Let $\mathcal{L}\subseteq\mathbb{Z}^n$ be an integer lattice (not necessarily full-rank). Is it guaranteed that there is a vector $\mathbf{v}\in\mathcal{L}$ such that $\|\mathbf{v}\|=\det(\mathcal{L})$?

If $\mathcal{L}$ is full-rank, then the answer is certainly yes. Let $\mathbf{B}$ be a matrix whose columns form a basis for $\mathcal{L}$. Then $\|\det(\mathbf{B})\|=\det(\mathcal{L})$. Then the adjugate of $\mathbf{B}$, namely $\mathbf{B}^{adj}=\det(\mathbf{B})\mathbf{B}^{-1}$, is a matrix of integer entries. Therefore, the columns of $\mathbf{B}\cdot\mathbf{B}^{adj}=\det(\mathbf{B})\mathbf{I}_n$ are in $\mathcal{L}$. But the columns are just $\det(\mathbf{B})$ times the standard basis vectors. Each of these scaled standard basis vectors has norm $\|\det(\mathbf{B})\|=\det(\mathcal{L})$.

What about integer lattices that are not full rank?

(Note that the requirement that $\mathcal{L}\subseteq \mathbb{Z}^n$ is an integer lattice is important here. For arbitrary real lattices $\mathcal{L}\subseteq\mathbb{R}^n$, the statement is false. For example, the lattice $\frac{1}{2}\cdot \mathbb{Z}^n$ has determinant $2^{-n}<1/2$ for $n\geq 2$. However all vectors have the norm at least $1/2$.)

  • $\begingroup$ How is $\det(\mathcal{L})$ defined if $\mathcal{L}$ does not have full rank? $\endgroup$
    – arkeet
    Commented Jul 10 at 0:06
  • 1
    $\begingroup$ $\det(\mathbf{B}^T\cdot\mathbf{B})^{1/2}$ where the columns of $\mathbf{B}$ form a basis for $\mathcal{L}$. Geometrically, this is the volume of any fundamental parallelpiped of the lattice. $\endgroup$
    – AAA
    Commented Jul 10 at 1:18

1 Answer 1


Here is a counterexample: Let $\mathcal{L} \subseteq \mathbb{Z}^3$ have basis $\{(1,1,0),(0,1,1)\}$. Then $\det(\mathcal{L})^2 = 3$, but $\|\mathbb{v}\|^2$ is even for any $\mathbb{v} \in \mathcal{L}$, so $\|\mathbb{v}\| = \det(\mathcal{L})$ is impossible.


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