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.

Use the standard split weight-space decomposition in characteristic zero, in particular the usual complex setting: $L={⨁}_{\phi }{L}_{\phi }$, with generalized weights for the nilpotent algebra $H$. The characteristic-zero condition prevents nonzero weight shifts from wrapping around; the unrestricted positive-characteristic version would need additional hypotheses.

For fixed $h\in H$, let $D=\mathrm{ad}h$. On a bracket of vectors from $L_{\phi },L_{\psi }$, the derivation identity givesIterating it gives the binomial expansion of these two commuting actions. At power $2n-1$, each term kills at least one factor, because generalized weight vectors are killed by the $n$th power of the corresponding shifted operator. Therefore the bracket lies in the generalized eigenspace for $(\phi +\psi )(h)$ for every $h$. Stabilization of kernels by dimension $n$ giveswhere $L_{\eta }=0$ when $\eta$ is not a weight.

Nilpotence of $H$ gives $H\subseteq L_0$. If $H$ is a Cartan subalgebra and $L_0/H\ne 0$, the action of $H$ on this quotient consists of nilpotent operators. Engel's theorem supplies a nonzero coset $v+H$ killed by $H$, giving $[H,v]\subseteq H$ with $v\notin H$, a contradiction to self-normalization. Conversely, if $L_0=H$ and $v$ normalizes $H$, then $[h,v]\in H$ for each $h$. Applying further powers of $\mathrm{ad}h$ kills this vector because its restriction to $H$ is nilpotent. Hence $v\in L_0=H$. This proves the equivalence.

The operator $\mathrm{ad}e$ sends $L_{\phi }$ into $L_{\phi +\alpha }$. Repetition follows a consecutive $\alpha$-string; after $r$ steps it must leave the weight set, because $r$ is the largest number of weights in any such string. Thus $\boxed{(\mathrm{ad}e{)}^r=0}$ on every weight space and hence on $L$.

Finally put $E=\mathrm{ad}e$ and $F=\mathrm{ad}f$. Expand $(E+F{)}^n$ as the sum of all words of length $n$, without commuting $E$ and $F$. Acting on $L_{\gamma }$, each word follows weights $\gamma +i\alpha +j\beta$ with $i+j$ increasing by one at each step. Linear independence of $\alpha ,\beta$ in characteristic zero makes the successive weights distinct. A nonzero length-$n$ path would require $n+1$ different weight spaces, but $\dim L=n$. Every word therefore vanishes, giving

A Cartan subalgebra is a nilpotent Lie subalgebra equal to its normalizer of a Lie subalgebra. Write $H=L_{0,x}$ for a choice $x\in A$ attaining the minimum dimension, and let $h=\dim H$. For $y\in A$, the subalgebra $H$ is invariant under $\mathrm{ad}(x+ty)$ because $A\subseteq H$. On $L/H$, the operator induced by $\mathrm{ad}x$ is invertible: $H$ is its full generalized zero eigenspace, and the finite-dimensional Fitting decomposition separates that from the invertible part.

The determinant of the operator induced by $\mathrm{ad}(x+ty)$ on $L/H$ is a polynomial in $t$, nonzero at zero. Thus it is nonzero for all but finitely many $t\in K$. For these values, the generalized zero eigenspace $L_{0,x+ty}$ lies in $H$. Minimality of $\dim H$ among elements from $A$ forces equality. Hence $\mathrm{ad}(x+ty){|}_H$ is nilpotent for all but finitely many $t$.

Each nonleading coefficient of its characteristic polynomial is a polynomial in $t$. Since $K$ is infinite, a polynomial vanishing at all but finitely many elements is identically zero. The coefficient of degree $i$ in the matrix entries is homogeneous of degree $i$; its highest possible $t^i$ coefficient is the corresponding characteristic coefficient of $\mathrm{ad}y{|}_H$. All these vanish, so that operator has characteristic polynomial $T^h$ and is nilpotent by the Cayley-Hamilton theorem. ThusFor existence, choose an $L_{0,x}=H$ of smallest dimension among all $x\in L$. Since $x\in H$, it also minimizes the same set with $x$ restricted to $A=H$. Applying the proved statement makes every $\mathrm{ad}y{|}_H$ nilpotent. Engel's theorem makes $H$ nilpotent, and the allowed self-idealizer property makes it self-normalizing. Therefore it is a Cartan subalgebra, illustrating Cartan subalgebra existence from a minimal Engel subalgebra.

For the three-dimensional example, direct use of the brackets gives $C_L(z)=Kz$ for every nonzero $z$: the equations $[u,z]=0$ are the vanishing of all two-by-two minors of the two coordinate vectors, so they are proportional. Every nonzero nilpotent Lie algebra has a nonzero center, as seen from its last nonzero lower-central term. A nilpotent subalgebra $H$ with a central nonzero element $z$ must therefore satisfy $H\subseteq C_L(z)=Kz$, and has dimension one. The zero subalgebra is not self-normalizing. Thus all Cartan subalgebras have dimension one. For example $Ka$ is nilpotent and its normalizer is $Ka$, so such subalgebras do occur.