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When $u$ is independent of $y$ and $z=x+iy$,PutThe zero-curvature equation becomesSincewe obtain the Isospectral Lax equationTherefore, for every positive integer $n$, the cyclic property of the matrix trace givesThese are the trace invariants of a Lax equation.

The PDE from part (b) reduces toWith $\varphi =2u$, this isTo extract a nontrivial invariant, use $n=2$. Writing $p=u_x$, direct multiplication givesThe terms involving $\lambda$ are constant, and $p={\varphi }^{\prime }/2$. Henceis a first integral. Direct differentiation verifies it:

Put $H=A-B$. This is a nonzero Hermitian matrix. By the finite-dimensional spectral theorem, it has a unit eigenvector $v$ with a nonzero real eigenvalue $lambda$. Define the rank-one orthogonal projection matrixThe cyclic property of the matrix trace givesHence this $P$ distinguishes $A$ and $B$ through the required traces.

Substituting $\delta A_{\mu }=g^{-1}{\partial }_{\mu }\alpha -i[A_{\mu },\alpha ]$ into the curvature and collecting terms gives the covariant transformationEquivalently, $\delta F_{\mu \nu }^a=f^{abc}{F}_{\mu \nu }^b{\alpha }^c$ in the stated conventions. Since $F_{\mu \nu }^a{F}^{\mu \nu ,a}$ is proportional to $\mathrm{tr}(F_{\mu \nu }{F}^{\mu \nu })$, cyclicity of the matrix trace givesThus the Yang-Mills action is gauge-invariant.

For the convention in the question, a finite Yang-Mills gauge transformation acts covariantly on the field strength:or with $U$ and $U^{†}$ exchanged if the opposite convention is used for $D_{\mu }$. Cyclicity of the matrix trace givesso the Yang-Mills theory Lagrangian is gauge invariant.

The quadratic gauge-field operator has zero directions $A_{\mu }\sim A_{\mu }+D_{\mu }\alpha$ along each gauge orbit. It therefore has no inverse on the full field space. Gauge fixing removes this degeneracy and produces a propagator, while the Faddeev-Popov determinant accounts for the corresponding Jacobian.

Choose the orientation of a surface represented by a smooth Gauss map $N:S\to {\mathbb{S}}^2$. The shape operator at $p$ isIf its principal curvatures are $k_1,k_2$, then by definition

Let $E(x)=Ax+b$ be a proper Euclidean motion of Euclidean three-space, so $A\in SO(3)$. Orient $\tilde{S}=E(S)$ by transporting the normal:The chain rule gives the orthogonal conjugacyThe determinant and trace are invariant under similarity, so the Euclidean invariance of the shape operator yields

If the surface orientation is reversed, then $N$ and $S_p$ are multiplied by $-1$. Therefore the orientation reversal of surface curvature says

The matrix trace isUsing matrix multiplication and interchanging dummy indices,

The proposed noncyclic interchange is false. With matrix units,one hasTrace is invariant under cyclic permutations, but an arbitrary swap need not be cyclic.

Multiply the Karush-Kuhn-Tucker conditions on the right by $\hat{\Omega }$ and take the matrix trace:Symmetry and the defining property of the subgradient of the absolute value giveConsequently the last two terms in the objective sum to $p$, and hence

An open-boundary matrix product state with physical dimension $q$ and bond dimension $\chi$ iswhere the $A^i$ are $\chi \times \chi$ matrices and $|\ell ⟩,|r⟩$ are boundary vectors. For periodic boundary conditions, replace the boundary contraction by the matrix trace $\mathrm{Tr}(A^{i_1}\cdots A^{i_N})$.

The same matrices define the matrix product state transfer mapWhen ${\sum }_i(A^i{)}^{†}{A}^i=I$, the Stinespring dilationis an isometry. Repeatedly applying $V$ stores each Kraus label in a fresh physical register:Contracting the remaining virtual system with $⟨\ell |$ gives an MPS, while tracing over all recorded labels gives repeated application of the completely positive map $\mathcal{E}$. The MPS is therefore a coherent unravelling of the channel. Equivalently, retaining the Kraus-label registers realizes a purification of its output. A non-normalized MPS tensor gives the same construction with a general completely positive map; an appropriate canonical gauge normalizes the transfer map on its support.

