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Let $V$ be a finite-dimensional symplectic vector space over an algebraically closed field of characteristic zero. For $a\in \mathfrak{sp}(V)$, write its Jordan–Chevalley decomposition as $a=S+N$. Applying the symplectic adjoint to this decomposition and using $a^{\ast }=-a$, uniqueness gives $S^{\ast }=-S$ and $N^{\ast }=-N$. Thus both parts belong to the symplectic Lie algebra. The operators $\mathrm{ad}S$ and $\mathrm{ad}N$ commute. The former is diagonalizable on $\mathrm{End}(V)$ in a matrix unit basis for an eigenbasis of $S$, and the latter is a nilpotent linear map, as follows fromThe restrictions to the invariant symplectic Lie algebra retain these properties. Hence these restrictions form the Jordan–Chevalley decomposition of $\mathrm{ad}a$.

If $\mathrm{ad}a$ is diagonalizable, $\mathrm{ad}N=0$. For every $v\in V$, the rank-one map $T_v(w)=\omega (v,w)v$ lies in $\mathfrak{sp}(V)$, since the two terms in $\omega (T_{v}w,u)+\omega (w,T_{v}u)$ cancel by alternation. For $v\ne 0$ its image is $kv$, so commuting with $T_v$ forces $N$ to preserve $kv$. A linear map preserving every line is scalar: compare its action on two linearly independent vectors and on their sum. Since $N$ is a nilpotent linear map, that scalar is zero. Thus $a=S$ is diagonalizable. Conversely, if $a$ is diagonalizable on $V$, its operator commutator action is diagonalizable on $\mathrm{End}(V)$ and therefore on its invariant symplectic Lie algebra. This proves both directions, including $V=0$.

Over an algebraically closed field of characteristic zero, every three-dimensional perfect Lie algebra is isomorphic to the sl2 Lie algebra. Indeed the trace of a commutator gives $\mathrm{tr}(\mathrm{ad}x)=0$ for all $x$, because each element is a sum of Lie brackets. If every $\mathrm{ad}x$ were a nilpotent linear map, the Engel theorem would make the Lie algebra nilpotent, contradicting its constant nonzero lower central series of a Lie algebra. Some $\mathrm{ad}x$ therefore has a nonzero eigenvalue $\lambda$. Its other eigenvalues are $0,-\lambda$, since it kills $x$ and has zero trace. They are distinct, so its eigenvectors give a basis $h,e,f$ after scaling $x$ to $h=2x/\lambda$, with $[h,e]=2e$ and $[h,f]=-2f$. The Jacobi identity puts $[e,f]$ in the zero eigenspace $kh$. Its coefficient cannot vanish, since otherwise the derived algebra would lie in $ke\oplus kf$. Rescaling $f$ gives $[e,f]=h$. Thus this basis is an sl2 triple, proving the claimed Lie algebra isomorphism.

A finite-dimensional Lie algebra representation in which every representing endomorphism is nilpotent has a nonzero vector annihilated by the entire Lie algebra. Applying this to the Adjoint representation is the key step in Engel theorem.

Every linear endomorphism of a finite-dimensional vector space admits the displayed direct sum of vector spaces. It separates the part on which repeated application is invertible from the part eventually sent to zero.

Proof. Stabilization of images of powers of a linear map gives $\mathrm{im}{h}^n=\mathrm{im}{h}^{2n}$. The rank-nullity theorem and nested null spaces give $\ker h^n=\ker h^{2n}$. If $v=h^{n}u$ also satisfies $h^{n}v=0$, then $u\in \ker h^{2n}=\ker h^n$, so $v=0$. Thus the displayed summands have zero intersection. Their dimensions sum to $n$ by the rank-nullity theorem, proving the decomposition.

Both summands are preserved by $h$. On $\mathrm{im}{h}^n$, stabilization makes $h$ a surjective linear map, hence invertible in finite dimension. On $\ker h^n$, its $n$th power is zero, so the restriction is a nilpotent linear map.

For a linear map $T:V\to V$, set $K_i=\ker T^i$ and $K_0=0$. Since $T(K_{i+1})\subseteq K_i$ and $T(K_i)\subseteq K_{i-1}$, the rule $v+K_i\mapsto Tv+K_{i-1}$ defines a linear map $K_{i+1}/K_i\to K_i/K_{i-1}$. It is injective: if $Tv\in K_{i-1}$, then $T^{i}v=0$, so $v\in K_i$. In finite dimension, this provesThe linear map need not be surjective: for $T=0$ on a nonzero vector space, its $i=1$ instance is $0\to V$. For a nilpotent linear map, these dimensions are the increments in the kernel dimensions of powers of a nilpotent map.

