The exercise goes like this:
Let $\Phi$ be an irreducible root system. Prove that each $\lambda_i$ is of the form $\sum_j q_{ij}\alpha_j$, where all $q_{ij}$ are positive rational numbers.
Here, as discussed in Sect. 13.1, $\Delta := \{\alpha_1, \ldots, \alpha_n \}$ is a base of $\Phi$, and $\{\lambda_i\}$ is the dual basis of $\{\alpha_1^\vee, \ldots, \alpha_n^\vee \}$.
Following Humphrey’s hint, I have proved that all $q_{ij}$ are nonnegative rational numbers and all $q_{ii}$ are positive via the previous exercise, observing that the basis $\{\alpha_1^\vee, \ldots, \alpha_n^\vee \}$ is obtuse (i.e. their pairwise inner product is nonpositive).
Question 1: Then Humphrey asked us to show:
If $q_{ij}>0$ and $(\alpha_j, \alpha_k) < 0$, then $q_{ik}>0$.
I got stuck here and do not know how to show this.
My attempts: I can only express $q_{ij}$ as
$$
q_{ij} = \dfrac{2}{(\alpha_j, \alpha_j)} (\lambda_i, \lambda_j),
$$
using the dual basis relation and I cannot see how the condition $(\alpha_j, \alpha_k) < 0$ can be used to illustrate the claim above.
Question 2: How can the above claim imply the result that all $q_{ij}$ are positive?
My attempts: As for each fixed $i$, we have $q_{ii}>0$. Then for any $k \neq i$, if $(\alpha_i, \alpha_k)<0$, then indeed $q_{ik}>0$. But what about the case $(\alpha_i, \alpha_k)=0$? If I could find some $j \neq i$ such that both $(\alpha_j, \alpha_k)<0$ and $(\alpha_i, \alpha_j)<0$ holds, then using $q_{ij}>0$ as a bridge, we are done. But why such a $j$ exists? Then I got stuck here. The condition that $\Phi$ is irreducible has not been used. So maybe here we have to apply this? (As Humphrey gave this exercise, it seems that it is not encouraged to use the classification theorem of irreducible root systems (and the table of their Cartan matrices)).
Thank you all for commenting and answering!
Best Answer
Question 1: Suppose that $q_{ik}=0$. Then $0\leq (\lambda_{i},\alpha_{k})=\sum\limits_{r\neq k}q_{ir}(\alpha_{r},\alpha_{k})$. Since for each simple root $\alpha_{r}\neq \alpha_{k}$ we have $(\alpha_{r},\alpha_{k})\leq 0$, the sum $\sum\limits_{r\neq k}q_{ir}(\alpha_{r},\alpha_{k})$ is also non-positive. This implies that $q_{ir}(\alpha_{r},\alpha_{k})=0$ for each $r\neq k$. In particular, $q_{ij}(\alpha_{j},\alpha_{k})=0$, which contradicts with the condition. Hence $q_{ik}>0$.
Question 2: Since $\Phi$ is irreducible, its coxter graph is connected. So for any two simple roots $\alpha_{i},\alpha_{j}$, we can find a set of simple roots $\alpha_{i_{0}}=\alpha_{i},\alpha_{i_{1}},\alpha_{i_{2}},\cdots,\alpha_{i_{r}}=\alpha_{j}$, such that $(\alpha_{i_{s}},\alpha_{i_{s+1}})<0$. So we can use the result in Question 1: $q_{ii}>0,(\alpha_{i_{0}},\alpha_{i_{1}})<0$ implies that $q_{i_{0},i_{1}}>0$. Repeat this argument and we obtain that $q_{i_{0}i_{r}}=q_{ij}>0$.