A cycle graph is not finite type

The matrix $A=2I-\mathrm{Adj}(G)$ is symmetric and satisfies $a_{ii}=2$, $a_{ij}=-1$ for adjacent $i\mathrm{\ne }j$ and $a_{ij}=0$ otherwise; its diagram, as in [L2], is the cycle $G$ on $m\ge 3$ vertices.

The nonzero vector $x=(1,\dots ,1)$ satisfies $Ax=0$, because every row has diagonal entry $2$ and exactly two entries $-1$. Equivalently $x^{T}Ax=2m-2m=0$. Hence $A$ is not positive definite; moreover every diagonal conjugate $DA{D}^{-1}$ has the nonzero null vector $Dx$, so no symmetric diagonal conjugate can be positive definite.

By [L1] a finite-type Cartan matrix must be symmetrizable to a positive definite matrix; $A$ is not. Moreover, [L2] makes $A$ the only Cartan matrix with this unoriented simple-edge cycle: on each edge the nonpositive integral entries have product $1$, so both are $-1$. Thus the cycle is not a finite-type Dynkin diagram even though it is finite and connected. This explicit family suffices to refute the claimed statement; no broader tree assertion is needed.