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To clarify a little more, $SL(n,\mathbb{Z})$ is the group of $n\times n$ matrices with integer entries and determinant equal to 1. It acts on $\mathbb{Z}^n$ as usual by left multiplication. Let $v\in \mathbb{Z}^n$ be a vector such that $gcd(v_1,\ldots,v_n)=1$. Is it true that there exists $M \in SL(n,\mathbb{Z})$ such that $Mv = (1,0,\ldots,0)$?

user1086467
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  • For $n=2$ it is easy to show - did you try it? – Dietrich Burde Apr 11 '25 at 17:28
  • Yes, and for $n=3$ it works as well. But I wonder if there is a proof for any $n$. – user1086467 Apr 11 '25 at 17:36
  • This seems to be true according to lemma 3 of the article https://www.researchgate.net/publication/259370319_On_the_dynamics_of_the_linear_action_of_SLn_Z . – Ito Yukis Apr 11 '25 at 17:50
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    Are you familiar with Smith normal forms? The theory (and algorithms!!) around it give you this immediately. I answered essentially this same question here a few years ago. The framing is a bit different, but the arguments to apply. Not voting to close as a duplicate yet because that would not be kosher under the circumstances. – Jyrki Lahtonen Apr 12 '25 at 04:27
  • There is also this. – Jyrki Lahtonen Apr 12 '25 at 04:29
  • I was not aware of the Smith normal forms... thanks for sharing that! Also, I'm sorry I did not find those posts, which indeed solve the question. I was using more the language of transitive actions and I guess that is the reason I did not came across them. – user1086467 Apr 12 '25 at 08:22

2 Answers2

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Yes. Write your integer vector $v$ as a row. A sequence of column operations takes your vector to the row $(1,0,0...,0).$ Put them as matrices in order, this is accomplished by the product of elementary matrices, call that $A.$

So far, $vA = (1,0,0...,0).$ Next, make a unimodular matrix $B$ with the top row equal to your desired target row vector $w.$ Finally $$ vAB = w$$

How to construct $A:$ the basic step is Bezout: suppose the 1,2 positions in your row start out $(x,y, ?, ?, ...)$ and $g = \gcd(x,y),$ finally $$ ax +by = g$$ Make a unimodular matrix with lower right identity matrix and the upper left 2 by 2 being

$$ \left( \begin{array}{rr} a & -\frac{y}{g} \\ b & \frac{x}{g} \\ \end{array} \right) $$ The result of multiplying the original row on the right by my block diagonal matrix is $$ (g,0, ??, ??, )$$ where the gcd of the entire row is still $1$

Repeat the same step with columns (1,3), then (1,4) and so on.

Oh, in the special case that $x=0$ the 2 by 2 matrix is

$$ \left( \begin{array}{rr} 0 & -1 \\ 1 & 0 \\ \end{array} \right) $$

If both $x,y=0$ you just switch the first element $x$ with the first nonzero element.

Will Jagy
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Since you are using the group theory tag, here is a proof using elementary group theory. Consider the subgroup $H< G=\mathbb Z^n$ generated by the vector $v$. Since $v$ is a primitive integer vector, the quotient $Q=G/H$ is a torsion-free abelian group of rank $n-1$, hence, is isomorphic to $\mathbb Z^{n-1}$ (by the classification of finitely generated abelian groups). Hence, the exact sequence $$ 0\to H\to G\to Q\to 0 $$ splits and you get a right inverse $r: Q\to G$ to the quotient homomorphism $G\to Q$. Hence, we obtain an isomorphism $G\to H\oplus Q= \mathbb Z \oplus \mathbb Z^{n-1}$. Thus, we can complete $v$ to a basis $v_1=v, v_2,...,v_n$ of the free abelian group $G$. Hence, we obtain an automorphism $\phi: G\to G$ sending $v_1,...,v_n$ to the standard basis. The matrix of this automorphism is your matrix $M$. By switching $v_2$ to $-v_2$ if necessary (here one uses the unstated assumption that $n\ge 2$), you get $\det M=1$.

Of course, this proof is kind of backwards since proving the classification theorem for finitely generated abelian groups involves number-theoretic arguments similar to the ones in Will's question.

Moishe Kohan
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