For a nilpotent linear map with minimal polynomial $x^m$, set $K_i=\ker T^i$ and $d_i=\dim K_i$, with $d_0=0$. ThenFor strictness, choose $w$ with $T^{m-1}w\ne 0$. The vector $T^{m-1-i}w$ belongs to $K_{i+1}\setminus K_i$ for $0\le i<m$. For the second inequality, the rank-nullity theorem givesThe first image is contained in the second because $T(K_{i+1})\subseteq K_i$.

If the Nilpotent Jordan block sizes are $s_1,\dots ,s_r$, thenThus successive increments count the Jordan blocks reaching a given size, and $(d_i-d_{i-1})-(d_{i+1}-d_i)$ counts blocks of size exactly $i$, with $d_{m+1}=d_m$. The nullities determine the entire Jordan normal form. Conversely any positive nonincreasing integer increments summing to $\dim V$ are realized by taking the corresponding numbers of each Nilpotent Jordan block.

A square matrix is nilpotent when one of its positive powers is zero. It represents a nilpotent linear map in a chosen basis; changing basis preserves nilpotence. Over an algebraically closed field, its Jordan normal form has only zero-eigenvalue blocks.

Let $P=b(T)q(T)$ and $Q=a(T)p(T)$. By part (ii), these are complementary projection operators with images $U=\ker (T-\lambda I{)}^{\alpha }$ and $W=\ker (T-\mu I{)}^{\beta }$. Thus $D=\lambda P+\mu Q$ acts as $\lambda I$ on $U$ and $\mu I$ on $W$.

The linear operator $T$ preserves both summands because it commutes with its own polynomials. Therefore $N=T-D$ preserves them too, and its restrictions satisfyPut $k=\max (\alpha ,\beta )$. Both restrictions of $N^k$ vanish, so the direct sum givesHence $T-D$ is a nilpotent linear map. The operator $D$ is diagonalizable, since its two summands are its eigenspaces, and it commutes with $T$. This illustrates the Jordan–Chevalley decomposition in the case of two distinct eigenvalues.

Use the monomial basis $e_k=x^{n-k}{y}^k$, $0\le k\le n$. Direct differentiation giveswith the endpoint terms zero. Repeated application therefore gives $P^{n+1}=Q^{n+1}=0$, while $P^n{e}_n=n!e_0\ne 0$ and $Q^n{e}_0=n!e_n\ne 0$. Each nilpotency index is exactly $n+1$, soNextand hence $(PQ-QP)e_k=(n-2k)e_k$. These $n+1$ real eigenvalues are distinct. A polynomial annihilates this operator exactly when it vanishes at all of them, givingThe commutator is diagonal in the displayed monomial basis. Since $n\ge 1$, $P,Q$ are nonzero nilpotent linear map; a diagonalizable nilpotent map would have a diagonal matrix with all eigenvalues zero and thus be the zero map. Therefore only $PQ-QP$ is diagonalizable. The calculation realizes raising and lowering on binary homogeneous polynomials.

For a perfect Lie algebra $\mathfrak{s}$, every element is a sum of Lie brackets. The Adjoint representation satisfies $\mathrm{ad}[x,y]=[\mathrm{ad}x,\mathrm{ad}y]$, and the trace of a commutator is zero. ThereforeSome $\mathrm{ad}x$ must have a nonzero eigenvalue. Otherwise, over the algebraically closed field $k$, every $\mathrm{ad}x$ would be a nilpotent linear map, and the Engel theorem would imply that $\mathfrak{s}$ is a nilpotent Lie algebra. This contradicts perfection: the lower central series of a Lie algebra ${\gamma }_1=\mathfrak{s}$, ${\gamma }_{r+1}=[\mathfrak{s},{\gamma }_r]$ would equal $\mathfrak{s}\ne 0$ at every step instead of terminating.

Choose such an $x$. Since $[x,x]=0$, zero is an eigenvalue of $\mathrm{ad}x$. Its trace is zero and its dimension is three, so the remaining eigenvalues are $\lambda ,-\lambda$ with $\lambda \ne 0$. These three eigenvalues are distinct in characteristic zero, hence $\mathrm{ad}x$ is diagonalizable. Put $h=2x/\lambda$, and choose nonzero eigenvectors $e,f$ withThe zero eigenspace is $kh$, so $h,e,f$ form a basis. Applying the Jacobi identity givesThus $[e,f]=ch$ for some $c\in k$. If $c=0$, all Lie brackets of basis vectors lie in $⟨e,f⟩$, contradicting $[\mathfrak{s},\mathfrak{s}]=\mathfrak{s}$. Therefore $c\ne 0$, and replacing $f$ by $f/c$ gives $[e,f]=h$.

