[proofplan] We use the positive definite symmetric form attached to the finite type Cartan matrix. If a simple cycle existed, we place a carefully normalized positive vector on the vertices of that cycle and compute its quadratic form. Each diagonal term contributes $1$, while each edge of the cycle contributes at most $-1$ because Cartan off-diagonal entries are negative integers. The cycle therefore produces a nonzero vector with nonpositive quadratic form, contradicting positive definiteness. [/proofplan]

[step:Pass from the Cartan matrix to a positive definite symmetric matrix]By the standard positive-definite symmetrization characterization of [finite type Cartan matrices](/page/Cartan%20Matrix), since $A$ is symmetrizable of finite type, there are positive [real numbers](/page/Real%20Numbers) $d_i > 0$ for $i \in I$ such that the matrix \begin{align*} B := (b_{ij})_{i,j \in I}, \qquad b_{ij} := d_i a_{ij}, \end{align*} is symmetric and positive definite. Since $A$ is a generalized Cartan matrix, $a_{ii}=2$ for every $i \in I$, $a_{ij} \leq 0$ for $i \neq j$, and $a_{ij}=0$ iff $a_{ji}=0$. Hence \begin{align*} b_{ii} = 2d_i > 0 \end{align*} for every $i \in I$, and if $\{i,j\}$ is an edge of $G(A)$, then $a_{ij}<0$, $a_{ji}<0$, and \begin{align*} b_{ij} = d_i a_{ij} = d_j a_{ji} < 0. \end{align*}[/step]

[guided]The finite type hypothesis is used through positive definiteness. We use the standard positive-definite symmetrization characterization of [finite type Cartan matrices](/page/Cartan%20Matrix): because $A$ is symmetrizable of finite type, symmetrizability gives positive numbers $d_i > 0$ and turns the generally nonsymmetric Cartan matrix $A$ into the symmetric matrix \begin{align*} B := (b_{ij})_{i,j \in I}, \qquad b_{ij} := d_i a_{ij}. \end{align*} The finite type condition is precisely the assertion, in this characterization, that this symmetric matrix is positive definite. The generalized Cartan matrix conditions give the sign pattern we need. For every vertex $i \in I$, \begin{align*} b_{ii} = d_i a_{ii} = 2d_i > 0. \end{align*} For distinct vertices $i \neq j$, an edge $\{i,j\}$ means $a_{ij}a_{ji} \neq 0$. Since off-diagonal Cartan entries are nonpositive integers and $a_{ij}=0$ iff $a_{ji}=0$, this implies $a_{ij}<0$ and $a_{ji}<0$. Therefore \begin{align*} b_{ij} = d_i a_{ij} = d_j a_{ji} < 0. \end{align*} Thus edges contribute negative cross terms in the quadratic form associated to $B$.[/guided]

[step:Assume a simple cycle and build a normalized vector on its vertices] Suppose, for contradiction, that $G(A)$ contains a simple cycle. Let $m \geq 3$ be its length, and write its distinct vertices as \begin{align*} i_1, i_2, \dots, i_m \in I, \end{align*} with indices read modulo $m$, so that $\{i_k,i_{k+1}\}$ is an edge of $G(A)$ for every $k \in \{1,\dots,m\}$. Define a vector $x = (x_i)_{i \in I} \in \mathbb{R}^{I}$ by \begin{align*} x_i := \begin{cases} (2d_i)^{-1/2}, & \text{if } i \in \{i_1,\dots,i_m\},\\ 0, & \text{otherwise.} \end{cases} \end{align*} The vector $x$ is nonzero because the cycle has at least one vertex and each $d_i$ is positive. [/step]

[step:Estimate the quadratic form from the cycle edges]The quadratic form associated to $B$ at $x$ is \begin{align*} x^\top Bx &= \sum_{i \in I} b_{ii}x_i^2 + 2\sum_{\substack{\{i,j\}\subset I\\ i\neq j}} b_{ij}x_ix_j, \end{align*} where the second sum is over unordered two-element subsets of $I$; the summand is well-defined because $B$ is symmetric. The diagonal contribution from the vertices of the cycle is \begin{align*} \sum_{i \in I} b_{ii}x_i^2 = \sum_{k=1}^{m} (2d_{i_k})(2d_{i_k})^{-1} = m. \end{align*} For each cycle edge $\{i_k,i_{k+1}\}$, symmetry of $B$ gives \begin{align*} 2b_{i_k i_{k+1}}x_{i_k}x_{i_{k+1}} &= 2d_{i_k}a_{i_k i_{k+1}} \cdot \frac{1}{\sqrt{2d_{i_k}}} \cdot \frac{1}{\sqrt{2d_{i_{k+1}}}} \\ &= \frac{d_{i_k}a_{i_k i_{k+1}}}{\sqrt{d_{i_k}d_{i_{k+1}}}} \\ &= -\sqrt{a_{i_k i_{k+1}}a_{i_{k+1}i_k}}. \end{align*} Because $a_{i_k i_{k+1}}$ and $a_{i_{k+1}i_k}$ are negative integers, their product is a positive integer, so \begin{align*} -\sqrt{a_{i_k i_{k+1}}a_{i_{k+1}i_k}} \leq -1. \end{align*} All additional edge terms among the selected vertices are also nonpositive, and all nonedge terms are zero. Therefore \begin{align*} x^\top Bx \leq m + \sum_{k=1}^{m}(-1) = 0. \end{align*}[/step]

