Problem 1246

Prove that for $V=\left\{\sum_{i=1}^{\infty} k_{i} A_{i} \mid\right.$ the integer sequence $\left\{k_{i}\right\}$ has only finitely many non-zero terms $\}$, it is easy to verify that $V$ has the following property:
For any $x, y \in V, k, l \in \mathbf{Z}, k x+l y \in V$.
(*)

We need to prove: There must exist $x_{1}, x_{2} \in V$, such that
$$V=\left\{k_{1} x_{1}+k_{2} x_{2} \mid k_{1}, k_{2} \in \mathbf{Z}\right\}$$

If $V=\{(0,0)\}$, the conclusion is obviously true.
If $V \neq\{(0,0)\}$, let $x_{1}=(a, b)$ be the non-zero vector in $V$ with the smallest norm, and let $L=\left\{t x_{1} \mid t \in \mathbf{R}\right\}$.

Assume $t x_{1} \in V$, and $t$ is not an integer. We can set $t=m+s, m \in \mathbf{Z}, 0<s<1$.
For any $x \in V$, there exist real numbers $k_{1}, k_{2}$, such that
$$x=k_{1} x_{1}+k_{2} x_{2}$$

If $k_{2} \notin \mathbf{Z}$, set $k_{2}=u+v, u \in \mathbf{Z}, 0<v<1$.
By property (*), $k_{1} x_{1}+v x_{2} \in V$.
Since $k_{1} x_{1}+v x_{2}=\left(k_{1} a+v c, k_{1} b+v d\right)$, and
$$0<\left|a\left(k_{1} b+v d\right)-b\left(k_{1} a+v c\right)\right|=v|a d-b c|<|a d-b c| .$$

This contradicts the choice of $x_{2}$!
Therefore, $k_{2} \in \mathbf{Z}$. Combining this with property (*), we know that $k_{1} x_{1} \in V$, and thus $k_{1} \in \mathbf{Z}$.
So, $V=\left\{k_{1} x_{1}+k_{2} x_{2} \mid k_{1}, k_{2} \in \mathbf{Z}\right\}$.
Since $x_{1}, x_{2} \in V$, there exist positive integers $N$ and integers $m_{1, i}, m_{2, i} (i=1, 2, \cdots, N)$, such that
$$x_{1}=\sum_{i=1}^{N} m_{1, i} A_{i}, x_{2}=\sum_{i=1}^{N} m_{2, i} A_{i}$$

Thus, for any vector $A_{n}$, there exist integers $k_{1}, k_{2}$, such that
$$A_{n}=k_{1} x_{1}+k_{2} x_{2}=\sum_{i=1}^{N}\left(k_{1} m_{1, i}+k_{2} m_{2, i}\right) A_{i}$$