Exercise 3.5.16* (Condition to complete a row $a_1,a_2,\ldots,a_n$ to obtain a matrix $A \in \mathrm{SL}_n(\mathbb{Z})$)

Proof. Suppose that there is a $n\times n$ matrix $A={(a_{\mathit{ij}})}_{1\le i,j\le n}$ with integral elements and determinant $1$ whose first row is $a_1,a_2,\dots ,a_n$, so that $a_{1j}=a_j$ for $1\le j\le n$. The development of the determinant following the first line is written

where $c_i={(-1)}^{1+j}{m}_{1j}$ is the cofactor of $a_j=a_{1j}$ in the matrix $A$ (and $m_{1j}$ the corresponding minor).

Then the relation $a_1{c}_1+a_2{c}_2+\cdots +a_n{c}_n=1$ shows that $a_1\wedge a_2\wedge \cdots \wedge a_n=1$.

We prove the converse by induction. For $n=2$, this is the result of Problem 3. Since $a_1\wedge a_2=1$, there exist integers $u,v$ such that $a_{1}v-a_{2}u=1$, thus $\left|\begin{array}{cc}\hfill a_1\hfill & \hfill a_2\hfill \\ \hfill u\hfill & \hfill v\hfill \end{array}\right|=1$, so

To introduce the method, we treat the case $n=3$, before showing the general case. It it is just for explaining things, so you can skip this part and go to the general method.

Suppose that $\gcd (a,b,c)=1$. Put $d=a\wedge b$. Then $d\wedge c=a\wedge b\wedge c=1$.

$\text{∙}$ If $d=0$, then $a=b=0$, and $c=𝜀=\pm 1$. Then we can take

$\text{∙}$ Suppose now that $d\ne 0$. Put $a^{\prime }=a∕d,b^{\prime }=b∕d$. Then $a^{\prime }\wedge b^{\prime }=1$. The case $n=2$ shows that there are integers $u,v$ such that $\left|\begin{array}{cc}\hfill a^{\prime }\hfill & \hfill b^{\prime }\hfill \\ \hfill u\hfill & \hfill v\hfill \end{array}\right|=1$, which gives $\left|\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill u\hfill & \hfill v\hfill \end{array}\right|=d$.

Then we search integers $\lambda ,\mu ,\nu$ such that

Here

thus

This shows that $A\in {\mathrm{SL}}_3(\mathbb{Z})$ if and only if $(-\mathit{\lambda v}+\mathit{\mu u})c+\mathit{\nu d}=1$. To prove the existence of such integers $\lambda ,\mu ,\nu$, we choose the integers $k,\nu$ such that

Since $d\wedge c=a\wedge b\wedge c=1$, such an integer $k$ exists. Moreover $u\wedge v=1$ (because $a^{\prime }v-b^{\prime }u=1$), therefore there are integers $\lambda ,\mu$ such that

(Note that it is useless to solve this equation by the Bézout’s algorithm, because $(a∕d)v-(b∕d)u=1$, thus $k=(\mathit{ka}∕d)v-(\mathit{kb}∕d)u$, so $\lambda =-\mathit{ka}∕d,\mu =-\mathit{kb}∕d$ is a solution.)

Then, by equations (1), (2), and (3),

so that $A=\left(\begin{array}{ccc}\hfill a\hfill & \hfill b\hfill & \hfill c\hfill \\ \hfill u\hfill & \hfill v\hfill & \hfill 0\hfill \\ \hfill \lambda \hfill & \hfill \mu \hfill & \hfill \nu \hfill \end{array}\right)\in {\mathrm{SL}}_3(\mathbb{Z})$.

For instance, take $(a,b,c)=(6,15,10)$. Then $d=5\wedge 15=3$, and $3\cdot 6-1\cdot 15=3=d$, so we can take $u=1,v=3$. Equation (2) is $10k+3\nu =1$, so we can take $k=1,\nu =-3$. Then $\lambda =-\mathit{ka}∕d=-2,\mu =-\mathit{kb}∕d=-5$, so that

Now we treat the general case. Suppose that for some integer $n\ge 2$, and for any $a_1^{\prime },a_2^{\prime },\dots a_n^{\prime }$ such that $a_1^{\prime }\wedge a_2^{\prime }\wedge \cdots \wedge a_n^{\prime }=1$, we can find a matrix $A^{\prime }={(a_{\mathit{ij}}^{\prime })}_{1\le i,j\le n}\in {\mathrm{SL}}_3(\mathbb{Z})$ such that $a_{11}^{\prime }=a_1^{\prime },a_{22}^{\prime }=a_2^{\prime },\dots ,a_n^{\prime }=a_{1n}^{\prime }$.

Take now $n+1$ integers $a_1,a_2\dots ,a_n,a_{n+1}$ such that $a_1\wedge a_2\wedge \cdots \wedge a_{n+1}=1$. Put $d=a_1\wedge a_2\wedge \cdots \wedge a_n$.

If $d=0$, then $a_1=a_2=\cdots =a_n=0$, and $a_{n+1}=𝜀=\pm 1$. We can take

such that $\det <!--\; FUNCTION\; APPLICATION\; -->(A)=1$.

If $d\ne 0$, put $a_1^{\prime }=a_1∕d,a_2^{\prime }=a_2∕d,\dots ,a_n^{\prime }=a_n∕d$. Then $a_1^{\prime }\wedge a_2^{\prime }\wedge \cdots \wedge a_n^{\prime }=1$. If we apply the induction hypothesis to $a_1^{\prime },a_2^{\prime },\dots ,a_n^{\prime }$, we obtain a matrix

Therefore

satisfies $\det <!--\; FUNCTION\; APPLICATION\; -->(B)=d$. We prove the existence of integers ${\lambda }_1,{\lambda }_2,\dots ,{\lambda }_n,{\lambda }_{n+1}$ such that

By expanding $\det <!--\; FUNCTION\; APPLICATION\; -->(A)$ along the last column, we obtain

where

Here $\det <!--\; FUNCTION\; APPLICATION\; -->(B)=d=a_1{c}_1+a_2{c}_2+\cdots +a_n{c}_n$, where $c_j={(-1)}^{1+j}{m}_{1j}$ is the cofactor of $a_j=a_{1j}$ in $B$, and

Therefore equation (4) gives

Since $d\wedge a_{n+1}=a_1\wedge a_2\wedge \cdots \wedge a_n\wedge a_{n+1}=1$, there exist integers $k$ and ${\lambda }_{n+1}$ such that

Moreover, we can find ${\lambda }_1,{\lambda }_2,\dots ,{\lambda }_n$ such that

Indeed, $1=(a_1∕d)c_1+(a_2∕d)c_2+\cdots +(a_n∕d)c_n$, so $c_1\wedge c_2\wedge \cdots \wedge c_n=1$. More explicitly, since $a_1{c}_1+a_2{c}_2+\cdots +a_n{c}_n=d$, we obtain $(k{a}_1∕d)c_1+(k{a}_2∕d)c_2+\cdots +(k{a}_n∕d)c_n=k$, so a solution is

Then equations (5), (6) and (7) show that

for this choice of ${\lambda }_1,\dots ,{\lambda }_{n+1}$, and so the induction is done.

There is a $n\times n$ matrix with integral elements and determinant $1$ whose first row is $a_1,a_2,\dots ,a_n$, if and only if $\gcd (a_1,a_2,\dots ,a_n)=1$. □