We have constructed an sl2 triple that is a basis. The linear map to the sl2 Lie algebra sendingpreserves all Lie brackets of basis vectors and is bijective. HenceThis is the classification of three-dimensional perfect Lie algebras without appealing to a classification of all simple Lie algebras.

We use the supplied weight decomposition $V={⨁}_{\lambda \in \Psi (\mathfrak{a})}{V}_{\lambda }$. There is a small distinction to justify: the definition of a toral subalgebra uses the Adjoint representation, while these weight spaces use the defining representation. For the symplectic Lie algebra, adjoint semisimplicity in the symplectic Lie algebra implies that each $a\in \mathfrak{a}$ is diagonalizable on $V$ as well. Here is a direct verification.

Define the symplectic adjoint $a^{\ast }$ by $\omega (av,w)=\omega (v,a^{\ast }w)$. The symplectic Lie algebra condition is $a^{\ast }=-a$. Write the Jordan–Chevalley decomposition in $\mathrm{End}(V)$ as $a=S+N$, with $S$ diagonalizable, $N$ a nilpotent linear map, and $SN=NS$. Uniqueness of the Jordan–Chevalley decomposition, applied to $a^{\ast }=-a$, gives $S^{\ast }=-S$ and $N^{\ast }=-N$. Thus $S,N\in \mathfrak{sp}(V)$. On $\mathrm{End}(V)$, $\mathrm{ad}S$ is diagonalizable by using a matrix unit basis in an eigenbasis of $S$, and $\mathrm{ad}N$ is a nilpotent linear map, sincevanishes for large $m$. These operators commute and preserve $\mathfrak{sp}(V)$, so their restrictions are the Jordan–Chevalley decomposition of $\mathrm{ad}a$. Because $\mathrm{ad}a$ is diagonalizable, its nilpotent part is zero: $\mathrm{ad}N=0$ on $\mathfrak{sp}(V)$.

To conclude that $N=0$ without an assumption on the center of a Lie algebra, observe that each map $T_v(w)=\omega (v,w)v$ lies in $\mathfrak{sp}(V)$. If $v\ne 0$, its image is $kv$, because $\omega$ is a nondegenerate bilinear form. Since $N$ commutes with $T_v$, it preserves $kv$. Hence $Nv\in kv$ for every $v$, forcing $N$ to be a scalar linear map: compare its action on two linearly independent vectors and their sum. A scalar nilpotent linear map is zero. For $V=0$ the assertion is immediate. This proves the needed diagonalizability. Commuting diagonalizable linear maps admit simultaneous diagonalization, giving the displayed weight decomposition.

Now let $v\in V_{\lambda }$ and $w\in V_{\mu }$. For every $a\in \mathfrak{a}$, preservation of the alternating bilinear form givesIf $\lambda +\mu \ne 0$ as a linear functional, choose $a$ on which it is nonzero; then $\omega (v,w)=0$. ThereforeFor $v\ne 0$ in $V_{\lambda }$, the fact that $\omega$ is a nondegenerate bilinear form supplies a $w\in V$ with $\omega (v,w)\ne 0$. Decomposing $w$ into weight spaces, only its $V_{-\lambda }$ component can contribute. Thus $V_{-\lambda }\ne 0$, and $\Psi (\mathfrak{a})=-\Psi (\mathfrak{a})$. The same argument on both sides shows that $\omega :V_{\lambda }\times V_{-\lambda }\to k$ is a perfect pairing, so $\dim V_{\lambda }=\dim V_{-\lambda }$. On $V_0$, the restriction of $\omega$ is a nondegenerate bilinear form that is an alternating bilinear form, so $\dim V_0$ is even.

Choose one representative ${\lambda }_i$ from each pair of nonzero weights of a representation, for $1\le i\le r$. HereThe linear map $\mathfrak{a}\to k^r$ sending $a$ to $({\lambda }_1(a),\dots ,{\lambda }_r(a))$ is injective: an element in its kernel acts by zero on every nonzero weight space, and also on $V_0$, so it is the zero element of $\mathfrak{a}\subseteq \mathrm{End}(V)$. Taking dimensions yieldsThe final inequality follows because each pair contributes at least two to $\dim V=2n$. Equivalently, the nonzero weights of a representation span the dual space ${\mathfrak{a}}^{\ast }$; they need not be linearly independent at this stage.

Write $K_i=\ker T^i$, with $K_0=\{0\}$. The minimal polynomial $x^m$ means $T^m=0$ and $T^{m-1}\ne 0$. Choose $w$ with $T^{m-1}w\ne 0$. For each $0\le i<m$, set $w_i=T^{m-1-i}w$. ThenThus $w_i\in K_{i+1}\setminus K_i$. Also $K_i\subseteq K_{i+1}$ because applying $T$ to zero still gives zero. Therefore the kernel dimensions of powers of a nilpotent map increase strictly until the nilpotency index:The same reasoning covers $m=1$, when the nilpotent linear map is zero on a nonzero vector space.

