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?
Problem on the determinant of a Matrix
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1For $1\leq n-1$ 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. – 2012-04-01
2 Answers
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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1@J.D. Does going back the way you came and then invoking Bezout's not count as a derivation? :P – 2012-04-01
Hint $a_1=(2,2,2,..., 2)$.