# How can $(n_1, n_2)$ and $(m_1,m_2)$ Z-linear span $Z^2$?

What conditions should hold for the four integers $n_1, n_2, m_1, m_2$ so that $(n_1, n_2)$ and $(m_1,m_2)$ can Z-linear span $Z^2$?

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They span iff $n_1m_2-n_2m_1=\pm 1$. –  i. m. soloveichik Aug 30 '12 at 20:01

This has a nice geometric proof. Note $\rm\: u,v\:$ is a $\rm\:\mathbb Z$-basis of $\rm\: \mathbb Z^2$ iff $\rm\: \mathbb Z^2$ is tiled by the fundamental parallelogram $\rm\,P\,$ with sides $\rm\:u,v,\:$ i.e. iff $\rm\,P\,$ intersects $\,\Bbb Z^2$ only at its $4$ vertices, i.e. the intersection has $\,\color{#C00}0\,$ $\rm\color{#C00}Interior$ points and $\:\color{#0A0}4$ $\rm\color{#0A0}Boundary$ points, i.e., by Pick's area formula, $$\rm \iff area\ P\, =\, \color{#C00}I + \color{#0A0}B/2 - 1\, =\, \color{#C00}0 + \color{#0A0}4/2 - 1\, =\, 1,\quad I,\,B\, =\, \#Interior,\, \#Boundary\ points$$

By analytic geometry $\rm\: area\ P\: =\: |det(u,v)|,\:$ so $\rm\: u,v\:$ is a $\rm\:\mathbb Z$-basis of $\rm\: \mathbb Z^2 \iff |det(u,v)| = 1\:.$

Remark $\$ Pick applied his area formula in an analogous manner to give a beautiful geometric proof of the Bezout linear representation of the gcd.

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The condition you are after is that

$$n_1m_2-n_2m_1=\pm 1$$

this is an if and only if statement.

PROOF. Suppose that the equation holds. Then

$$m_2(n_1,n_2)-n_2(m_1,m_2)=(\pm 1,0)$$ $$m_1(n_1,n_2)-n_1(m_1,m_2)=(0,\pm 1)$$

which clearly spans $\mathbb{Z}^2$.

Now suppose that $(n_1,n_2),(m_1,m_2)$ span $\mathbb{Z}^2$. Then $\exists a_1,b_1\in \mathbb{Z}$ such that $a_1(n_1,n_2)+b_1(m_1,m_2)=(1,0)$ and similarly $\exists a_2,b_2\in \mathbb{Z}$ such that $a_2(n_1,n_2)+b_2(m_1,m_2)=(0,1)$. So we have

$$\left(\begin{array}{cc}a_1 & b_1 \\a_2 & b_2\end{array}\right)\left(\begin{array}{cc}n_1 & n_2 \\m_1 & m_2\end{array}\right)=\left(\begin{array}{cc}1 & 0 \\0 & 1\end{array}\right)$$

Taking determinents we have that both matrices have determinent $\pm 1$. Since they must both be values in $\mathbb{Z}$.

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