[proofplan] We choose the diagonal matrix $D$ so that multiplication by $D$ cancels the denominator in each row of the Cartan matrix. With this choice, $DA$ becomes exactly the Gram matrix of the ordered simple roots with respect to the Euclidean inner product on $V$. Symmetry follows from symmetry of the inner product, and positive definiteness follows because the simple roots are linearly independent. [/proofplan] [step:Choose the diagonal symmetrising matrix from the root lengths] Define the diagonal matrix $D = (d_{ij}) \in \mathbb{R}^{n \times n}$ by \begin{align*} d_{ij} = \begin{cases} \frac{(\alpha_i,\alpha_i)}{2}, & i=j,\\ 0, & i \neq j. \end{cases} \end{align*} Since each root $\alpha_i$ is nonzero and $(\cdot,\cdot)$ is a Euclidean inner product, we have $(\alpha_i,\alpha_i)>0$ for every $1 \leq i \leq n$. Hence every diagonal entry of $D$ is a positive real number. [/step] [step:Identify $DA$ with the Gram matrix of the simple roots] For $1 \leq i,j \leq n$, the $(i,j)$-entry of $DA$ is \begin{align*} (DA)_{ij} &= d_{ii}a_{ij} \\ &= \frac{(\alpha_i,\alpha_i)}{2}\frac{2(\alpha_i,\alpha_j)}{(\alpha_i,\alpha_i)} \\ &= (\alpha_i,\alpha_j). \end{align*} Thus $DA$ is the Gram matrix $G = (g_{ij}) \in \mathbb{R}^{n \times n}$ of the ordered simple roots, where \begin{align*} g_{ij} = (\alpha_i,\alpha_j). \end{align*} Since the Euclidean inner product is symmetric, $g_{ij}=g_{ji}$ for all $i,j$, so $DA=G$ is symmetric. [guided] The purpose of the diagonal matrix $D$ is to multiply the $i$th row of $A$ by the positive scalar $(\alpha_i,\alpha_i)/2$. This is exactly the scalar needed to cancel the denominator in the definition of the Cartan matrix. For every pair of indices $1 \leq i,j \leq n$, we compute \begin{align*} (DA)_{ij} &= d_{ii}a_{ij} \\ &= \frac{(\alpha_i,\alpha_i)}{2}\frac{2(\alpha_i,\alpha_j)}{(\alpha_i,\alpha_i)} \\ &= (\alpha_i,\alpha_j). \end{align*} Therefore $DA$ is not merely symmetric by construction; it is the Gram matrix of the vectors $\alpha_1,\dots,\alpha_n$. The symmetry is then inherited from the inner product: \begin{align*} (DA)_{ij} = (\alpha_i,\alpha_j) = (\alpha_j,\alpha_i) = (DA)_{ji}. \end{align*} [/guided] [/step] [step:Prove positive definiteness using linear independence of the simple roots] Let $c=(c_1,\dots,c_n) \in \mathbb{R}^n$ be nonzero. Define the vector $\beta \in V$ by \begin{align*} \beta = \sum_{i=1}^{n} c_i\alpha_i. \end{align*} The simple roots $\alpha_1,\dots,\alpha_n$ are linearly independent, so $\beta \neq 0$. Using the Gram matrix identity from the previous step, we obtain \begin{align*} c^\top(DA)c &= \sum_{i=1}^{n}\sum_{j=1}^{n} c_i(DA)_{ij}c_j \\ &= \sum_{i=1}^{n}\sum_{j=1}^{n} c_i(\alpha_i,\alpha_j)c_j \\ &= \left(\sum_{i=1}^{n} c_i\alpha_i,\sum_{j=1}^{n} c_j\alpha_j\right) \\ &= (\beta,\beta). \end{align*} Since $(\cdot,\cdot)$ is positive definite and $\beta \neq 0$, we have $(\beta,\beta)>0$. Hence $c^\top(DA)c>0$ for every nonzero $c \in \mathbb{R}^n$, so $DA$ is positive definite. [guided] To prove that a real symmetric matrix is positive definite, we must show that its quadratic form is positive on every nonzero vector. Let $c=(c_1,\dots,c_n) \in \mathbb{R}^n$ be nonzero, and form the corresponding linear combination of simple roots: \begin{align*} \beta = \sum_{i=1}^{n} c_i\alpha_i \in V. \end{align*} The simple roots are linearly independent, so the coefficient vector $c \neq 0$ implies that this linear combination is not the zero vector. Thus $\beta \neq 0$. Now evaluate the quadratic form of $DA$. Since the previous step identified $(DA)_{ij}$ with $(\alpha_i,\alpha_j)$, we have \begin{align*} c^\top(DA)c &= \sum_{i=1}^{n}\sum_{j=1}^{n} c_i(DA)_{ij}c_j \\ &= \sum_{i=1}^{n}\sum_{j=1}^{n} c_i(\alpha_i,\alpha_j)c_j \\ &= \left(\sum_{i=1}^{n} c_i\alpha_i,\sum_{j=1}^{n} c_j\alpha_j\right) \\ &= (\beta,\beta). \end{align*} The final expression is strictly positive because $(\cdot,\cdot)$ is a Euclidean inner product and $\beta$ is nonzero. Therefore \begin{align*} c^\top(DA)c > 0 \end{align*} for every nonzero $c \in \mathbb{R}^n$, which is precisely positive definiteness of $DA$. [/guided] [/step] [step:Conclude symmetrisability and positive definiteness] The diagonal matrix $D$ has positive real diagonal entries, and the matrix $DA$ is symmetric positive definite. Therefore the Cartan matrix $A$ is symmetrisable by a positive diagonal matrix, with symmetrisation $DA$ positive definite. [/step]