 Solution

A reduced crystallographic root system in a real Euclidean space is a finite spanning subset $\Phi$ not containing zero, with $\Phi \cap \mathbb{R}\alpha =\{\alpha ,-\alpha \}$, closed under $s_{\alpha }(v)=v-2(v,\alpha )\alpha /(\alpha ,\alpha )$, and satisfying $2(\beta ,\alpha )/(\alpha ,\alpha )\in \mathbb{Z}$ for all roots. A base of a root system is a vector-space basis $\Delta \subseteq \Phi$ such that every root has integer coefficients of one sign in that basis.

Choose $h$ not orthogonal to any root and call roots with $(h,\alpha )>0$ positive. Let $\Delta$ be the positive roots that cannot be written as a sum of two positive roots. Repeated decomposition terminates because each summand has strictly smaller positive $h$-value in a finite set. Thus every positive root is a nonnegative integer combination of $\Delta$.

For distinct $\alpha ,\beta \in \Delta$, $(\alpha ,\beta )\le 0$: otherwise the allowed difference-root lemma would make either $\alpha -\beta$ or $\beta -\alpha$ positive, decomposing one of them. In a real linear relation among $\Delta$, move negative coefficients to the other side, obtaining $u=v$ as positive combinations with disjoint supports. Then $\Vert u{\Vert }^2=(u,v)\le 0$, so $u=0$. But a nonempty positive combination has $(h,u)>0$, a contradiction. Thus the elements are linearly independent. They span every root and hence $E$, proving existence of a base of a root system. The difference-root lemma is used only for distinct roots; that qualification excludes the zero difference.

For the displayed set, there are eight coordinate roots, twenty-four roots with two nonzero unit entries, and sixteen half-coordinate roots: forty-eight in total. It spans ${\mathbb{R}}^4$, is reduced, and has squared root lengths one and two. Signed coordinate permutations preserve it. Reflections in coordinate roots are sign changes, and reflections in roots $e_i\pm e_j$ are signed coordinate permutations. Every half-coordinate root is obtained by a signed permutation from $h=(1/2,1/2,1/2,1/2)$, so it remains to check its reflectionIt sends $e_i$ to $e_i-h$, a half-coordinate root; it fixes $e_i-e_j$ and sends $e_i+e_j$ to the negative sum of the other two coordinate vectors. For a half-coordinate root, the sum of its four signs is $0,\pm 2$ or $\pm 4$; the formula sends it, respectively, to itself, a coordinate root, or its negative when parallel to $h$. Thus reflection closure holds. For a long root, the required Cartan integer is an integer dot product; for a coordinate root it is twice a half-integer; for $h$ it is ${\sum }_i{\beta }_i\in \mathbb{Z}$. Symmetry gives the remaining cases. Therefore this is the F4 root system.

In the printed base order, the Gram matrix isUsing the convention $a_{ij}=2({\alpha }_i,{\alpha }_j)/({\alpha }_j,{\alpha }_j)$, the Cartan matrix isThe row-coroot convention transposes this matrix. The Dynkin diagram is the chain $1-2\Rightarrow 3-4$, with the double bond pointing toward the shorter third root:

The associated symmetric positive definite form, normalized as a squared length, isIndeed it equalswhich is strictly positive for nonzero $x$. The alternative Gram normalization $Q/2$ is the same positive definite form up to a positive scalar.