# Show that every dominant fundamental weight is decomposed as a positive rational sum of roots.

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.

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 $$q_{ik} > 0$$.