An idempotent $e$ gives a projective module $Ae$, since $A=Ae\oplus A(1-e)$. It is primitive when it cannot be written as a sum of two nonzero orthogonal idempotents. This is equivalent to indecomposability of $Ae$: a projection onto a direct summand is an idempotent of ${\mathrm{End}}_A(Ae)\cong (eAe{)}^{\mathrm{op}}$, and conversely such an idempotent splits $Ae$. Starting with $1$, split any nonprimitive summand repeatedly. The process terminates because each splitting increases the number of nonzero vector-space summands in a finite-dimensional algebra. Thuswith orthogonal primitive idempotents and indecomposable projective modules. Each summand is generated by its idempotent.

Let $D={\mathrm{Hom}}_k(-,k)$. A Frobenius algebra is a finite-dimensional algebra with $A\cong D(A)$ as left modules over $A$, where $(a\varphi )(x)=\varphi (xa)$. Equivalently there is a linear functional $\lambda :A\to k$ for which $(x,y)\mapsto \lambda (xy)$ is nondegenerate. A symmetric algebra satisfies the stronger bimodule isomorphism $A\cong D(A)$, with $(a\varphi b)(x)=\varphi (bxa)$; equivalently the functional can be chosen with $\lambda (xy)=\lambda (yx)$. This is the associative-algebra meaning of symmetric, rather than the symmetric algebra of a vector space.

For a finite-dimensional module,The socle is its largest semisimple submodule, and the radical of a module gives its largest semisimple quotient. With $J=J(A)$ one has $\mathrm{Rad}M=JM$ and $\mathrm{Soc}M=\{m:Jm=0\}$: $A/J$ is semisimple, and $J$ annihilates every simple module.

Now let $P=Ae$ for a primitive idempotent, allowing the field $k$ to be nonsplit. Symmetry identifies $Ae$ with $D(eA)$ through the pairingIt is nondegenerate on these corners. If $v=ve\ne 0$, choose $a$ with $\lambda (av)\ne 0$; then $\lambda (eav)=\lambda (ave)=\lambda (av)\ne 0$. The other variable is treated similarly. The pairing respects the left $A$-action, soThe right-module quotient $eA/eAJ$ is simple. A primitive idempotent remains primitive modulo the nilpotent radical: idempotents lift through that radical, so a decomposition in the quotient would lift to one in $eAe$. In the semisimple algebra $A/J$, the left simple $Ae/JAe$ and the dual of the right simple $eA/eAJ$ belong to the same matrix block. There is exactly one left simple isomorphism class in that block, even over a nonalgebraically closed field. Hence the head-socle identity for a symmetric algebra isThis is an isomorphism of left modules, and need not be canonical.

For the modular examples put $r=(123)$, $s=(12)$, so $r^3=s^2=1$ and $srs=r^{-1}$. The coefficient-of-identity functional on the group algebra is symmetric and nondegenerate, so the identity just proved applies.

In characteristic two, the central idempotents are $b_0=1+r+r^2$ and $b_1=1-b_0$. The first block has basis $b_0,b_{0}s$ and is $k{C}_2\cong k[u]/(u^2)$. For the second, choose a primitive cube root $\omega$ and useThe two $r$-eigenlines are interchanged by $s$, proving irreducibility. Polynomial interpolation in $r$ gives diagonal matrix units, and multiplying by $s$ gives the off-diagonal units. Thus the four-dimensional block maps onto $M_2(k)$ and is isomorphic to it:There are two simple modules, the trivial one-dimensional module $T$ and the two-dimensional module $V$; the sign module equals $T$. Their projective covers are $P_T=k[u]/(u^2)$, with layers $T,T$, and $P_V=V$. The regular module is $P_T\oplus V^{\oplus 2}$, and the Cartan matrix of a group algebra is $\mathrm{diag}(2,1)$. Every module is a sum of $T,P_T,V$: the first block is classified by Jordan chains for a nilpotent linear map of lengths one and two, while the matrix block is semisimple. Thus there are exactly three indecomposable isomorphism classes. This describes the 2-modular blocks of S3.

