Suppose $(a,b) \in \mathbb{Z}_n \times \mathbb{Z}_n$ has order $n$.
Prove that there exists $(u,v) \in \mathbb{Z}_n \times \mathbb{Z}_n$ such that every element of $\mathbb{Z}_n \times \mathbb{Z}_n$ equals $k_1(a,b) + k_2(u,v)$ for some $k_1, k_2 \in \mathbb{Z}$.
This is final part of question I asked earlier.
I know very little linear algebra, so i'm not sure how to go about proving that $(a,b)$ and $(u,v)$ form a basis for $\mathbb{Z}_n \times \mathbb{Z}_n$ any direction or help would be greatly appreciated, thaks
Since $(a, b)$ has order $n$, $\gcd (a, b, n) = 1$. In general, if $\gcd (a, b, n) = d$ then the order of $(a, b)$ is at most $n/d$ (in fact equal but we don't need this) since $n$ divides $(n/d) \cdot (a, b)$.
Thus there are integers $x, y, z$ such that $ax + by + nz = 1$. Now we can take $(u, v) = (-y, x)$. One way to see this is to reinterpret $ax + by = 1 \pmod{n}$ as computing the determinant of the matrix $$\begin{bmatrix} a & -y\\ b & x \end{bmatrix}$$ over $\mathbb{Z}/n$. Or we could just use the inverse directly, for any $(p, q)$: $(p, q) = (px + qy) \cdot (a, b) + (aq - bp) \cdot (-y, x) \pmod{n}$ .