Given a representation $R$ of some Lie algbra $g$, we can compute the corresponding representation $R'$ (in general reducible) for some subgroup Lie algebra $ g \supset g'$ by utilizing the weights in the Dynkin basis. (This means, we use the fundamental weights as a basis for our weight space. )
If the representation $R$ is given in terms of such a list of weights, we can find the subgroup representation by dividing the weights into several components, e.q. for $su(5) \supset su(3) \times su(2)$ we have $(1,0,1,0) \rightarrow (1,0) (1)$. (This corresponds to deleting the third node from the $su(5)$ Dynkin diagram. Doing this for all weights yields a new list of weights, where we can read of the corresponding subgroup representations. In the example above our weight $(1,0,1,0)$ tells us that in terms of the subgroup this belongs to $3$ of $su(3)$ and $2$ of $su(2)$.
My problem is understanding why this only works in the Dynkin basis. In contrast, if we use the simple roots as a basis for the weights of our representation we have, for example, no zeroes in the weights and we can't simply recognize $1$ dimensional representations of the subgroups.