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Optimization of a quadratic form over binary choices, here using signs $x_i\in \{-1,1\}$. Lifting $x{x}^T$ turns the objective into a matrix trace; dropping the rank-one condition gives a semidefinite relaxation of binary quadratic optimization.

For $F=dA+A\wedge A$, the gauge-theory Bianchi identity is $DF=0$ and the variation is $\delta F=D\delta A$. The graded product rule for the covariant exterior derivative gives the displayed boundary integral. Thus a constant-coefficient Yang-Mills theta term leaves the bulk Yang-Mills equations unchanged, but can change the boundary variational condition. Its four-form density is exactly gauge-invariant by conjugation of $F$ and cyclicity of the matrix trace, even if the gauge parameter is nonzero on the boundary. Gauge changes of a local Chern-Simons 3-form must not be confused with gauge changes of this globally defined curvature density.

For an equal-part biregular graph with part size $n$ and degree $\delta n$, let $M$ be its biadjacency matrix and $H=M-\delta J$, where $J$ is the all-ones matrix. The constant row and column sums make $H$ annihilate constants on both sides. Since $M{M}^{\mathsf{T}}={\delta }^{2}nJ+H{H}^{\mathsf{T}}$ and the two summands have zero products, their squared matrix trace splits as displayed. The bipartite four-cycle density counts the squared matrix trace of $M{M}^{\mathsf{T}}$ divided by $n^4$.

For a matrix-valued gauge field $A$, let $F=dA+A\wedge A$ be its gauge field strength; products below combine matrix multiplication with the wedge product of differential forms. The five-form can be written $Q_5=\mathrm{Tr}(A(dA{)}^2+\frac{3}{2}{A}^{3}dA+\frac{3}{5}{A}^5)$. Applying the exterior derivative, using its graded product rule and the graded cyclic property of the matrix trace, gives $d{Q}_5=\mathrm{Tr}(F^3)$. The six-form is an invariant polynomial in the gauge curvature, while its local potential depends on the connection.

A variation of the connection one-form has $\dot{F}=D_A\dot{A}$. The Bianchi identity and graded cyclic matrix trace imply ${\partial }_t\mathrm{Tr}(F\wedge F)=2d\mathrm{Tr}(\dot{A}\wedge F)$. Thus the Chern-Weil theory four-form changes by an exact differential form, and its integral on a closed four-cycle is connection-independent. For a general unitary bundle, include the corresponding trace-product term in the Second Chern form.

Applying a scalar function separately to each matrix entry: $(f[X]{)}_{ij}=f(X_{ij})$. This differs from spectral matrix functional calculus. In particular, $\arcsin [Y]$ in Gaussian hyperplane rounding is entrywise, and its contribution to a matrix trace can be restricted to selected blocks by the support of the other matrix.

For a proper Euclidean motion of Euclidean three-space $E(x)=Ax+b$, transport the unit normal by $\tilde{N}(E(p))=AN(p)$. The corresponding shape operators are orthogonally conjugate:Their determinant and trace, hence Gaussian curvature and mean curvature, are unchanged.

For an integer $0\le k\le n$, the fantope is the convex hull of rank-$k$ orthogonal projection matrices in ${\mathbb{R}}^n$. Equivalently, it consists of real symmetric matrices with eigenvalues in $[0,1]$ and matrix trace $k$. To prove the equivalence, apply the spectral theorem for real symmetric matrices and express the eigenvalue vector as a convex combination of the zero-one extreme points of the capped simplex. The sum of the largest eigenvalues is its support function.

A gamma-matrix trace is a matrix trace of products of gamma matrices. In four spacetime dimensions, $\mathrm{tr}({\gamma }^{\mu }{\gamma }^{\nu })=4{\eta }^{\mu \nu }$ and $\mathrm{tr}({\gamma }^{\mu }{\gamma }^{\nu }{\gamma }^{\rho }{\gamma }^{\sigma })=4({\eta }^{\mu \nu }{\eta }^{\rho \sigma }-{\eta }^{\mu \rho }{\eta }^{\nu \sigma }+{\eta }^{\mu \sigma }{\eta }^{\nu \rho })$.