In characteristic three put $u=r-1$. The ideal $I=uA$ satisfies $I^3=0$, and $A/I\cong k{C}_2\cong k\oplus k$, so $J(A)=I$. The only simples are the trivial and sign modules $S_{+},S_{-}$. Put $e_{\pm }=(1\pm s)/2$. The projectives $P_{\pm }=A{e}_{\pm }$ have dimension three. A convenient radical generator is $v=r-r^{-1}$:The basis $e_{\pm },v{e}_{\pm },v^2{e}_{\pm }$ gives the radical layersThus $A=P_{+}\oplus P_{-}$ and its Cartan matrix of a group algebra isThe elements $e_{-}v{e}_{+}$ and $e_{+}v{e}_{-}$ give one arrow in each direction between the two simples. All paths of length three vanish, and the six paths of lengths zero, one and two are linearly independent. Hence $A$ is the path algebra of that two-vertex oriented cycle modulo the ideal of paths of length three. Its connected quiver also proves there is a single block of S3 in characteristic three.

Any module is a $\mathbb{Z}/2$-graded vector space given by the $s$-eigenspaces, with $v$ interchanging the grades and $v^3=0$. Constructing Jordan chains with homogeneous bases in the kernel and image filtrations and homogeneous lifts decomposes it into graded chains. Each chain has length $1,2$ or $3$ and is determined by its top sign. It is a uniserial module and hence indecomposable. Consequently there are exactly six indecomposables, the length-$\ell$ quotients $P_{\pm }/J^{\ell }{P}_{\pm }$, $\ell =1,2,3$.

The Cayley-Hamilton theorem says that every square complex matrix satisfies its own characteristic polynomial: if ${\chi }_A(t)=t^n+c_{n-1}{t}^{n-1}+\cdots +c_0$, thenFor a $2\times 2$ diagonalisable matrix, write $A=SD{S}^{-1}$ with $D=\mathrm{diag}({\lambda }_1,{\lambda }_2)$. The characteristic polynomial is $(t-{\lambda }_1)(t-{\lambda }_2)$, including repeated roots. Evaluating it at $D$ gives zero in each diagonal position, and polynomial evaluation respects matrix similarity: ${\chi }_A(A)=S{\chi }_A(D)S^{-1}=0$. This proves the requested case.

For the final assertion, let $\lambda$ be any complex eigenvalue of $B$, with nonzero eigenvector $\mathbf{v}$. From $B^k=0$ we obtain $0=B^k\mathbf{v}={\lambda }^k\mathbf{v}$, so $\lambda =0$. The characteristic polynomial of this nilpotent linear map is therefore $t^n$. Applying the full Cayley-Hamilton theorem yields $\boxed{B^n=0}$. This argument does not assume $B$ is diagonalisable.

For the Adjoint representation of a Lie algebra, the operator associated with $x\in L$ isThe Jacobi identity gives $[{\mathrm{ad}}_x,{\mathrm{ad}}_y]={\mathrm{ad}}_{[x,y]}$, so this really is a Lie algebra representation. Define the lower central series of a Lie algebra by $L^1=L$ and $L^{r+1}=[L,L^r]$. A nilpotent Lie algebra is one for which $L^{c+1}=0$ for some integer $c$. If this holds, ${\mathrm{ad}}_x$ carries $L^r$ into $L^{r+1}$, and consequently $({\mathrm{ad}}_x{)}^{c}L=0$. This proves one implication without any condition on the field.

For the converse, we prove the linear form of the Engel theorem: if $A\subseteq \mathrm{End}(V)$ is a finite-dimensional Lie algebra consisting entirely of nilpotent linear maps, with $V\ne 0$ finite-dimensional, then there is a nonzero vector annihilated by all of $A$. This argument works over any field.

Induct on $\dim A$. The assertion is immediate when $A=0$. First observe that if $b^m=0$, then its commutator action on $\mathrm{End}(V)$ is nilpotent, becauseLet $B$ be a proper Lie subalgebra of $A$. The commutator action of $B$ on $A/B$ therefore consists of nilpotent linear maps. Its image has dimension at most $\dim B<\dim A$, so induction supplies a nonzero coset $a+B$ annihilated by $B$. Thus $[B,a]\subseteq B$, and the normalizer $N_A(B)=\{a\in A:[a,B]\subseteq B\}$ strictly contains $B$. This is the Engel normalizer lemma.

Choose $B$ maximal among proper Lie subalgebras. Its normalizer must be $A$, so $B$ is a Lie algebra ideal. Moreover $\dim (A/B)=1$: otherwise the inverse image of a one-dimensional subalgebra of $A/B$ would lie strictly between $B$ and $A$. By induction,Since $B$ is an ideal, $W$ is $A$-invariant: $b(av)=a(bv)+[b,a]v=0$. Take $x\in A\setminus B$. Its restriction to $W$ is nilpotent and hence has a nonzero kernel. A nonzero vector in that kernel is annihilated by $B$ and $x$, and therefore by all of $A$. This completes the induction.

