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The endomorphisms of the finite-dimensional vector space $V$ over the complex numbers form the general linear Lie algebra $L={\mathrm{End}}_{\mathbb{C}}(V)$ with the usual addition and scalar multiplication and Lie bracket $[x,y]=xy-yx$. This commutator is bilinear and antisymmetric, and expanding the six products verifies the Jacobi identity.

For a Lie subalgebra $\mathfrak{g}=L_1$, being an abelian Lie algebra means $[\mathfrak{g},\mathfrak{g}]=0$. A nilpotent Lie algebra has ${\gamma }_1=\mathfrak{g}$, ${\gamma }_{j+1}=[\mathfrak{g},{\gamma }_j]$ eventually zero; a solvable Lie algebra has ${\mathfrak{g}}^{(0)}=\mathfrak{g}$, ${\mathfrak{g}}^{(j+1)}=[{\mathfrak{g}}^{(j)},{\mathfrak{g}}^{(j)}]$ eventually zero. These are the lower central series of a Lie algebra and derived series of a Lie algebra, respectively. Nilpotence is a condition on the Lie bracket, and does not require every member to be a nilpotent endomorphism: a nonzero scalar multiple of the identity spans an abelian Lie algebra.

A flag of a vector space is an increasing chain of vector subspaces. The flag we construct is a complete flag, $0=V_0\subset V_1\subset \cdots \subset V_d=V$, where $\dim V_i=i$, and each $V_i$ is an invariant subspace for $\mathfrak{g}$. We first prove the common-eigenvector assertion in the Lie theorem, by induction on $\dim \mathfrak{g}$; the zero algebra is immediate. For nonzero solvable $\mathfrak{g}$, its derived algebra is proper, so there is a codimension-one ideal of a Lie algebra $\mathfrak{k}$ containing $[\mathfrak{g},\mathfrak{g}]$. Write $\mathfrak{g}=\mathfrak{k}\oplus \mathbb{C}x$. By induction there are $v\ne 0$ and a linear functional $\lambda$ on $\mathfrak{k}$ such that $kv=\lambda (k)v$ for every $k\in \mathfrak{k}$.

Let $W$ be the cyclic subspace spanned by $v,xv,x^{2}v,\dots$. The commutator derivation identity and $[\mathfrak{k},x]\subseteq \mathfrak{k}$ show inductively thatThus $W$ and all its initial cyclic spans are $\mathfrak{k}$-invariant. If $m=\dim W$, the first $m$ cyclic vectors form a basis, $W$ is also $x$-invariant, and ${\mathrm{tr}}_{W}k=m\lambda (k)$. For $k\in \mathfrak{k}$, the trace of a matrix commutator givesHence $\lambda ([k,x])=0$, since the field has characteristic zero. The nonzero common weight spaceis $x$-invariant: $k(xw)=x(kw)+[k,x]w=\lambda (k)xw$. The restriction of $x$ to $V_{\lambda }$ has an eigenvector, because $\mathbb{C}$ is an algebraically closed field. This is a common eigenvector for $\mathfrak{g}$. Its line is invariant, and repeating the argument on the quotient vector space gives the complete invariant flag. Equivalently, this proves simultaneous triangularization of a Lie algebra representation.

In a basis adapted to this complete flag, every member of $\mathfrak{g}$ is upper triangular, so every member of its derived algebra is strictly upper triangular. Products of $d$ strictly upper triangular matrices vanish, and each iterated Lie bracket of $d$ such matrices is a sum of these products. Consequently the derived algebra is a nilpotent Lie algebra. We may therefore take $L_2=[L_1,L_1]$: it is an ideal, and the quotient Lie algebra $L_1/L_2$ is abelian. This also covers $\dim V=0$.

A finite-dimensional Lie algebra over the complex numbers $L$ is a semisimple Lie algebra when its solvable radical is zero, equivalently when it has no nonzero solvable ideals. Its Killing form isThe cyclic property of the trace makes this bilinear form symmetric and gives its invariance of a bilinear form on a Lie algebra:It follows that $R=\{x:B_L(x,L)=0\}$ is an ideal of a Lie algebra. For $x\in R$, $\mathrm{ad}x$ induces the zero map on $L/R$. Therefore, for $x,y\in R$, the matrix trace splits over the invariant subspace $R$ and the quotient to give $B_R(x,y)=B_L(x,y)=0$. In particular $B_R([R,R],R)=0$. The Cartan solvability criterion implies that $R$ is solvable. Since $L$ is semisimple, $R=0$: the Killing form is nondegenerate.