For an anti-Hermitian $SU(N)$ connection one-form, normalize the Chern-Simons three-form by $1/(8{\pi }^2)$ and use $A^g=g^{-1}Ag+g^{-1}dg$. Its change is $-d\mathrm{Tr}(dg{g}^{-1}\wedge A)/(8{\pi }^2)-\mathrm{Tr}(g^{-1}dg{)}^3/(24{\pi }^2)$. On a closed three-manifold the first term integrates to zero, while the latter is an integer with the fundamental matrix trace. Hence the integral is defined modulo integers; integer-level exponentials are invariant under large Yang-Mills gauge transformations.

The virtual similarity transformation $A^i\mapsto X{A}^i{X}^{-1}$ leaves every periodic MPS amplitude invariant by cyclicity of the matrix trace. The fundamental theorem of matrix product states says that, up to an overall phase, this is the only ambiguity for injective uniform tensors.

On a closed manifold, the positive Laplace-Beltrami operator has a smooth heat kernel and its heat semigroup is a trace-class operator for positive time. Expanding in a smooth orthonormal eigenbasis gives $K_t(x,y)={\sum }_j{e}^{-t{\lambda }_j}{\varphi }_j(x)\overline{{\varphi }_j(y)}$. Elliptic regularity bounds each derivative of ${\varphi }_j$ polynomially in ${\lambda }_j$, while the exponential dominates those powers, so the expansion converges with all derivatives. Integrating the diagonal and using the unit norm of each eigenfunction proves the formula. For a bundle operator, insert the fibre matrix trace inside the integral.

The unique equilibrium has $u=\mu$ and $f=0$. Its determinant is $-f_v/ϵ>0$ and trace is $-1+(f_v-\gamma f_u)/ϵ$. Thresholds are obtained from the two displayed equations, and a positive-trace interval exists only if the fast normal derivative rises above $ϵ$. At a nondegenerate trace crossing, the linear oscillation frequency is $\sqrt{-f_v/ϵ}$. The sign of a nonlinear Hopf coefficient is needed to classify the local periodic branch; in a bounded positive planar flow, a unique repelling equilibrium also allows a Poincaré-Bendixson existence argument for a periodic orbit.

The inertia tensor packages the moments of inertia about every axis through a specified point $O$ into one symmetric matrix. An axis still needs a direction: if $\mathbf{n}$ is a unit vector, its scalar moment of inertia is $I_{O,\mathbf{n}}={\mathbf{n}}^T{\mathbf{I}}_O\mathbf{n}$. The same mass layout can therefore be easy to spin about one direction and harder about another. The inertia tensor also determines rotational kinetic energy and relates angular velocity to angular momentum in the appropriate rigid body model.

Take position vectors $\mathbf{r}=(x,y,z)$ relative to $O$, expressed in an orthonormal basis. For point masses $m_a$, or a continuous distribution with mass density $\rho$ and $dm=\rho dV$,Here ${\delta }_{ij}$ is the Kronecker delta. Equivalently, using the identity matrix $1$ and an outer product,Every entry has SI units $\mathrm{kg}{\mathrm{m}}^2$. The diagonal entries are axial moments of inertia; the off-diagonal entries carry a minus sign in this convention. Some texts instead define the integrals without the prefixed minus sign $\int xydm$, $\int xzdm$, $\int yzdm$ as their named off-diagonal quantities and put the minus signs into the displayed matrix. Check that convention before combining formulas. ^{[ref] [ref]}

The symmetric matrix is positive semidefinite, because its quadratic form isThe dot product and cross product identify this as a sum of squared distances when $\mathbf{u}$ is a unit vector. A zero moment of inertia occurs precisely when all the mass lies on that axis, up to sets of zero mass. A genuine three-dimensional body with positive mass density in a volume has strictly positive axial moments of inertia. Do not confuse this inertia tensor with the mass second-moment tensor $\mathbf{S}=\int \mathbf{r}{\mathbf{r}}^{T}dm$: they obey ${\mathbf{I}}_O=\mathrm{tr}(\mathbf{S})1-\mathbf{S}$, where $\mathrm{tr}$ denotes the matrix trace.

