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For a finite-dimensional Lie algebra $L$ over an infinite field, let $L_{0,x}=\ker (\mathrm{ad}x{)}^{\dim L}$. A member of minimum dimension among these generalized zero spaces is a Cartan subalgebra, using the standard fact that each $L_{0,x}$ is a self-normalizing Lie subalgebra. The key minimality argument is more general: if $A\subseteq H=L_{0,x}$ and $H$ minimizes dimension for $x\in A$, then $H\subseteq L_{0,y}$ for every $y\in A$.

Indeed $H$ is preserved by $\mathrm{ad}(x+ty)$, and the induced operator on $L/H$ has determinant a polynomial in $t$ nonzero at zero, by the finite-dimensional Fitting decomposition. For all but finitely many $t$, its generalized zero space lies in $H$ and hence equals $H$ by minimality. The restriction to $H$ is therefore nilpotent for all such $t$. Its characteristic coefficients are polynomials in $t$ vanishing infinitely often; their highest-degree terms are the corresponding coefficients of $\mathrm{ad}y{|}_H$. These vanish, so that restriction is nilpotent. For a global minimum set $A=H$; Engel's theorem makes $H$ nilpotent and self-normalization gives the result.

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.

The dimensions of the nested images of a linear mapcan strictly decrease at most $n$ times. If all the first $n$ inclusions are strict, $\mathcal{R}(h^n)=0$ and is stable. Otherwise an earlier equality occurs and (d) makes every subsequent image equal. Thus in all casesBy the rank-nullity theorem and inclusion of the null spaces, this also gives $\ker h^n=\ker h^{2n}$. If $v\in \mathcal{R}(h^n)\cap \ker h^n$, write $v=h^{n}u$. Then $0=h^{n}v=h^{2n}u$, so $u\in \ker h^{2n}=\ker h^n$ and $v=0$. Hence the intersection is zero. Finally the rank-nullity theorem givesZero intersection and this dimension sum proveThis is the finite-dimensional Fitting decomposition. It includes the zero-dimensional case, for which $V=0$ and the assertion is immediate.

For a linear endomorphism $h$ of an $n$-dimensional vector space, the images of a linear map formed by its powers decrease and stabilize by exponent $n$. Indeed, $h^{m+k}v=h^m(h^{k}v)$ proves inclusion. If equality occurs at one step, applying $h$ to both sides proves equality at the next step, and induction gives equality forever after. Before stabilization, every inclusion is strict and lowers dimension by at least one. Starting at dimension $n$ allows at most $n$ such decreases.

By the rank-nullity theorem, the increasing null spaces of the same powers stabilize at the same time. This gives a short proof of the finite-dimensional Fitting decomposition.