[proofplan]
The finite root system direction is obtained by symmetrizing the Cartan matrix with the squared lengths of the simple roots; the resulting matrix is the Gram matrix of the simple roots, hence positive definite. Conversely, a positive diagonal symmetrizer gives a Euclidean inner product on the real [vector space](/page/Vector%20Space) with basis $\alpha_1,\dots,\alpha_n$. The generalized Cartan matrix then defines orthogonal reflections preserving the root lattice, and positive definiteness forces the generated reflection group to be finite. The orbit of the simple roots under this finite crystallographic reflection group is a finite reduced crystallographic root system whose Cartan matrix is exactly $A$.
[/proofplan]
[step:Recover a positive definite symmetrization from a finite root system]
Assume that $A$ is the Cartan matrix of a finite reduced crystallographic root system $\Phi$ in a Euclidean space $E$ with inner product $(\cdot,\cdot)_E$, and let $\Delta=\{\alpha_1,\dots,\alpha_n\}$ be the corresponding simple system. Thus
\begin{align*}
a_{ij}=\frac{2(\alpha_j,\alpha_i)_E}{(\alpha_i,\alpha_i)_E}
\end{align*}
for all $1 \leq i,j \leq n$.
Define $d_i=\frac{1}{2}(\alpha_i,\alpha_i)_E>0$ and let $D=\operatorname{diag}(d_1,\dots,d_n)$. Then
\begin{align*}
(DA)_{ij}
=d_i a_{ij}
=\frac{(\alpha_i,\alpha_i)_E}{2}\frac{2(\alpha_j,\alpha_i)_E}{(\alpha_i,\alpha_i)_E}
=(\alpha_j,\alpha_i)_E.
\end{align*}
Hence $DA$ is the Gram matrix of the ordered basis $\alpha_1,\dots,\alpha_n$ of $\operatorname{span}_{\mathbb{R}}\Phi$. Gram matrices of bases in a Euclidean space are symmetric positive definite, so $DA$ is symmetric positive definite.
[/step]
[step:Build the Euclidean space and the simple reflections from the symmetrizer]
Conversely, suppose that $D=\operatorname{diag}(d_1,\dots,d_n)$ has $d_i>0$ and that $B:=DA$ is symmetric positive definite. Let $V$ be the real vector space with basis $\alpha_1,\dots,\alpha_n$. Define a [bilinear form](/page/Bilinear%20Form)
\begin{align*}
(\cdot,\cdot)_B:V \times V &\to \mathbb{R}
\end{align*}
by prescribing
\begin{align*}
(\alpha_i,\alpha_j)_B=B_{ij}=d_i a_{ij}.
\end{align*}
Since $B$ is symmetric positive definite, $(\cdot,\cdot)_B$ is a Euclidean inner product on $V$.
For each $1 \leq i \leq n$, define a [linear map](/page/Linear%20Map)
\begin{align*}
s_i:V &\to V
\end{align*}
on the basis $\alpha_1,\dots,\alpha_n$ by
\begin{align*}
s_i(\alpha_j)=\alpha_j-a_{ij}\alpha_i.
\end{align*}
Because $a_{ii}=2$, we have $s_i(\alpha_i)=-\alpha_i$. For arbitrary $1 \leq i,j \leq n$,
\begin{align*}
\frac{2(\alpha_j,\alpha_i)_B}{(\alpha_i,\alpha_i)_B}
=
\frac{2d_j a_{ji}}{2d_i}
=
\frac{d_j a_{ji}}{d_i}
=
a_{ij},
\end{align*}
where the final equality uses the [symmetry condition](/theorems/1360) $d_i a_{ij}=d_j a_{ji}$. Therefore
\begin{align*}
s_i(v)=v-\frac{2(v,\alpha_i)_B}{(\alpha_i,\alpha_i)_B}\alpha_i
\end{align*}
for every $v \in V$. Thus $s_i$ is the orthogonal reflection across the hyperplane $\alpha_i^\perp=\{v \in V:(v,\alpha_i)_B=0\}$.
[guided]
The matrix $B=DA$ is positive definite, so it can be used as a Gram matrix. We therefore define the real vector space $V$ with basis $\alpha_1,\dots,\alpha_n$ and declare the inner products of basis vectors to be
\begin{align*}
(\alpha_i,\alpha_j)_B=d_i a_{ij}.
\end{align*}
Symmetry of $B$ makes this bilinear form symmetric, and positive definiteness of $B$ makes it a Euclidean inner product.
Now define $s_i$ by the formula forced by the Cartan matrix:
\begin{align*}
s_i(\alpha_j)=\alpha_j-a_{ij}\alpha_i.
\end{align*}
To identify this map geometrically, compare it with the orthogonal reflection formula. Since $a_{ii}=2$, we have
\begin{align*}
(\alpha_i,\alpha_i)_B=d_i a_{ii}=2d_i.
\end{align*}
Also, symmetry of $B=DA$ gives $d_i a_{ij}=d_j a_{ji}$. Hence
\begin{align*}
\frac{2(\alpha_j,\alpha_i)_B}{(\alpha_i,\alpha_i)_B}
=
\frac{2d_j a_{ji}}{2d_i}
=
\frac{d_j a_{ji}}{d_i}
=
a_{ij}.