Apply this result successively to $V$, then its quotient by a common annihilated line, and so on. We obtain a complete flag $0=V_0\subset V_1\subset \cdots \subset V_n=V$ with $A{V}_i\subseteq V_{i-1}$. Equivalently every element of $A$ is strictly upper triangular in one common basis; any product of $n$ such operators is zero.

Now take $V=L$ and $A=\mathrm{ad}(L)$. The hypothesis supplies exactly the nilpotence needed above. Every iterated bracket of length $n+1$, where $n=\dim L$, is an application of $n$ adjoint operators and vanishes. These brackets span $L^{n+1}$, giving $L^{n+1}=0$. The zero algebra is already nilpotent. Thus $L$ is nilpotent if and only if every ${\mathrm{ad}}_x$ is nilpotent. The common flag is essential: separate nilpotence of unrelated operators would not imply that their mixed products vanish.

Over the complex numbers, every finite-dimensional representation of a solvable Lie algebra has a basis in which every representing matrix is upper triangular. The Lie theorem is often stated first as the existence of a common eigenvector in every nonzero finite-dimensional Lie algebra representation of a solvable Lie algebra. Applying that assertion successively to quotient representations gives an invariant complete flag, and hence the upper triangular form. The same proof works over any algebraically closed field of characteristic zero.

We prove the common eigenvector assertion by induction on $\dim \mathfrak{g}$, writing the action as $av$. The zero Lie algebra is immediate. If $\mathfrak{g}\ne 0$ is solvable, its derived series of a Lie algebra shows that $[\mathfrak{g},\mathfrak{g}]\ne \mathfrak{g}$. Choose a codimension-one ideal of a Lie algebra $\mathfrak{a}$ containing $[\mathfrak{g},\mathfrak{g}]$, and choose $x\notin \mathfrak{a}$. By induction there are $v\ne 0$ and a linear functional $\lambda :\mathfrak{a}\to \mathbb{C}$ such that $av=\lambda (a)v$ for all $a\in \mathfrak{a}$.

Let $W$ be the span of $v,xv,x^{2}v,\dots$. If $m=\dim W$, the first $m$ of these vectors form a basis, and $W$ is $x$-invariant. We claim that for each $a\in \mathfrak{a}$,For $j=0$ this is the definition of $v$. For the induction step, use $ax=xa+[a,x]$ and $[a,x]\in \mathfrak{a}$. Applying the induction hypothesis to both $a$ and $[a,x]$ proves the claim. Consequently $W$ is $\mathfrak{a}$-invariant, and every $a\in \mathfrak{a}$ acts on $W$ by an upper triangular matrix with all diagonal entries $\lambda (a)$.

Since both $x$ and $a$ preserve $W$, the matrix trace of their commutator on $W$ is zero. The claim applied to $[x,a]\in \mathfrak{a}$ givesHere characteristic zero is essential: $m\ne 0$ in the field, so $\lambda ([x,a])=0$. Now the common weight spaceis nonzero and $x$-invariant. Indeed, for $w\in E_{\lambda }$,An endomorphism of a nonzero finite-dimensional complex vector space has an eigenvector, so choose an eigenvector of $x$ in $E_{\lambda }$. It is a common eigenvector for $\mathfrak{g}=\mathfrak{a}+\mathbb{C}x$. This proves the Lie theorem.

For the printed matrices in characteristic $p$, $[x,y]=x$, and there is no common eigenvector. Index the standard basis by $e_0,\dots ,e_{p-1}$. The cyclic entry in the PDF givesThe $p$ diagonal eigenvalues of $y$ are distinct in $k$. Thus every eigenvector of $y$ is a scalar multiple of a single $e_j$. Since $p\ge 2$, $x{e}_j$ is never a scalar multiple of $e_j$, proving the assertion even when $k$ is not an algebraically closed field.

For $1\le j\le p-1$,At the cyclic boundary,Hence $[x,y]=x$. The two-dimensional Lie subalgebra $kx+ky$ has derived algebra $kx$, whose own derived algebra is zero, so it is solvable. It nevertheless has no common eigenvector, including after extending $k$ to its algebraic closure. This is a failure of Lie theorem in positive characteristic. In the proof above, the obstruction is precisely that $\dim W=p$ can vanish as a scalar in $k$.

The derived algebra of a complex solvable Lie algebra is nilpotent. First suppose $\mathfrak{g}\subseteq \mathfrak{gl}(V)$. By the Lie theorem, put every element of $\mathfrak{g}$ in upper triangular form. The diagonal of a commutator of upper triangular matrices is zero, so $\mathfrak{d}=[\mathfrak{g},\mathfrak{g}]$ consists of strictly upper triangular matrices. The Lie algebra of all such matrices is a nilpotent Lie algebra: if $F_r$ consists of matrices whose entries vanish whenever $j-i<r$, thenTherefore the lower central series of a Lie algebra of $\mathfrak{d}$ reaches zero. Alternatively, every element of $\mathfrak{d}$ is a nilpotent linear map, and the Engel theorem states that a finite-dimensional Lie subalgebra consisting of nilpotent linear maps is a nilpotent Lie algebra.

