Let $\Gamma=(\Gamma_0,\Gamma_1)$ be a multigraph, that is, $\Gamma_0$ is a (finite) set of vertices and $\Gamma_1$ is a (finite) multiset of edges. This means we allow multiple edges between two vertices. We also allow loops, that is, edges from a vertex $i$ to the same vertex $i$. Given two vertices $i,j\in\Gamma_0$ let $n_{ij}$ be the number of edges between $i$ and $j$ (if $i=j$ then $n_{ii}$ is the number of loops at $i$). Note that $n_{ij}=n_{ji}$.
We define a $\mathbb{Z}$-bilinear form on the abelian group $\mathbb{Z}^{\Gamma_0}$ by the formula $$ (x,y) = \sum_{i\in \Gamma_0}(2-n_{ii})x_i y_i - \sum_{i,j\in\Gamma_0}n_{ij}x_ix_j. $$ The quadratic form $$ q(x)=\frac{(x,x)}{2} $$ is called the Tits form of $\Gamma$. Then a connected multigraph $\Gamma$ is said to be of affine type if $q$ is positive semidefinite but no positive definite.
Let $$ K = \{x\in \mathbb{Z}^{\Gamma_0} \; | \; q(x)=0\}. $$ I want to prove that if $\Gamma$ is connected and of affine type there exists a vector $\delta=(\delta_{i})_{i\in\Gamma_0}\in K$ such that $\delta_i>0$ for all $i\in\Gamma_0$ and $K=\mathbb{Z}\delta$. This is Lemma 6.4 in the book A Journey Through Representation Theory by Gruson and Serganova.
The proof is as follows: Let $0\neq \delta\in K$ and define $|\delta|=(|\delta_i|)_{i\in\Gamma_0}$. Then one has that $$ 0\leq q(|\delta|)\leq q(\delta)=0, $$ so $\delta\in K$. Let $$ \mathrm{supp}(\delta) = \{i\in\Gamma_0 \; | \; \delta_i=0\}. $$ Then $\mathrm{supp}(\delta) = \mathrm{supp}(|\delta|)$, so without loss of generality we can assume $\delta_i\geq 0$ for all $i\in I$. Now assume that $\delta_i=0$ for some $i$ and let $\varepsilon_i\in\mathbb{Z}^{\Gamma_0}$ be the vector with $(\varepsilon_i)_j=\delta_{i,j}$, the Kronecker symbol. Then among the the $i$'s with $\delta_i=0$ we choose one connected to at least one $j$ with $\delta_j>0$ (here we use that $\Gamma$ is connected). Then $(\delta,\varepsilon_i)<0$ and $$ q(\varepsilon_i+2\delta)<0, $$ which contradicts the fact that $\Gamma$ is of affine type. Thus $\mathrm{supp}(\delta)=\Gamma_0$.
Finally, let $\delta'\in\Gamma_0$ another element. Then $\mathrm{supp}(\delta)=\Gamma_0=\mathrm{supp}(\delta')$ and there exists a pair of integers $a,b$ such that $\mathrm{supp}(a\delta+b\delta')\neq \Gamma_0$. Now, in the proof Gruson and Serganova say that this implies that $a\delta+b\delta'=0$ (which, for example, follows if $a\delta+b\delta'\in K$), but I cannot see why this is the case. Can anyone see why we can conclude that $a\delta+b\delta'=0$?