Type of a root (sub)-system Let $E:=\{x\in \mathbb{R}^{l+1}:x_1+x_2+\cdots + x_{l+1}=0\}$ and let $\Phi\subseteq E$ denote its root system of type $A_l$ given the basis $\Delta=\{e_i-e_{i+1}, 1\leq i \leq l\}$ and with $\{e_i\}$ denoting the standard orthonormal basis of $\mathbb{R}^{l+1}$.
Let further $\Delta'\subseteq \Delta$ and consider $\Phi'=\text{span}_\mathbb{Z}(\Delta')\cap \Phi$ and $E'=\text{span}_\mathbb{R}(\Delta')\cap E$.
I want to determine the $\textit{type}$ of the root system $\Phi'\subseteq E'$.
Am I correct in assuming that one could consider the cut-out of the Cartan matrix of $\Phi$ corresponding to the Cartan matrix of $\Phi'$ in order to conclude that the Dynkin diagram of $\Phi'$ is given as a sum $\bigoplus A_{k}$ such that $\sum k= \dim(E')$ (and that all such choices of $k$ are possible)?
Thank you
 A: So if we take a subset of the simple roots $\Delta' \subset \Delta$ this corresponds to taking a subset of nodes of the Dynkin diagram. This gives a subdiagram where the nodes are this subset and the edges are only those edges connecting two nodes in this subset (i.e. no dangling edges). This diagram is then exactly the diagram of $\Phi'$. It is not too hard to prove this so I will leave that as an exercise. NB: we are only allowed to make these subdiagrams by removing nodes and the edges attached to them. We cannot just remove edges.
Thus by removing nodes from $A_l$ we can see that all the possible subdiagrams are of the form $A_{l_1}\times \cdots \times A_{l_k}$ for some $k$. Note that not all possible $l_i$'s will work and indeed we do not even have $l_1 + \cdots + l_k = l$. This should make sense because $A_l$ corresponds to $\mathfrak{sl}_{l+1}$ and what we are doing is dividing $\mathbb{C}^{l+1}$ into a direct sum of subspaces (They don't have to span the whole of $\mathbb{C}^{l+1}$) so we have $(l_1 + 1) + \cdots \cdots + (l_k + 1) = l_1 + \cdots + l_k + k \leq l +1$.
So some low dimensional examples. $A_2$ has only 1 proper, non-trivial subdiagram and that is of the form $A_1$. This corresponds to finding a copy of $\mathbb{C}^{2}$ inside $\mathbb{C}^{3}$. $A_3$ has three: $A_2$, $A_1 \times A_1$ and $A_1$ where we have removed one of the outer nodes, the middle node and two of the nodes respectively. These correspond to finding $\mathbb{C}^{3}\leq \mathbb{C}^{4}$, $\mathbb{C}^{2}\oplus\mathbb{C}^{2} = \mathbb{C}^{4}$ and $\mathbb{C}^{2}\leq \mathbb{C}^{4}$.