For an abstract complex solvable Lie algebra $\mathfrak{g}$, apply the preceding result to its Adjoint representation. The Lie algebra $\mathrm{ad}\mathfrak{d}=[\mathrm{ad}\mathfrak{g},\mathrm{ad}\mathfrak{g}]$ is nilpotent. Since $\ker (\mathrm{ad}{|}_{\mathfrak{d}})=\mathfrak{d}\cap Z(\mathfrak{g})$ is central in $\mathfrak{d}$, some term of the lower central series of a Lie algebra of $\mathfrak{d}$ lies in that central ideal; the next term is zero. Thus $\mathfrak{d}$ itself is nilpotent.

Conversely, every nilpotent Lie algebra is solvable, since its derived series of a Lie algebra is contained term by term in its lower central series of a Lie algebra. If $\mathfrak{d}$ is nilpotent, it is therefore solvable, and $\mathfrak{g}/\mathfrak{d}$ is Abelian. More directly, the derived series of a Lie algebra of $\mathfrak{g}$, after its first term, is the derived series of a Lie algebra of $\mathfrak{d}$. Consequently the derived algebra nilpotence criterion is

The form attached to $V$ is $B_V(x,y)={\mathrm{tr}}_V(xy)$. More generally, for a Lie algebra representation $\rho$, the Trace form of a Lie algebra representation isThe unqualified Killing form is the special case of the Adjoint representation,The distinction matters: a Trace form of a Lie algebra representation can be degenerate even when $\mathfrak{g}$ is semisimple, for example on the trivial Lie algebra representation.

The Trace form of a Lie algebra representation is bilinear and symmetric, because $\mathrm{tr}(AB)=\mathrm{tr}(BA)$. It is an invariant bilinear form on a Lie algebra:This follows by expanding both commutators and cyclically permuting factors under the matrix trace. Equivalently,Its radical of a bilinear form is an ideal of a Lie algebra, since if $B_V(y,-)=0$, then $B_V([x,y],z)=-B_V(y,[x,z])=0$. The Killing form is also preserved by every automorphism of a Lie algebra, because the corresponding adjoint operators are conjugate. On a complex finite-dimensional Lie algebra, the Cartan criterion for semisimplicity says that the Killing form is nondegenerate exactly when the Lie algebra is semisimple. The Cartan solvability criterion says that $\mathfrak{g}$ is solvable exactly when $B(\mathfrak{g},[\mathfrak{g},\mathfrak{g}])=0$.

We next construct the sl2 subalgebra associated with a root. Use the root-space decompositionFor $u\in {\mathfrak{g}}_{\beta }$, $v\in {\mathfrak{g}}_{\gamma }$, invariance of the Killing form givesThus $B({\mathfrak{g}}_{\beta },{\mathfrak{g}}_{\gamma })=0$ unless $\beta +\gamma =0$, and $B({\mathfrak{g}}_{\beta },\mathfrak{h})=0$ for nonzero $\beta$. Nondegeneracy of $B$ on $\mathfrak{g}$ now implies that $-\alpha$ is a root and that $B$ pairs ${\mathfrak{g}}_{\alpha }$ and ${\mathfrak{g}}_{-\alpha }$ nondegenerately.

Nondegeneracy of $B{|}_{\mathfrak{h}}$ defines a unique $t_{\alpha }\in \mathfrak{h}$ byChoose $x\in {\mathfrak{g}}_{\alpha }$ and $y\in {\mathfrak{g}}_{-\alpha }$ with $B(x,y)=1$. Their Lie bracket lies in the zero root space, namely $\mathfrak{h}$, andTherefore $[x,y]=t_{\alpha }$.

The essential nonisotropic root lemma is that $\alpha (t_{\alpha })\ne 0$. Suppose instead that it vanished. Then $[t_{\alpha },x]=[t_{\alpha },y]=0$, so $⟨x,y,t_{\alpha }⟩$ would be a solvable Lie algebra with derived algebra $\mathbb{C}{t}_{\alpha }$. Apply the Lie theorem to its action on $\mathfrak{g}$ by the Adjoint representation. The commutator $\mathrm{ad}{t}_{\alpha }=[\mathrm{ad}x,\mathrm{ad}y]$ is strictly upper triangular in a suitable basis, hence nilpotent. But $t_{\alpha }\in \mathfrak{h}$, so the root-space decomposition makes $\mathrm{ad}{t}_{\alpha }$ diagonalizable. A diagonalizable nilpotent linear map is zero. Thus $t_{\alpha }$ is central in $\mathfrak{g}$. The center of a Lie algebra of a semisimple Lie algebra is zero; equivalently a central element lies in the radical of the Killing form. This forces $t_{\alpha }=0$, contradicting $\alpha \ne 0$.

