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?