For completeness, the trace step in the Cartan solvability criterion is precisely the mechanism of the previous solution. For a complex matrix Lie algebra $\mathfrak{a}$ with $\mathrm{tr}(xy)=0$ for $x\in [\mathfrak{a},\mathfrak{a}]$, $y\in \mathfrak{a}$, set $U=[\mathfrak{a},\mathfrak{a}]$ and $W=\mathfrak{a}$. If $\beta \in M$ and $x=[a,b]$, then $\mathrm{tr}(x\beta )=\mathrm{tr}(a[b,\beta ])=0$, since $[b,\beta ]\in U$. Linearity and the trace orthogonality nilpotence lemma show that every member of $U$ is nilpotent. The Engel theorem makes $U$ nilpotent and hence $\mathfrak{a}$ solvable. Apply this to $\mathfrak{a}=\mathrm{ad}R$; the kernel of this Adjoint representation of a Lie algebra is the abelian center of $R$, so $R$ is solvable as claimed.

For an arbitrary complex Lie algebra, a Cartan subalgebra $H$ means a nilpotent Lie algebra that is self-normalizing: $N_L(H)=\{x:[x,H]\subseteq H\}=H$. This definition does not assume that $H$ is abelian. We prove that it is abelian when $L$ is semisimple.

Use the generalized-weight decomposition for a nilpotent Lie algebra for the action of $H$ on $L$. Its zero generalized weight space isWe have $H\subseteq L^0$, since $H$ is nilpotent. If $L^0/H\ne 0$, the Engel theorem gives a nonzero coset annihilated by every $h\in H$. Its representative $x$ satisfies $[H,x]\subseteq H$, contradicting $N_L(H)=H$. Thus $L^0=H$.

For a nonzero generalized weight $\lambda$, choose $h_0\in H$ with $\lambda (h_0)\ne 0$. The operator $\mathrm{ad}{h}_0$ is invertible on $L^{\lambda }$ and nilpotent on $H$. For $h\in H$, $z\in L^{\lambda }$, write $z=(\mathrm{ad}{h}_0{)}^{m}w$ with $m$ large enough that $(\mathrm{ad}{h}_0{)}^{m}h=0$. Invariance of the Killing form givesOn the other hand, $H$ is solvable, so the Lie theorem triangularizes its action on $L$. For $a,b,h\in H$, the matrix $\mathrm{ad}[a,b]$ is strictly upper triangular, while $\mathrm{ad}h$ is upper triangular. Thus $B_L([H,H],H)=0$. Together with $L=H\oplus {⨁}_{\lambda \ne 0}{L}^{\lambda }$, this yields $B_L([H,H],L)=0$. Nondegeneracy gives $[H,H]=0$.

Finally, if $x$ commutes with $H$, it normalizes $H$, hence lies in $H$. Any abelian subalgebra containing $H$ consists of such elements. Thus $H$ is a maximal abelian subalgebra, indeed $C_L(H)=H$.

We take algebras to be unital and modules to be unital. A finite-dimensional semisimple algebra over the complex numbers is one whose regular left module is a semisimple module, that is, a direct sum of simple modules. We use the basic finite-dimensional equivalences: this is equivalent to a zero Jacobson radical, and the Jacobson radical of any finite-dimensional algebra is a nilpotent ideal.

The Artin–Wedderburn theorem here says that there are positive integers $d_1,\dots ,d_r$ withThe nonisomorphic simple modules are the standard column modules of the factors, of dimensions $d_i$, and the factor sizes are unique up to reordering.

Here is a proof. Decompose the regular left module as ${}_{A}A={⨁}_{i=1}^r{S}_i^{\oplus n_i}$ with pairwise nonisomorphic simple modules $S_i$. These are all the simple modules: any simple module is generated by a nonzero vector and is therefore a quotient of the regular module. The Schur lemma gives ${\mathrm{Hom}}_A(S_i,S_j)=0$ for $i\ne j$ and ${\mathrm{End}}_A(S_i)=\mathbb{C}$. For the latter assertion, an endomorphism has an eigenvalue over $\mathbb{C}$, and the kernel of its difference from that scalar is a nonzero submodule, hence the whole simple module. ConsequentlyEvery regular-module endomorphism is right multiplication by its value at $1$, so ${\mathrm{End}}_A({}_{A}A)\cong A^{\mathrm{op}}$. Taking opposite algebras and using matrix transposition gives the asserted decomposition with $d_i=n_i$.

