[proofplan] We first prove that every positive root is a non-negative integral combination of simple roots by descending through decompositions into sums of positive roots; finiteness of $\Phi^+$ forces the process to stop. Negative roots then follow by multiplying the positive expansion by $-1$. The remaining point is [linear independence](/page/Linear%20Independence): a non-trivial real relation among simple roots is split into two non-negative combinations with disjoint supports, and the standard root-system sign property for simple roots makes the inner product of the two sides non-positive, contradicting positivity of the norm of their common value. [/proofplan]

[step:Decompose every positive root into simple roots]Let $\rho \in \Phi^+$. We prove that there are integers $n_\alpha \geq 0$, indexed by $\alpha \in \Delta$, such that \begin{align*} \rho = \sum_{\alpha \in \Delta} n_\alpha \alpha. \end{align*} If $\rho \in \Delta$, take $n_\rho = 1$ and $n_\alpha = 0$ for $\alpha \neq \rho$. If $\rho \notin \Delta$, then by the definition of $\Delta$ there exist $\beta,\gamma \in \Phi^+$ such that \begin{align*} \rho = \beta + \gamma. \end{align*} If either $\beta$ or $\gamma$ is not simple, decompose it again as a sum of two positive roots. This iterative process terminates because $\Phi^+$ is finite and, after choosing any linear functional $\lambda: E \to \mathbb{R}$ defining the positive system by $\Phi^+ = \{\alpha \in \Phi : \lambda(\alpha) > 0\}$, each decomposition \begin{align*} \eta = \eta_1 + \eta_2, \qquad \eta_1,\eta_2 \in \Phi^+, \end{align*} satisfies \begin{align*} 0 < \lambda(\eta_1) < \lambda(\eta), \qquad 0 < \lambda(\eta_2) < \lambda(\eta). \end{align*} Thus no infinite descending chain of positive roots can occur. Therefore the process ends in finitely many simple roots, giving the required expression for $\rho$ as a non-negative integral combination of elements of $\Delta$.[/step]

[guided]We want to show that simple roots generate all positive roots. Let $\rho \in \Phi^+$ be arbitrary. If $\rho$ is already simple, then the expansion is immediate: \begin{align*} \rho = 1 \cdot \rho + \sum_{\alpha \in \Delta \setminus \{\rho\}} 0 \cdot \alpha. \end{align*} Now suppose $\rho$ is not simple. By the definition of a simple root, this means precisely that $\rho$ can be written as a sum of two positive roots: \begin{align*} \rho = \beta + \gamma, \qquad \beta,\gamma \in \Phi^+. \end{align*} If $\beta$ or $\gamma$ is not simple, we repeat the same operation on that root. The only point needing justification is termination. Choose a linear functional $\lambda: E \to \mathbb{R}$ defining the positive system, so that \begin{align*} \Phi^+ = \{\alpha \in \Phi : \lambda(\alpha) > 0\}. \end{align*} Whenever a positive root $\eta$ decomposes as \begin{align*} \eta = \eta_1 + \eta_2, \qquad \eta_1,\eta_2 \in \Phi^+, \end{align*} we have \begin{align*} \lambda(\eta) = \lambda(\eta_1) + \lambda(\eta_2), \end{align*} with both summands strictly positive. Hence \begin{align*} 0 < \lambda(\eta_1) < \lambda(\eta), \qquad 0 < \lambda(\eta_2) < \lambda(\eta). \end{align*} So every child root has strictly smaller $\lambda$-value than the root being decomposed. Because $\Phi^+$ is finite, the set $\lambda(\Phi^+)$ is finite. A strictly decreasing sequence inside this finite set cannot continue indefinitely. Therefore repeated decomposition must stop after finitely many steps, and it can stop only at roots that are simple. Collecting the terminal simple roots gives \begin{align*} \rho = \sum_{\alpha \in \Delta} n_\alpha \alpha \end{align*} for some integers $n_\alpha \geq 0$.[/guided]

[step:Extend the expansion from positive roots to all roots] Let $\rho \in \Phi$. If $\rho \in \Phi^+$, the previous step gives \begin{align*} \rho = \sum_{\alpha \in \Delta} n_\alpha \alpha \end{align*} with all $n_\alpha \in \mathbb{Z}_{\geq 0}$. If $\rho \in -\Phi^+$, then $-\rho \in \Phi^+$, so there are integers $m_\alpha \geq 0$ such that \begin{align*} -\rho = \sum_{\alpha \in \Delta} m_\alpha \alpha. \end{align*} Multiplying by $-1$ gives \begin{align*} \rho = \sum_{\alpha \in \Delta} (-m_\alpha)\alpha, \end{align*} where all coefficients $-m_\alpha$ are non-positive integers. Thus every root is an integral combination of simple roots whose coefficients have one sign. [/step]

