Let $\phi: \Bbb Z \times \Bbb Z \to \Bbb Z$ be a group homomorphism such that $\phi ((1,0))=-2$ and $\phi ((0,1))=7$. Find $\ker(\phi)$.
I know that $\phi ((a,b))=-2a+7b$. In order to find the $\ker(\phi)$, it must happen that $-2a+7b=0$, so, $2a=7b$. But I don't know how to proceed, I am stuck at this point. An intuitive response is that $\ker(\phi)=2\Bbb Z \times 7\Bbb Z$, but I am not sure.
I really appreciate any hint.
Let us continue your reasoning: Let $(a,b) \in \mathbb{Z}^2$ $$\begin{align*} (a,b) \in \ker \phi & \iff -2a + 7b = 0 \iff 2a = 7b \end{align*}$$ But Gauss theorem tells you that $2|7b \implies 2 |b$, dually $7|2a \implies 7 | a$, so $$\begin{align*} (a,b) \in \ker \phi & \iff \exists n,m \in \mathbb{Z}, \, 7n = a, \, 2m = b, \, 2a = 7b \\ &\iff \exists n,m \in \mathbb{Z}, \, 7n = a, \, 2m = b, \, 2\cdot 7n = 7\cdot 2m\\ &\iff \exists n,m \in \mathbb{Z}, \, 7n = a, \, 2m = b, \, n = m\\ &\iff \exists n \in \mathbb{Z}, \, 7n = a, \, 2n = b \\ &\iff (a,b) \in \operatorname{range} (n \mapsto (7n,2n)) \end{align*}$$
In category theory we would say that the kernel of $\phi$ is $f:\mathbb{Z} \to \mathbb{Z}^2$ defined by $f(1) = (7,2)$. But in more practical terms, $\ker \phi = \{(7n,2n), n \in \mathbb{Z}\}$.
What I meant with my comment regarding linear algebra was how to get from a cartesian equation describing an hyperplane ($2x = 7y$) to the description via a spanning vector (here (7,2)). Also it had to be dimension one!