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I'm studying Humphreys's Introduction to Lie Algebras and Representation Theory, and I'm working on the following problem from the book (Chapter 13, problem 8):

Let $\Phi$ be irreducible. Prove that each $\lambda_i$ is of the form $\sum_jq_{ij}\alpha_J$, where all $q_{ij}$ are positive rational number.

Here, $\lambda_i$ is a fundamental dominant weight, and every $\alpha$ is simple. The book gives a hint, too:

Deduce from a previous exercise that all $q_{ij}$ are nonnegative. From $(\lambda_i,\lambda_i) > 0$ obtain $q_{ij}>0$. Then show that if $q_{ij}> 0$ and $(\alpha_j,\alpha_k)<0$, then $q_{ik}>0$.

I've managed to show the first part of the hint, and I can complete the problem using it, but I can't figure out how to show that, if $q_{ij}>0$ and $(\alpha_j,\alpha_k)<0$, then $q_{ik}>0$.

Any help would be really appreciated.

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Assume that $q_{ij} > 0$ and $(\alpha_j, \alpha_k) < 0$. We know from Humphreys' problem 7 that $q_{ik} \ge 0$. If $q_{ik} = 0$, then we have $$0 \le \delta_{ik} (\alpha_k, \alpha_k)/2 = (\lambda_i, \alpha_k) = \sum_{m\neq k} q_{im} (\alpha_m, \alpha_k) \le 0$$ hence $q_{im} = 0$ or $(\alpha_m, \alpha_k) = 0$ for all $m\neq k$. This contradicts our assumptions on $m = j$. Hence, it must be that $q_{ik} > 0$.

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