For the injective translationally invariant case, the fundamental theorem of matrix product states states that two tensors $A=\{A^i\}$ and $B=\{B^i\}$ of the same minimal bond dimension generate the same periodic MPS for every sufficiently large length if and only iffor every physical index $i$, with one invertible matrix $X$. Equality as normalized rays permits the phase $e^{i\theta }$; equality as vectors for all lengths restricts the resulting factor $e^{iN\theta }$ accordingly. The converse is immediate from cyclicity of the matrix trace:

For the nontrivial direction, block enough sites that both tensors are injective and definewith ${\Gamma }_B^L$ defined similarly. Injectivity means that ${\Gamma }_A^L$ and ${\Gamma }_B^L$ have trivial kernels. Equality of all sufficiently long periodic states implies equality of the local support spaces $\mathrm{im}{\Gamma }_A^L=\mathrm{im}{\Gamma }_B^L$, so there is an invertible linear map $F$ on the virtual matrix algebra satisfying

Compare two adjacent blocks and contract arbitrary environments on their left and right. Because both block maps are injective, equality of the physical contractions forcesThus $F$ is a unital algebra automorphism of $M_{\chi }(\mathbb{C})$. By every automorphism of a full matrix algebra is inner, $F(X)=X_{0}X{X}_0^{-1}$ for an invertible $X_0$. Applying this relation to a block with one physical site exposed givesThis proves the theorem and identifies the freedom as the gauge equivalence of injective matrix product state tensors. For noninjective tensors, their canonical forms first split into injective blocks; equality then permits a permutation of equivalent blocks together with a similarity transformation and phase on each block.

The tensor product of the defining representation and its dual is the endomorphism representation. Splitting an endomorphism into its matrix trace and traceless part givesIn Dynkin labels, the two irreducible summands have highest weights $[0,0,0]$ and $[1,0,1]$.

The relation in part vi givesso the full-superspace term has gauge invariance. The chiral field-strength superfield transforms covariantly,Cyclicity of the matrix trace then gives $\mathrm{Tr}(W^{\prime }{}^{\alpha }{W}_{\alpha }^{\prime })=\mathrm{Tr}(W^{\alpha }{W}_{\alpha })$. Both superspace integrals, and hence the entire Lagrangian density, are invariant.

The continuous symmetry isStrictly, the diagonal central $U(1)$ acts trivially, so the faithful group is the quotient by that common center. Since $L$ and $R$ are unitary matrices,Every term is therefore unchanged by cyclicity of the matrix trace:and the same conjugation argument applies to $\mathrm{Tr}(M^{†}M)$ and $\mathrm{Tr}[(M^{†}M{)}^2]$.

Since $\delta F_{\mu \nu }=-ig[F_{\mu \nu },\alpha ]$, both $F_{\mu \nu }{F}^{\mu \nu }$ and $F_{\mu \nu }{}^{\star }{F}^{\mu \nu }$ vary by a commutator. Cyclicity of the matrix trace therefore makes both terms gauge invariant. In electric and magnetic variables,The Yang-Mills kinetic term is even under $C$, $P$, and $T$. The Yang-Mills theta term is even under $C$ because both fields change sign and transpose, but odd under $P$ and under $T$ because exactly one of $\mathbf{E},\mathbf{B}$ changes sign.

Expanding $F_{\mu \nu }$, using antisymmetry of ${ϵ}^{\mu \nu \rho \sigma }$, and cyclically rearranging the matrix trace giveswhere the Chern-Simons current isThe theta contribution to the action is therefore a boundary term. For variations that vanish at the spacetime boundary, its variation vanishes, so it contributes nothing to the classical Euler-Lagrange field equations. Its integral can still distinguish topologically nontrivial gauge configurations in the quantum theory.

For a determinant-one lift of a hyperbolic holonomy representation, the squared matrix trace of each fixed group word is a polynomial in the finitely many generator-matrix entries. The displayed identity identifies equal positive lengths with equal squared traces. Lift signs have no effect. Finite choices of length matchings therefore give finite unions of affine algebraic sets.

For real positive semidefinite matrices $A,B$, their product need not be symmetric, but its matrix trace is nonnegative:The equality uses the cyclic property of the matrix trace, and the last matrix is a positive semidefinite matrix. Consequently the Loewner order inequality $A⪯C$ implies $\mathrm{tr}(AB)\le \mathrm{tr}(CB)$ whenever $B⪰0$.

A semidefinite program optimizes a linear function subject to equalities between affine functions and positive semidefinite matrix inequalities. It generalizes linear programming: a diagonal matrix is a positive semidefinite matrix exactly when its diagonal entries are nonnegative. Matrix trace expresses the objective as $⟨C,X⟩=\mathrm{tr}(CX)$ for real symmetric matrices.

Ifthen every positive power has constant trace:The final equality is the cyclic property of the matrix trace.

The matrix trace of $[A,B]=AB-BA$ is zero, since $\mathrm{tr}(AB)={\sum }_{i,j}{A}_{ij}{B}_{ji}=\mathrm{tr}(BA)$. This obstructs a matrix with nonzero trace from being a single commutator.