First off, if is positive but not simple, then is a (positive) root for some simple . If for all , then the same argument we used when we showed is linearly independent would show that is linearly independent. But this is impossible because is already a basis.
So for some , and thus . It must be positive, since the height of must be at least . That is, at least one coefficient of with respect to must be positive, and so they all are.
In fact, every can be written (not uniquely) as the sum for a bunch of , and in such a way that each partial sum is itself a positive root. This is a great proof by induction on , for if then is in fact simple itself. If is not simple, then our argument above gives a so that for some with . And so on, by induction.
If is simple, then the reflection permutes the positive roots other than . That is, if , then as well. Indeed, we write
with all nonnegative. Clearly for some (otherwise ). But the coefficient of in must still be . Since this is positive, all the coefficients in the decomposition of are positive, and so . Further, it can’t be itself, because is the image .
In fact, this leads to a particularly useful little trick. Let be the half-sum of all the positive roots. That is,
then for all simple roots . The reflection shuffles around all the positive roots other than itself, which it sends to . This is a difference in the sum of , which the turns into .
Now take a bunch of (not necessarily distinct) and write . If , then there is some index that we can skip. That is,
Write for every from to , and . By our assumption, and . Thus there is some smallest index so that . Then , and we must have . But we know that . In particular,
And then we can write
From this we can conclude that if is an expression in terms of the basic reflections with as small as possible, then . Indeed, if , then
and we’ve just seen that in this case we can leave off as well as some in the expression for .