[proofplan] The diagonal entries follow immediately from the definition of $a_{ij}$. For distinct simple roots, the simple-root sign property rules out a positive Cartan integer: reflecting one simple root across the hyperplane orthogonal to the other would produce a root with both positive and negative simple-root coefficients. The vanishing condition is exactly orthogonality of the two roots, which is symmetric. Finally, multiplying the $i$-th row by half the squared length of $\alpha_i$ recovers the symmetric Gram matrix of the simple roots. [/proofplan]

[step:Compute the diagonal entries from the definition] Fix $i \in \{1,\dots,n\}$. Since $\alpha_i$ is a root, $\alpha_i \ne 0$, so $(\alpha_i,\alpha_i)>0$. By the definition of the Cartan entry, \begin{align*} a_{ii} = \frac{2(\alpha_i,\alpha_i)}{(\alpha_i,\alpha_i)} = 2. \end{align*} Thus $a_{ii}=2$ for every $1 \leq i \leq n$. [/step]

[step:Show that distinct simple roots have nonpositive Cartan entries]Let $i,j \in \{1,\dots,n\}$ with $i\ne j$. Because $\Phi$ is crystallographic, the Cartan integer \begin{align*} a_{ij}=\frac{2(\alpha_i,\alpha_j)}{(\alpha_i,\alpha_i)} \end{align*} belongs to $\mathbb{Z}$. It remains to prove $a_{ij}\leq 0$. Suppose, for contradiction, that $a_{ij}>0$. Since $a_{ij}\in \mathbb{Z}$, this gives $a_{ij}\geq 1$. Let $s_i:V\to V$ be the reflection in the hyperplane orthogonal to $\alpha_i$, defined by \begin{align*} s_i(v):=v-\frac{2(\alpha_i,v)}{(\alpha_i,\alpha_i)}\alpha_i. \end{align*} The root system axiom gives $s_i(\Phi)=\Phi$, so $s_i(\alpha_j)\in \Phi$. Computing the reflection gives \begin{align*} s_i(\alpha_j) = \alpha_j-\frac{2(\alpha_i,\alpha_j)}{(\alpha_i,\alpha_i)}\alpha_i = \alpha_j-a_{ij}\alpha_i. \end{align*} In the simple root basis $\Delta$, this root has coefficient $1$ on $\alpha_j$ and coefficient $-a_{ij}\leq -1$ on $\alpha_i$. Thus it is neither a nonnegative integer linear combination of simple roots nor a nonpositive integer linear combination of simple roots. This contradicts the defining sign property of a simple root basis, namely that every root in $\Phi$ is either a nonnegative integer linear combination of elements of $\Delta$ or a nonpositive integer linear combination of elements of $\Delta$. Therefore $a_{ij}\leq 0$. Combining this with $a_{ij}\in\mathbb{Z}$ gives \begin{align*} a_{ij}\in \mathbb{Z}_{\leq 0}. \end{align*}[/step]

[guided]Fix distinct indices $i$ and $j$. The crystallographic hypothesis is used first: it says that for any two roots $\alpha,\beta\in\Phi$, the number \begin{align*} \frac{2(\alpha,\beta)}{(\alpha,\alpha)} \end{align*} is an integer. Applying this with $\alpha=\alpha_i$ and $\beta=\alpha_j$ gives $a_{ij}\in\mathbb{Z}$. We now prove that this integer cannot be positive. Assume $a_{ij}>0$. Since $a_{ij}$ is an integer, this means $a_{ij}\geq 1$. Define the reflection \begin{align*} s_i:V&\to V\\ v&\mapsto v-\frac{2(\alpha_i,v)}{(\alpha_i,\alpha_i)}\alpha_i. \end{align*} By the root system axioms, $s_i$ preserves the root system, so $s_i(\alpha_j)\in\Phi$. Substituting $v=\alpha_j$ gives \begin{align*} s_i(\alpha_j) = \alpha_j-\frac{2(\alpha_i,\alpha_j)}{(\alpha_i,\alpha_i)}\alpha_i = \alpha_j-a_{ij}\alpha_i. \end{align*} Now we use the defining property of a simple root basis. Every root must have all simple-root coefficients either nonnegative or nonpositive. But the root $\alpha_j-a_{ij}\alpha_i$ has coefficient $1$ on $\alpha_j$ and coefficient $-a_{ij}\leq -1$ on $\alpha_i$. Since $i\ne j$, these are two different simple-root coordinates, one positive and one negative. Hence this vector cannot be a root when expressed relative to the simple basis $\Delta$, contradicting $s_i(\alpha_j)\in\Phi$. Therefore the assumption $a_{ij}>0$ is impossible. Since $a_{ij}$ is an integer, we conclude $a_{ij}\in\mathbb{Z}_{\leq 0}$.[/guided]

[step:Identify zero Cartan entries with orthogonality] Let $i,j \in \{1,\dots,n\}$ with $i\ne j$. Since $(\alpha_i,\alpha_i)>0$, the formula \begin{align*} a_{ij}=\frac{2(\alpha_i,\alpha_j)}{(\alpha_i,\alpha_i)} \end{align*} shows that \begin{align*} a_{ij}=0 \iff (\alpha_i,\alpha_j)=0. \end{align*} Similarly, since $(\alpha_j,\alpha_j)>0$, \begin{align*} a_{ji}=0 \iff (\alpha_j,\alpha_i)=0. \end{align*} The inner product is symmetric, so $(\alpha_i,\alpha_j)=(\alpha_j,\alpha_i)$. Hence \begin{align*} a_{ij}=0 \iff a_{ji}=0. \end{align*} [/step]

[step:Rescale rows by root lengths to obtain a symmetric matrix] For each $i\in\{1,\dots,n\}$, define \begin{align*} d_i := \frac{(\alpha_i,\alpha_i)}{2}. \end{align*} Since $\alpha_i\ne 0$ and the inner product is positive definite, $d_i>0$. Let \begin{align*} D:=\operatorname{diag}(d_1,\dots,d_n)\in \mathbb{R}^{n\times n}. \end{align*} For $1\leq i,j\leq n$, the $(i,j)$-entry of $DA$ is \begin{align*} (DA)_{ij} = d_i a_{ij} = \frac{(\alpha_i,\alpha_i)}{2}\cdot \frac{2(\alpha_i,\alpha_j)}{(\alpha_i,\alpha_i)} = (\alpha_i,\alpha_j). \end{align*} By symmetry of the inner product, \begin{align*} (DA)_{ij} = (\alpha_i,\alpha_j) = (\alpha_j,\alpha_i) = (DA)_{ji}. \end{align*} Therefore $DA$ is symmetric. This proves the existence of positive [real numbers](/page/Real%20Numbers) $d_1,\dots,d_n$ with the required property. [/step]