\end{align*}
Therefore the formula defining $s_i$ becomes
\begin{align*}
s_i(\alpha_j)
=
\alpha_j-\frac{2(\alpha_j,\alpha_i)_B}{(\alpha_i,\alpha_i)_B}\alpha_i,
\end{align*}
which is exactly the orthogonal reflection in the hyperplane perpendicular to $\alpha_i$. Since the basis vectors span $V$, the same reflection formula holds for every $v \in V$.
[/guided]
[/step]
[step:Show that the reflection group is finite]
Let $Q=\bigoplus_{i=1}^n \mathbb{Z}\alpha_i$ be the root lattice, and let $W \leq GL(V)$ be the subgroup generated by $s_1,\dots,s_n$. Since every $a_{ij}$ is an integer, each $s_i$ maps $Q$ into $Q$. Since $s_i^2=\operatorname{id}_V$, each $s_i$ maps $Q$ bijectively onto $Q$, so every element of $W$ preserves $Q$.
Each $s_i$ is orthogonal for $(\cdot,\cdot)_B$, hence every $w \in W$ preserves $(\cdot,\cdot)_B$. For each basis vector $\alpha_i$, the orbit $W\alpha_i$ lies in the set
\begin{align*}
\{\lambda \in Q : (\lambda,\lambda)_B=(\alpha_i,\alpha_i)_B\}.
\end{align*}
This set is finite because $(\cdot,\cdot)_B$ is positive definite and $Q$ is a discrete lattice in $V$. Therefore each orbit $W\alpha_i$ is finite.
The action of $W$ on the ordered basis $\alpha_1,\dots,\alpha_n$ determines each element of $W$ uniquely. Since each $\alpha_i$ has only finitely many possible images under $W$, the group $W$ is finite.
[/step]
[step:Construct the finite crystallographic root system]
Define
\begin{align*}
\Phi=\{w\alpha_i : w \in W,\ 1 \leq i \leq n\}\subset V.
\end{align*}
Since $W$ is finite, $\Phi$ is finite. Since $\alpha_1,\dots,\alpha_n \in \Phi$ and form a basis of $V$, the set $\Phi$ spans $V$.
For $\beta=w\alpha_i \in \Phi$, define the reflection
\begin{align*}
s_\beta:V &\to V,\\
v &\mapsto v-\frac{2(v,\beta)_B}{(\beta,\beta)_B}\beta.
\end{align*}
Because $w$ is orthogonal, $s_\beta=w s_i w^{-1}$. Hence $s_\beta(\Phi)=\Phi$.
The set $\Phi$ is reduced. Indeed, each $\alpha_i$ is primitive in the lattice $Q$, and every $w \in W$ is a lattice automorphism of $Q$, so every root $w\alpha_i$ is primitive in $Q$. If $\beta,\gamma \in \Phi$ and $\gamma=c\beta$ for some $c \in \mathbb{R}$, then primitivity of both lattice vectors forces $c=\pm 1$.
Finally, $\Phi$ is crystallographic. Let $\beta,\gamma \in \Phi$. Since $s_\beta=w s_i w^{-1}$ for some $w \in W$ and some $i$, the map $s_\beta$ preserves $Q$. Thus
\begin{align*}
s_\beta(\gamma)-\gamma
=
-\frac{2(\gamma,\beta)_B}{(\beta,\beta)_B}\beta
\end{align*}
lies in $Q \cap \mathbb{R}\beta$. Since $\beta$ is primitive in $Q$, $Q \cap \mathbb{R}\beta=\mathbb{Z}\beta$. Therefore
\begin{align*}
\frac{2(\gamma,\beta)_B}{(\beta,\beta)_B}\in \mathbb{Z}.
\end{align*}
Thus $\Phi$ is a finite reduced crystallographic root system.
The standard chamber theorem for finite crystallographic reflection groups says that, for a finite crystallographic reflection group generated by the simple reflections attached to a basis $\Delta=\{\alpha_1,\dots,\alpha_n\}$ with $a_{ij}\leq 0$ for $i\neq j$, the set $\Delta$ is a simple system for $\Phi$ and every root in $\Phi$ is an integral linear combination of the $\alpha_i$ with coefficients either all nonnegative or all nonpositive. This is the only external finite-reflection-group input used here; it is not yet linked in the wiki.
[/step]
[step:Compute the Cartan matrix of the constructed root system]
With respect to the simple system $\Delta=\{\alpha_1,\dots,\alpha_n\}$, the Cartan integer of the constructed root system is
\begin{align*}
\frac{2(\alpha_j,\alpha_i)_B}{(\alpha_i,\alpha_i)_B}
=
\frac{2d_j a_{ji}}{2d_i}
=
\frac{d_j a_{ji}}{d_i}
=
a_{ij}.
\end{align*}
Thus the Cartan matrix of $\Phi$ with respect to $\Delta$ is precisely $A$. This proves that $A$ is the Cartan matrix of a finite reduced crystallographic root system, completing the equivalence.
[/step]
