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Suppose we have $n-1$ linearly independent vectors $a_1, \ldots,a_{n-1} \in \mathbb{Z^n}$. Is it possible to find another vector $a_n\in \mathbb{Z^n}$ such that the determinant of the matrix $M$ whose row vectors are $a_1,\ldots,a_n$ is 1?

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    $\begingroup$ For $1\leq n-1$ let $a_i\in\mathbb{Z}^n$ be the vector with $2$ in the $i$-th position and $0$ everywhere else. Can you find an $a_n$ in this case? Why or why not? Also, note for the sake of intuition: the determinant of a matrix (aka linear transformation) can be interpreted as the volume of the image of the unit cube. $\endgroup$ – William Apr 1 '12 at 17:48
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Let $\rm M_k$ be the matrix formed by cutting out the $\rm k$th column of the $\rm a_i$ row vectors and putting them together. Then the condition that $\rm \det M=1$ imposed on $\rm \vec{v}=a_n$ is equivalent to

$$\rm (\det M_1)\,v_1-(\det M_2)\,v_2+\cdots\pm (\det M_n)\,v_n=1.$$

By Bezout's lemma, this has a solution in $\rm \vec{v}$ if and only if $\rm \gcd(\det M_1,\cdots,\det M_n)=1$.

This is not always the case. Take e.g. in $\rm n=3$, with $\rm v,w\ne0$ and $\rm u\ne0,\pm1$,

$$\rm a_1^T=\begin{pmatrix}u \\ 0 \\ \rm v\end{pmatrix} \quad and \quad a_2^T=\begin{pmatrix} 0 \\ \rm u \\ \rm w \end{pmatrix}$$

where we have $\rm \gcd(\det M_1, \det M_2,\det M_3)=\gcd(-uv,uw,u^2)\ne 1$. More generally, for arbitrary $\rm n$ following you's suggestion, let $\rm a_i=x_i\,\vec{e}_i$ with $\rm \gcd(x_1,\cdots,x_n)\ne1$. The resulting matrix will be diagonal, so its determinant is the product of the diagonal entries, which must be $>1$ in magnitude if any of the $\rm x_i$ are.

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  • $\begingroup$ Follow up question: We can rewrite the constraint as $$({\rm det}M_1\quad -{\rm det}M_2\quad \ldots \quad \pm {\rm det} M_n \quad -1) \begin{pmatrix}v_1 \\ v_2 \\ \vdots\\v_n \\ 1\end{pmatrix}= 0,$$ or $Mu = 0.$ Since ${\rm rank}(M^{T}M) = 1,$ the nullity $> 0.$ Can't we solve $M^{T}M u = 0$ for $u$ (i.e. finding the nullspace basis of $M^{T}M$ and forming $u$)? Or is the fact the last entry in $u$ is $1,$ put conditions on finding $u$? $\endgroup$ – user2468 Apr 1 '12 at 18:28
  • $\begingroup$ @J.D. We want $u\in\Bbb{Z}^n$ no? $\endgroup$ – anon Apr 1 '12 at 18:46
  • $\begingroup$ Yes. Indeed $u \in \mathbb{Z}^{n}$. My source of confusion: I can not derive the condition $\gcd({\rm det}M_1,\ldots,{\rm det}M_n) = 1$ from my formulation. $\endgroup$ – user2468 Apr 1 '12 at 18:48
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    $\begingroup$ @J.D. Does going back the way you came and then invoking Bezout's not count as a derivation? :P $\endgroup$ – anon Apr 1 '12 at 18:54
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Hint $a_1=(2,2,2,..., 2)$.

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