Writing $c=\alpha (t_{\alpha })$, defineThe root-space decomposition and $[x,y]=t_{\alpha }$ giveThe three vectors are linearly independent because they lie in the distinct summands ${\mathfrak{g}}_{\alpha }$, $\mathfrak{h}$, and ${\mathfrak{g}}_{-\alpha }$. Their span is therefore a copy of the sl2 Lie algebra.

The weight lattice consists of the functionals integral on all coroots. With $H_{\alpha }$ the coroot above, the weight lattice iswhere the fundamental weights satisfy ${\omega }_i(H_{{\alpha }_j})={\delta }_{ij}$ for the simple roots ${\alpha }_j$. Here $P$ lies in the real span of the roots, viewed inside ${\mathfrak{h}}^{\ast }$.

The classification of finite-dimensional sl2 representations says that every finite-dimensional complex sl2 Lie algebra representation is a direct sum of irreducibles $L_m$, $m\in {\mathbb{Z}}_{\ge 0}$, on which the standard $H$ has eigenvalues $m,m-2,\dots ,-m$. Restrict any finite-dimensional Lie algebra representation of $\mathfrak{g}$ to each sl2 subalgebra associated with a root. If $v$ has weight $\lambda$, then $H_{\alpha }v=\lambda (H_{\alpha })v$, so $\lambda (H_{\alpha })$ is an integer. Thus every weight lies in $P$. The same restrictions show that the commuting simple coroots act diagonalizably, justifying the simultaneous weight-space decomposition.

For $\mathfrak{sp}(4)$, the roots are $\pm 2{\epsilon }_1$, $\pm 2{\epsilon }_2$, and $\pm {\epsilon }_1\pm {\epsilon }_2$. Work on ${\mathbb{C}}^4$ with the alternating bilinear form having matrixThe symplectic Lie algebra isUsing the matrix units $E_{ij}$, take the Cartan subalgebraDefine ${\epsilon }_i(a_1{h}_1+a_2{h}_2)=a_i$. A regular diagonal element of $\mathfrak{h}$ has centralizer precisely $\mathfrak{h}$, and every element of $\mathfrak{h}$ acts diagonalizably. Thus it is a Cartan subalgebra. The requested Cartan decomposition is the root-space decompositionChoose positive roots ${\epsilon }_1-{\epsilon }_2$, $2{\epsilon }_2$, ${\epsilon }_1+{\epsilon }_2$, $2{\epsilon }_1$. The symplectic root sl2 triple are given explicitly byFor the negative root spaces, use the corresponding $Y_{\alpha }$. These eight root vectors, together with $h_1,h_2$, form a basis: the block description above has dimension $4+3+3=10$, and the ten listed vectors are independent. Finally, the matrix unit identityverifies $[X_{\alpha },Y_{\alpha }]=H_{\alpha }$ for every row. The diagonal differences verify $[H_{\alpha },X_{\alpha }]=2X_{\alpha }$ and $[H_{\alpha },Y_{\alpha }]=-2Y_{\alpha }$. Thus each row supplies a basis of the required sl2 subalgebra associated with a root.

Put $A=\mathrm{im}\alpha$. The restriction $\beta {|}_A$ has image $\mathrm{im}(\beta \alpha )$ and kernel $A\cap \ker \beta$. By the rank-nullity theorem,Since its image is contained in $\mathrm{im}\beta$, this proves the upper bound. Also $\dim (A\cap \ker \beta )\le \dim \ker \beta =\dim V-r(\beta )$, giving the rank inequality for a composition:Examples on $U=V=W={\mathbb{R}}^2$ distinguish the inequalities. Taking $\alpha =\beta =I$ gives equality in both. Taking $\alpha =\mathrm{diag}(1,0)$ and $\beta =\mathrm{diag}(0,1)$ gives composition rank zero, strictly below the upper bound 1 and equal to the lower bound 0. Taking $\alpha =\beta =\mathrm{diag}(1,0)$ gives rank 1, strictly above the lower bound 0 and equal to the upper bound 1.

For the final nilpotent linear map, repeated application of the lower bound gives $r({\alpha }^k)\ge 2n-2k$. Since ${\alpha }^n={\alpha }^{n-k}{\alpha }^k=0$, another application givesThe two bounds coincide, yieldingIn fact the same formula includes $k=0,n$. For $n=1$ the requested index range is empty and the assumptions force $\alpha =0$.