Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $Z^2$?

share|cite|improve this question
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.

share|cite|improve this answer

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 $\mathbb{Z}$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.