Rational numbers can be defined as those numbers $a \in \mathbb{R}$ for which there exists an integer $v \in \mathbb{Z}$ such that $av \in \mathbb{Z}$. Let us consider the following higher-dimensional extension of this definition.

Definition: A matrix $A \in \mathbb{R}^{n \times n}$ is rational if there exists a vector $v \in \mathbb{Z}^n \setminus \{0\}$ such that $Av \in \mathbb{Z}^n$.

I would like to show the following.

Conjecture: If $A$ is rational, then $A^T$ is also rational.

My most promising proof attempt has been to note that singular matrices are rational according to the above definition and singular matrices are closed under transposition. So maybe it is possible to adapt that proof to the situation here?


How about $$A=\pmatrix{\sqrt2&\sqrt2\\\sqrt3&\sqrt3}?$$

  • $\begingroup$ If anyone else is wondering, this counterexample can easily be made non-singular: just add 1 to the top left entry, then $A [1 \,\, -1]^T = [1 \,\, 0]^T$ but $A^T$ is still not rational since there is no way to make the last entry integral. $\endgroup$ – gTcV Apr 18 at 6:20

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