1/2$, and the $\|l_i\|$ are all equal. Therefore equality holds throughout, and $\Lambda_0$ is the hypercubical lattice ${\bf Z}^4$ scaled by the common value of $\|l_i\|$, while $\Lambda_0$ is the span of $$ \begin{pmatrix} 1 & 0 & 0 & \frac{1}{2} \\ 0 & 1 & 0 & \frac{1}{2} \\ 0 & 0 & 1 & \frac{1}{2} \\ 0 & 0 & 0 & \frac{1}{2} \end{pmatrix} $$ scaled by the same factor.
It remains to prove the Proposition. We show that $M$ is a weighted average of $\|x'\|^2$ over $2^n$ of the candidate vectors $x'$, whence at least one of them must have $\|x'\|^2 \leq M$.
We illustrate the technique with the critical special case that $c_i = 1/2$ for each $i$. Then we average over the $2^n$ vectors $x' = \sum_{i=1}^n \frac12 \epsilon_i l_i$ with each $\epsilon_i = 1$ or $-1$. Then $$ \|x'\| = \frac14 \sum_{i=1}^n \sum_{j=1}^n \epsilon_i \epsilon_j (l_i,l_j). $$ Each of the $n$ terms with $i=j$ contributes $\frac14 \|l_i\|$, while each of the cross-terms with $i \neq j$ averages to zero. Hence the average of $\|x'\|$ is $\frac14 \sum_{i=1}^n \|l_j\|^2$, which is the value of $M$ in this case.
In general we average over the $2^n$ vectors $x' = \sum_{i=1}^n c'_i l_i$ where each $c'_i$ is either $c_i$ or $c_i-1$, taken with weights $w(x') = \prod_{i=1}^n (1-|c'_i|)$. Each factor $1-|c'_i|$ is $1-c_i$ or $c_i$ respectively, so $\sum_{x'} w(x') = 1$ and each $w(x') \geq 0$. Moreover, for each $i$ the weighted average of the $l_i$ coefficients vanishes, and the weighted average of their squares is $$ (1-c_i) c_i^2 + c_i (1-c_i)^2 = c_i (1-c_i). $$ Therefore in the expansion of the weighted sum $$ \sum_{x'} w(x') \|x'\|^2 = \sum_{x'} w(x') \sum_{i=1}^n \sum_{j=1}^n c'_i c'_j (l_i,l_j) $$ each $i=j$ term contributes $c_i (1-c_i) \|l_i\|^2$ and each $i \neq j$ cross-term contributes zero. Hence the weighted sum is $\sum_{i=1}^n c_i (1-c_i) \|l_i\|^2 = M$, and the inequaity is proved.
In the case of equality, $c_i (1-c_i) = \min(c_i^2, (1-c_i)^2)$ for each $i$, and all $x'$ of nonzero weight have the same norm. The first condition implies that each $c_i \in \{0, 1/2, 1\}$, and the second quickly forces the orthogonality of any two $l_i$ for which $c_i = 1/2$. $\Box$