[guided]We now compute the quadratic form and isolate the contribution of the assumed cycle. Since $x_i=0$ away from the cycle, only vertices in $\{i_1,\dots,i_m\}$ can contribute. The diagonal terms are normalized to be exactly $1$ each: \begin{align*} \sum_{i \in I} b_{ii}x_i^2 = \sum_{k=1}^{m} b_{i_k i_k}x_{i_k}^2 = \sum_{k=1}^{m} (2d_{i_k})(2d_{i_k})^{-1} = m. \end{align*} Now take one edge of the cycle, say $\{i_k,i_{k+1}\}$. Its cross term in $x^\top Bx$ is \begin{align*} 2b_{i_k i_{k+1}}x_{i_k}x_{i_{k+1}}. \end{align*} Using $b_{ij}=d_i a_{ij}$ and the definition $x_i=(2d_i)^{-1/2}$ on cycle vertices, we get \begin{align*} 2b_{i_k i_{k+1}}x_{i_k}x_{i_{k+1}} &= 2d_{i_k}a_{i_k i_{k+1}} \cdot \frac{1}{\sqrt{2d_{i_k}}} \cdot \frac{1}{\sqrt{2d_{i_{k+1}}}} \\ &= \frac{d_{i_k}a_{i_k i_{k+1}}}{\sqrt{d_{i_k}d_{i_{k+1}}}}. \end{align*} Since $B$ is symmetric, $d_{i_k}a_{i_k i_{k+1}}=d_{i_{k+1}}a_{i_{k+1}i_k}$. Both Cartan entries are negative on an edge, so the preceding expression is the negative square root: \begin{align*} 2b_{i_k i_{k+1}}x_{i_k}x_{i_{k+1}} = -\sqrt{a_{i_k i_{k+1}}a_{i_{k+1}i_k}}. \end{align*} The product $a_{i_k i_{k+1}}a_{i_{k+1}i_k}$ is a positive integer, hence its square root is at least $1$. Therefore every cycle edge contributes at most $-1$. There may be extra edges among the chosen vertices, for example chords of the cycle. Those only make the quadratic form smaller, because their cross terms are also nonpositive. Nonedges contribute zero because $a_{ij}=a_{ji}=0$. Hence \begin{align*} x^\top Bx \leq m + \sum_{k=1}^{m}(-1) = 0. \end{align*}[/guided]

[step:Contradict positive definiteness and conclude acyclicity]The vector $x$ is nonzero, while $B$ is positive definite. Therefore positive definiteness requires \begin{align*} x^\top Bx > 0. \end{align*} This contradicts the estimate $x^\top Bx \leq 0$. Hence $G(A)$ contains no simple cycle. The theorem assumes the Dynkin diagram is connected, so its underlying graph is connected. Thus, by the graph-theoretic characterization of a [tree](/page/Tree), the connected finite graph $G(A)$ is a tree. In particular, the underlying graph of the connected finite type Dynkin diagram contains no cycle.[/step]

[guided]We now use the point of passing to $B$: positive definiteness forbids a nonzero vector from having nonpositive quadratic form. The vector $x \in \mathbb{R}^{I}$ was defined by \begin{align*} x_i := \begin{cases} (2d_i)^{-1/2}, & \text{if } i \in \{i_1,\dots,i_m\},\\ 0, & \text{otherwise,} \end{cases} \end{align*} and it is nonzero because the assumed cycle has vertices and each $d_i$ is positive. Since $B$ is positive definite, the defining property of positive definiteness gives \begin{align*} x^\top Bx > 0. \end{align*} But the previous estimate, obtained by adding the diagonal contribution $m$ and the at-most-$-1$ contribution from each of the $m$ cycle edges while observing that all remaining selected-vertex cross terms are nonpositive or zero, gives \begin{align*} x^\top Bx \leq 0. \end{align*} These two inequalities are incompatible. Therefore the assumption that $G(A)$ contains a simple cycle is false, so $G(A)$ contains no simple cycle. The theorem assumes that the Dynkin diagram is connected, so its underlying graph $G(A)$ is connected. By the graph-theoretic characterization of a [tree](/page/Tree), a finite connected graph with no simple cycle is a tree. Consequently the underlying graph of the connected finite type Dynkin diagram contains no cycle, which is the asserted conclusion.[/guided]