[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*}
[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]
[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]
