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A Dynkin diagram has one vertex for each simple root. Vertices $i,j$ are joined by $a_{ij}{a}_{ji}$ bonds, where $a_{ij}$ are the off-diagonal Cartan integers. No bond means orthogonality; one, two and three bonds mean angles $2\pi /3,3\pi /4,5\pi /6$. A multiple-bond arrow points to the shorter root. Thus the Dynkin diagram determines the Cartan matrix, including the relative lengths inside each connected component.

An irreducible root system corresponds to a connected Dynkin diagram. The classification of finite crystallographic Dynkin diagrams givesHere $C_2$ is already represented by $B_2$, and $B_1,C_1$ by $A_1$. General systems are orthogonal direct sums of these, with separate component scales. The following diagrams specify all the types; dots denote a continued chain, not additional vertices.

For orientation, the An Dynkin diagram is a single chain, the Bn Dynkin diagram has a terminal double bond pointing to its terminal short root, and the Cn Dynkin diagram reverses that arrow. The Dn Dynkin diagram has arm lengths $1,1,l-3$. The En Dynkin diagram has arm lengths $1,2,2$; $1,2,3$; or $1,2,4$. The F4 Dynkin diagram has a central double bond and the G2 Dynkin diagram a triple bond.

The positivity behind this finite list is the positive-definite Gram matrix of the simple roots. In particular the underlying graph is a tree: if it contained a cycle, the sum of its normalized roots would have squared norm at most zero, since each bonded inner product is at most $-1/2$. This contradicts linear independence. Restricting the possible bonds and branches by positive definiteness gives precisely the listed chains and trees. Conversely their Cartan matrices are positive definite after symmetrization and their reflection-generated root sets give the indicated finite root systems.

Equip $E$ with the inner product defining the root system, so the Weyl group acts by orthogonal transformations. As usual, the root system spans its ambient space $E$. Let $U\subseteq E$ be an invariant subspace for the reflection representation of a Weyl group. Its orthogonal complement is also invariant.

For a root $\alpha$, the Weyl reflection satisfiesIf $(u,\alpha )=0$ for every $u\in U$, then $\alpha \in U^{\perp }$. Otherwise choose $u\in U$ with $(u,\alpha )\ne 0$. Both $u$ and $s_{\alpha }(u)$ belong to $U$, and their difference is a nonzero multiple of $\alpha$, so $\alpha \in U$. HenceThe two sets are mutually orthogonal and individually closed under their root reflections. If both were nonempty, this would decompose $R$ into two orthogonal root systems, contradicting that it is an irreducible root system. All roots therefore lie in one of the two subspaces. Because they span $E$, either $U=E$ or $U^{\perp }=E$, the latter giving $U=0$. Thus the reflection representation is an irreducible representation.

The reflection representation is the action of the Weyl group on the real span of its root system, generated by its Weyl reflections. For an irreducible root system, it is an irreducible representation: if $U$ is an invariant subspace, each root either lies in $U^{\perp }$ or, by subtracting a reflected vector of $U$, lies in $U$. These alternatives partition the roots into orthogonal sets. Irreducibility and spanning force $U=0$ or the whole space.

For an irreducible root system spanning $V$, the span $U$ of a nonzero root's Weyl group orbit is invariant, as is $U^{\perp }$. A root reflection preserving both subspaces must have its normal entirely in one of them: otherwise its rank-one correction mixes the subspaces. Thus a proper nonzero $U$ would partition all roots into mutually orthogonal subsets, contradicting irreducibility. Consequently some orbit root is nonorthogonal to any given root. Since Weyl reflections preserve lengths, the pairwise integrality restrictions for nonorthogonal roots also bound length ratios of orthogonal roots. All squared-length ratios lie in $\{1,2,1/2,3,1/3\}$; there cannot be three distinct lengths, since ratios two and three relative to a shortest root would yield the forbidden ratio $3/2$.