[step:Show distinct simple roots have non-positive inner product]Let $\alpha,\beta \in \Delta$ with $\alpha \neq \beta$. We claim that \begin{align*} (\alpha,\beta) \leq 0. \end{align*} For a finite root system, the standard root-string property implies that if $(\alpha,\beta) > 0$, then $\alpha - \beta \in \Phi$. Since $\Phi = \Phi^+ \sqcup (-\Phi^+)$, either $\alpha - \beta \in \Phi^+$ or $\beta - \alpha \in \Phi^+$. If $\alpha - \beta \in \Phi^+$, then \begin{align*} \alpha = (\alpha - \beta) + \beta \end{align*} expresses the simple root $\alpha$ as a sum of two positive roots, contradiction. If $\beta - \alpha \in \Phi^+$, then \begin{align*} \beta = (\beta - \alpha) + \alpha \end{align*} expresses the simple root $\beta$ as a sum of two positive roots, contradiction. Hence $(\alpha,\beta) > 0$ is impossible, so $(\alpha,\beta) \leq 0$.[/step]

[guided]The key geometric fact about finite root systems is the root-string sign property: for distinct roots $\alpha,\beta \in \Phi$, if their inner product is positive, then the difference $\alpha - \beta$ is again a root. We apply this to two distinct simple roots $\alpha,\beta \in \Delta$. Assume for contradiction that \begin{align*} (\alpha,\beta) > 0. \end{align*} Then the root-string property gives \begin{align*} \alpha - \beta \in \Phi. \end{align*} Because $\Phi^+$ is a positive system, every root is either positive or the negative of a positive root. Therefore either \begin{align*} \alpha - \beta \in \Phi^+ \end{align*} or \begin{align*} -(\alpha - \beta) = \beta - \alpha \in \Phi^+. \end{align*} In the first case, \begin{align*} \alpha = (\alpha - \beta) + \beta \end{align*} writes $\alpha$ as a sum of two positive roots, namely $\alpha-\beta$ and $\beta$. This contradicts $\alpha \in \Delta$, since simple roots are exactly the positive roots that cannot be decomposed as sums of two positive roots. In the second case, \begin{align*} \beta = (\beta - \alpha) + \alpha \end{align*} writes $\beta$ as a sum of two positive roots, contradicting $\beta \in \Delta$. Both alternatives are impossible, so the assumption $(\alpha,\beta)>0$ must be false. Hence \begin{align*} (\alpha,\beta) \leq 0. \end{align*}[/guided]

[step:Split a linear relation into disjoint non-negative parts] Suppose that there is a real linear relation \begin{align*} \sum_{\alpha \in \Delta} c_\alpha \alpha = 0, \end{align*} where $c_\alpha \in \mathbb{R}$ and only finitely many $c_\alpha$ are non-zero. Define \begin{align*} \Delta_+ &:= \{\alpha \in \Delta : c_\alpha > 0\}, \\ \Delta_- &:= \{\alpha \in \Delta : c_\alpha < 0\}. \end{align*} Then \begin{align*} \sum_{\alpha \in \Delta_+} c_\alpha \alpha = \sum_{\beta \in \Delta_-} (-c_\beta)\beta. \end{align*} Let $v \in E$ denote this common vector: \begin{align*} v := \sum_{\alpha \in \Delta_+} c_\alpha \alpha = \sum_{\beta \in \Delta_-} (-c_\beta)\beta. \end{align*} The coefficients in both sums are non-negative, and the indexing sets $\Delta_+$ and $\Delta_-$ are disjoint. [/step]

