I'm confused about the following comment in Knapp's Lie Groups 2ed, page 397. Here, $\Delta$ is a root system associated to a complex semisimple Lie algebra, $\alpha, \beta$ are orthogonal roots and we also have that $\alpha \pm \beta \in \Delta$ (that is, they are not strongly orthogonal).
I believe that the phrase simple component here means irreducible component. Since $\alpha \pm \beta$ are roots, $\alpha$ and $\beta$ must lie in the same irreducible component of $\Delta$. Further, since $\Delta$ comes from a semisimple Lie algebra, this irreducible component is a reduced root system (and hence the comment about only two root lengths).
Question: What set of simple roots within (an irreducible component of) $\Delta$ is the author using to obtain the Cartan matrix/Dynkin diagram with double line?
Attempt: I'm guessing there might be a way to show the existence of a simple system $\Pi \subset \Delta$ containing both $\alpha$ and $\beta - \alpha$. This would show that \begin{equation} 2\frac{(\alpha, \beta -\alpha)}{|\alpha|^2} \cdot 2 \frac{(\alpha, \beta-\alpha)}{|\beta-\alpha|^2} = 2, \end{equation} thereby giving the double line. But I don't see how to construct this $\Pi$.
For example, if someone can provide a proof/reference showing that one can choose a simple system containing both $\alpha$ and $\beta - \alpha$, I will be happy.
To put my comments into a proper answer:
We can see that the number of lines between two nodes in the diagram is given by $$N := \langle\alpha_i, \alpha_j\rangle\langle\alpha_j, \alpha_i\rangle = 2\frac{(\alpha_i, \alpha_j)}{|\alpha_i|^2} \cdot 2 \frac{(\alpha_j, \alpha_i)}{|\alpha_j|^2} = 4\frac{(\alpha_i, \alpha_j)^2}{|\alpha_i|^2|\alpha_j|^2} = 4 cos^2 (\theta).$$ Where $\theta$ is the angle between them. Since we have $ \langle\alpha_i, \alpha_j\rangle,\langle\alpha_j, \alpha_i\rangle,N $ all integers and from the above they are between $-3$ and $3$ ($N$ is also positive) there are only a few possibilities. Additionally we have (as long as $\alpha_i, \alpha_j$ are not orthogonal): $$ \frac{\langle\alpha_i, \alpha_j\rangle}{\langle\alpha_j, \alpha_i\rangle} = \frac{|\alpha_i|^2}{|\alpha_j|^2}.$$
Comparing the possibilities, it is clear that $\frac{|\alpha_i|^2}{|\alpha_j|^2}$ is in fact equal to $N$ whenever $N$ is non-zero (assuming $\alpha_i$ is the longer root). So now we just need the fact that if there are two roots somewhere in our system with $\frac{|\alpha_i|^2}{|\alpha_j|^2} = 2$ then there are two non orthogonal simple roots (in any choice of simple roots) with the same ratio of lengths.
In fact it should be clear that a set of simple roots should automatically have at least one root of each length so we just need a pair that are non-orthogonal. But this is also immediate since the Dynkin diagram is connected and we have only two lengths present: At some point between our two originally chosen nodes we must have a transition from long to short.