TITLE: How to prove that $B^\vee$ is a base for coroots?
QUESTION [10 upvotes]: Let $\Phi$ be a root system in a real inner product space $E$. Define $\alpha^\vee = \frac{2\alpha}{(\alpha, \alpha)}$. Then the set $\Phi^\vee = \{\alpha^\vee: \alpha \in \Phi \}$ is also a root system.
Let $B$ be a base for the root system $\Phi$, ie. $B$ is a basis for $E$ and each $\alpha \in \Phi$ can be written as $\alpha = \sum_{\beta \in B} {k_\beta} \beta$ such that $k_\beta$ are integers of same sign.

Question: How to prove that $B^\vee = \{ \alpha^\vee: \alpha \in B\}$ is a base for the root system $\Phi^\vee$?

It is easy to see that $B^\vee$ is a basis for $E$. For the other property, here is what I have so far. Let $\alpha = \sum_{\beta \in B} {k_\beta} \beta$ be a root. Then
$$\alpha^\vee = \sum_{\beta \in B} \frac{k_\beta (\beta, \beta)}{(\alpha, \alpha)} \beta^\vee$$
so to prove the claim, it is necessary to prove that $\frac{k_\beta (\beta, \beta)}{(\alpha, \alpha)}$ is an integer. I don't see how that follows. In the case where $(\alpha, \beta) = 0$ the proportion $\frac{(\beta, \beta)}{(\alpha, \alpha)}$ could be anything, so I am a bit confused.

REPLY [8 votes]: Here is a different solution, which is perhaps the one hinted at in the book by Humphreys. See the comments in the answer by mezhang. The notation and terminology is that used by Humphreys in Chapter III of his book.
Now $\Delta = \Delta(\gamma)$ for some regular element $\gamma$ in the Euclidean space $E$ where the root system $\Phi$ resides. Since $\alpha$ and $\alpha^\vee$ determine the same hyperplanes, $\gamma$ is regular also with respect to $\Phi^\vee$. Thus $\Delta^\vee(\gamma)$ forms a base for $\Phi^\vee$. Here $\Delta^\vee(\gamma)$ is the set of roots of $\Phi^\vee$ that are indecomposable with respect to $\gamma$. 
We prove that $\Delta^\vee \subseteq \Delta^\vee(\gamma)$, which gives equality since both sets are bases for $E$. Let $\alpha^\vee \in \Delta^\vee$. If $\alpha^\vee = \beta_1^\vee + \beta_2^\vee$ for $\beta_i \in (\Phi)^+(\gamma)$, then $\alpha$ is a linear combination of two distinct positive roots, with positive coefficients. But this is not possible since $\Delta$ is a base. Thus $\alpha^\vee$ is indecomposable, ie. $\alpha^\vee \in \Delta^\vee(\gamma)$.