[step:Use the inner product sign condition to force the relation to vanish]Using the two expressions for $v$, we compute \begin{align*} (v,v) &= \left( \sum_{\alpha \in \Delta_+} c_\alpha \alpha, \sum_{\beta \in \Delta_-} (-c_\beta)\beta \right) \\ &= \sum_{\alpha \in \Delta_+}\sum_{\beta \in \Delta_-} c_\alpha(-c_\beta)(\alpha,\beta). \end{align*} For every $\alpha \in \Delta_+$ and $\beta \in \Delta_-$, the roots $\alpha$ and $\beta$ are distinct because $\Delta_+ \cap \Delta_- = \varnothing$. By the previous step, \begin{align*} (\alpha,\beta) \leq 0. \end{align*} Since $c_\alpha > 0$ and $-c_\beta > 0$, every summand satisfies \begin{align*} c_\alpha(-c_\beta)(\alpha,\beta) \leq 0. \end{align*} Therefore \begin{align*} (v,v) \leq 0. \end{align*} But the inner product on $E$ is positive definite, so $(v,v) \geq 0$, with equality only when $v=0$. Hence $v=0$. If $\Delta_+$ were non-empty, then \begin{align*} v = \sum_{\alpha \in \Delta_+} c_\alpha \alpha \end{align*} would express the zero vector as a positive linear combination of positive roots. Choose the same defining functional $\lambda: E \to \mathbb{R}$ for $\Phi^+$. Applying $\lambda$ gives \begin{align*} 0 = \lambda(v) = \sum_{\alpha \in \Delta_+} c_\alpha \lambda(\alpha), \end{align*} where each term is strictly positive, a contradiction. Thus $\Delta_+ = \varnothing$. The same argument applied to \begin{align*} v = \sum_{\beta \in \Delta_-} (-c_\beta)\beta \end{align*} gives $\Delta_- = \varnothing$. Therefore all $c_\alpha = 0$, so $\Delta$ is linearly independent over $\mathbb{R}$.[/step]

[guided]We now prove linear independence. Suppose \begin{align*} \sum_{\alpha \in \Delta} c_\alpha \alpha = 0 \end{align*} is a real linear relation. Separate the positive and negative coefficients by defining \begin{align*} \Delta_+ &:= \{\alpha \in \Delta : c_\alpha > 0\}, \\ \Delta_- &:= \{\alpha \in \Delta : c_\alpha < 0\}. \end{align*} Moving the negative-coefficient terms to the other side gives \begin{align*} \sum_{\alpha \in \Delta_+} c_\alpha \alpha = \sum_{\beta \in \Delta_-} (-c_\beta)\beta. \end{align*} Let $v \in E$ be the common vector: \begin{align*} v := \sum_{\alpha \in \Delta_+} c_\alpha \alpha = \sum_{\beta \in \Delta_-} (-c_\beta)\beta. \end{align*} Why introduce $v$? Because we can compute its squared norm using the left expression in the first slot and the right expression in the second slot. Bilinearity of the inner product gives \begin{align*} (v,v) &= \left( \sum_{\alpha \in \Delta_+} c_\alpha \alpha, \sum_{\beta \in \Delta_-} (-c_\beta)\beta \right) \\ &= \sum_{\alpha \in \Delta_+}\sum_{\beta \in \Delta_-} c_\alpha(-c_\beta)(\alpha,\beta). \end{align*} The sets $\Delta_+$ and $\Delta_-$ are disjoint, so every pair $\alpha \in \Delta_+$ and $\beta \in \Delta_-$ consists of distinct simple roots. From the previous step, \begin{align*} (\alpha,\beta) \leq 0. \end{align*} Also $c_\alpha > 0$ and $-c_\beta > 0$. Hence every summand in the double sum is non-positive: \begin{align*} c_\alpha(-c_\beta)(\alpha,\beta) \leq 0. \end{align*} Therefore \begin{align*} (v,v) \leq 0. \end{align*} On the other hand, the inner product on $E$ is positive definite, so \begin{align*} (v,v) \geq 0, \end{align*} with equality if and only if $v=0$. Combining the two inequalities gives $v=0$. It remains to rule out the possibility that $v=0$ is a non-empty positive combination of positive roots. Let $\lambda: E \to \mathbb{R}$ be a defining functional for the positive system, so $\lambda(\alpha)>0$ for every $\alpha \in \Phi^+$. If $\Delta_+$ were non-empty, applying $\lambda$ to \begin{align*} 0 = v = \sum_{\alpha \in \Delta_+} c_\alpha \alpha \end{align*} would yield \begin{align*} 0 = \lambda(v) = \sum_{\alpha \in \Delta_+} c_\alpha \lambda(\alpha). \end{align*} Each term $c_\alpha\lambda(\alpha)$ is strictly positive, which is impossible. Hence $\Delta_+ = \varnothing$. The same argument applied to \begin{align*} 0 = v = \sum_{\beta \in \Delta_-} (-c_\beta)\beta \end{align*} shows $\Delta_- = \varnothing$. Thus there are no positive or negative coefficients, so every $c_\alpha$ is zero. This proves that $\Delta$ is linearly independent over $\mathbb{R}$.[/guided]

[step:Conclude that the simple roots form a base] We have shown that every root $\rho \in \Phi$ is an integral linear combination of elements of $\Delta$, with all coefficients non-negative when $\rho \in \Phi^+$ and all coefficients non-positive when $\rho \in -\Phi^+$. We have also shown that $\Delta$ is linearly independent over $\mathbb{R}$. Therefore $\Delta$ is a base of the root system $\Phi$. [/step]