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Retain the first $N\ge 2$ number states of a quantum harmonic oscillator. Let ${\Pi }_N$ be the orthogonal projection onto their span and compress $X_N={\Pi }_{N}X{\Pi }_N$, $P_N={\Pi }_{N}P{\Pi }_N$, restricted to this span. The compressed annihilation operator $a_N$ satisfiesFor $n<N-1$, the two products give $n+1$ and $n$; on the top state they give $0$ and $N-1$. Thus, with $\ell =\sqrt{\hslash /(m\omega )}$,On every quantum superposition of $|0⟩,\dots ,|N-2⟩$, this commutator acts exactly as $i\hslash I_N$. For a normalized retained state with coefficients $c_n$,Small probability at the cutoff must therefore be judged relative to $1/N$ for this expectation. For $N=8$, the diagonal of $[X_N,P_N]/(i\hslash )$ is $(1,1,1,1,1,1,1,-7)$, with zero trace.

This is unavoidable for finite matrices: trace of a matrix commutator gives $\mathrm{tr}(XP-PX)=0$, whereas $\mathrm{tr}(i\hslash I_N)=i\hslash N\ne 0$. Compression does not preserve products: ${\Pi }_{N}XP{\Pi }_N$ can use intermediate states outside the retained span, while $X_N{P}_N$ cannot. Accordingly, compressing the full oscillator Hamiltonian operator gives ${\Pi }_{N}H{\Pi }_N=\hslash \omega (a_N^{†}{a}_N+I_N/2)$, but building it from the compressed squares givesThe two prescriptions agree on lower number states and disagree at the boundary. In the infinite oscillator, the canonical commutation relation holds on the common invariant dense subspace of finite number state combinations. Its unbounded position and momentum operators do not admit the finite trace argument; writing the formal identity does not assert that both products act on every vector of the Hilbert space.

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 $\dim V=0$.

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.