For a matrix algebra $M_d(\mathbb{C})$, the matrix units show that its only simple module is ${\mathbb{C}}^d$: the spaces $E_{jj}S$ are isomorphic through $E_{ij}$, and a nonzero vector in one of them generates one copy of the column module. Simplicity makes that copy all of $S$. In a product algebra, the mutually orthogonal central idempotents decompose every module into its factor modules; a simple module uses exactly one factor. This proves the module assertion and also uniqueness, since the primitive central idempotents and the dimensions of their simple modules determine the factors.

Write a nonzero complex number as $w=r{e}^{i\theta }$. Matching modulus and complex argument gives $r^4=2\sqrt{2}$ and $4\theta =\pi /4+2\pi k$. Thus the four distinct fourth roots areThese angles differ by $\pi /2$, so the values are distinct. Every further integer $k$ repeats one of them; this is multiplication by the fourth roots of unity.

Expanding the commutator immediately gives $[A,A]=0$, $[A,B]=-[B,A]$, and $[A,\lambda B]=\lambda [A,B]$. For its matrix trace, writeConsequently every matrix commutator is traceless: $\boxed{\mathrm{tr}[A,B]=0}$. This is the trace of a matrix commutator identity, valid for arbitrary square matrices over the complex numbers.

The exponential acts on the Fock vacuum: $|\psi ⟩=e^{\zeta {\sum }_n{\psi }_n{a}_n^{†}}|0⟩$. For bosons, ${\psi }_n$ are commuting complex numbers. Since $[a_n,{\sum }_m{\psi }_m{a}_m^{†}]={\psi }_n$, commuting $a_n$ through the exponential gives $a_n|\psi ⟩={\psi }_n|\psi ⟩$. These are unnormalized bosonic coherent states.

For fermions, the labels are independent odd Grassmann variables, which anticommute with one another and with the fermionic operators. For one mode, the fermionic coherent state is $|\psi ⟩=(1-\psi a^{†})|0⟩$. The canonical anticommutation relations give $a(-\psi a^{†})|0⟩=\psi |0⟩=\psi |\psi ⟩$, using ${\psi }^2=0$. The even factors $1-{\psi }_n{a}_n^{†}$ commute between modes, so the same argument applies to every $n$. A Grassmann eigenvalue is a formal extension of the state space, not an ordinary complex eigenvalue of the nilpotent annihilation operator.

With dual state $⟨\psi |=⟨0|e^{\zeta {\sum }_n{a}_n{\bar{\psi }}_n}$, both cases obey $⟨\psi |{\psi }^{\prime }⟩=e^{{\sum }_n{\bar{\psi }}_n{\psi }_n^{\prime }}$.

Write the complex number as $z=x+iy$. Since $iz=-y+ix$, the modulus of the complex exponential givesThis equals one precisely when $y=0$. The quadratic formula shows that the imaginary part of either root is $\pm \sqrt{4-b^2}/2$. Consequently the roots are real exactly at the two endpoints:Both endpoints satisfy the condition. The repeated roots are respectively $1$ and $-1$; their multiples by $i$ have exponentials on the unit circle.

A counterclockwise planar rotation matrix has the formTo recover this form from preservation of the Euclidean norm, let $\mathbf{u},\mathbf{v}$ be the columns of $R$. Applying the norm condition to the two coordinate unit vectors gives $|\mathbf{u}|=|\mathbf{v}|=1$. Applying it to their sum givesso $\mathbf{u}\cdot \mathbf{v}=0$. The columns are an orthonormal basis, and $R^{T}R=I$: $R$ is an orthogonal matrix.

Write $\mathbf{u}=(\cos \theta ,\sin \theta {)}^T$. There are two perpendicular unit vectors available for $\mathbf{v}$, namely $\pm (-\sin \theta ,\cos \theta {)}^T$. The first gives determinant $1$, the second $-1$. The positive determinant hypothesis selects the first. A norm-preserving planar map with positive determinant is a rotation.

For the commutation condition, writeDirect matrix multiplication givesEquality forces $c=-b$ and $d=a$, so, with $u=a$ and $v=-b$,If $u=v=0$, take $\lambda =0$ and $R=I$. Otherwise set $\lambda =\sqrt{u^2+v^2}>0$ and choose $\cos \theta =u/\lambda$, $\sin \theta =v/\lambda$. ThenThis describes the centralizer of a planar quarter-turn. Its nonzero elements act as a rotation followed by a uniform scaling, just as multiplication by a nonzero complex number does.