Matrix trace

The trace is the sum of diagonal entries,Index interchange proves $\mathrm{tr}(AB)=\mathrm{tr}(BA)$ and hence cyclic invariance under any cyclic permutation of a product.

Let $U\subseteq W\subseteq {\mathrm{End}}_{\mathbb{C}}(V)$ be vector subspaces and $M=\{T:[T,W]\subseteq U\}$. If $\alpha \in M$ and $\mathrm{tr}(\alpha \beta )=0$ for all $\beta \in M$, then $\alpha$ is a nilpotent endomorphism.

To prove it, let $\beta$ act by $\overline{\lambda }$ on the generalized eigenspace of $\alpha$ of eigenvalue $\lambda$. Polynomial interpolation on eigenvalue differences, together with adjoint compatibility of additive Jordan decomposition, expresses $\mathrm{ad}\beta$ as a polynomial in $\mathrm{ad}\alpha$ with zero constant term. Since $\mathrm{ad}\alpha$ maps $W$ into $U$ and preserves $U$, this gives $\beta \in M$. Taking the matrix trace on each generalized eigenspace yields $0={\sum }_{\lambda }\dim (V_{\lambda })|\lambda {|}^2$, so every eigenvalue is zero. This is the linear-algebra step behind the Cartan solvability criterion; $U,W$ need not be Lie subalgebras.

Over $\mathbb{R}$ or $\mathbb{C}$, every linear functional on all $n\times n$ matrices satisfying $f(AB)=f(BA)$ is a scalar multiple of the trace. The matrix units show that the functional vanishes off the diagonal and has equal values on every diagonal unit. The normalization $f(I)=n$ fixes the scalar to one.

A traceless matrix is a square matrix whose trace is zero. Such matrices form the kernel of the trace linear functional.

Every commutator of square matrices is traceless by cyclicity of the trace. Conversely, off-diagonal matrix units and differences of diagonal units are themselves commutators and span all traceless matrices. This proves equality with the linear span of commutators, without needing a theorem about representation as one commutator.

For square matrices whose product is defined, $\mathrm{tr}(A_1{A}_2\cdots A_n)$ is unchanged by a cyclic permutation of the factors. In particular, every commutator has trace zero.