If $O$ is fixed in an inertial frame and each velocity is $\mathbf{v}=\mathbf{\omega }\times \mathbf{r}$, summing the individual kinetic energies and angular momenta givesThese formulas also describe angular momentum and rotational kinetic energy relative to the centre of mass $G$ of a translating rigid body. In that case the total kinetic energy in the inertial frame is $K_{\mathrm{total}}=\frac{1}{2}M|{\mathbf{V}}_G{|}^2+\frac{1}{2}{\mathbf{\omega }}^T{\mathbf{I}}_G\mathbf{\omega }$. For a general moving point $O$, the body's actual velocities include ${\mathbf{V}}_O$, so ${\mathbf{I}}_O\mathbf{\omega }$ alone need not be its angular momentum about $O$. ^{[ref] [ref]}

Changing the orthonormal basis changes components, rather than the physical distribution: if the columns of an orthogonal matrix $Q$ are the new basis vectors expressed in the old basis, then ${\mathbf{I}}_O^{\prime }=Q^T{\mathbf{I}}_{O}Q$. Changing the reference point is a different operation, governed by the tensor parallel-axis theorem. Both principal axes and principal moments of inertia must therefore be quoted with their reference point.

For complex invertible matrices $A,B$, $AB=-BA$ implies $B^{-1}AB=-A$ and $A^{-1}BA=-B$. Invariance of matrix trace under matrix similarity gives both zero traces. Invariance of determinant gives $\det A=(-1{)}^n\det A$, and invertibility forces $n$ even. Equivalently, eigenvalues occur in pairs $\lambda ,-\lambda$ with equal algebraic multiplicity, and zero is excluded. Both invertibility assumptions matter: $A=\mathrm{diag}(1,0,0)$ and $B=\mathrm{diag}(0,1,0)$ anticommute in dimension three but are singular and have nonzero traces.

For a real symmetric matrix $X$ and $1\le k\le n$,Indeed $Y=Q{Q}^T$ is a rank-$k$ orthogonal projection matrix in the fantope, and projecting onto the eigenvectors of the largest eigenvalues attains the support function maximum. Equivalently, the largest matrix trace of a compression to a $k$-dimensional vector subspace is the sum of the largest eigenvalues.

A lake supplies a melt-opened subglacial channel while its changing hydrostatic head changes effective pressure and material creep closure. In the displayed scaled model the positive equilibrium is $S={\mu }^{3/4}$, $N={\mu }^{1/12}$. Its Jacobian has trace ${\mu }^{1/4}/3>0$ and determinant $4{\mu }^{7/6}$. The instability becomes oscillatory when $\mu >1/1728$. Growing linear oscillations signal departure from steady drainage; they do not prove unbounded physical lake discharge after nonlinear effects and lake depletion become important.

The Laplacian takes the divergence of a scalar field's gradient. For a real-valued function $f$ on an open region of Euclidean space, with continuous partial derivatives through order two, its expression in a Cartesian coordinate system isThus the Laplacian, also called the Laplace operator, adds the coordinate second derivatives; equivalently it is the matrix trace of the Hessian matrix. Its output is another scalar field.^[ref]

Three original examples distinguish shape from this sum. The bowl $f(x,y)=x^2+y^2$ has $\Delta f=2+2=4$. The saddle $g(x,y)=x^2-y^2$ instead has $\Delta g=2-2=0$: opposite second derivatives cancel. It is a harmonic function, although its graph of a function is not a plane. For the Gaussian function $h(x,y)=e^{-r^2}$, with $r^2=x^2+y^2$, differentiation givesIts Laplacian is negative inside the circle of radius one, zero on that circle, and positive outside it. The Gaussian function itself is positive throughout: its value and its Laplacian describe different properties.

The sum does not favor a particular orientation. Under a change of Cartesian coordinate system represented by an orthogonal matrix $Q$, the Hessian matrix becomes $Q^{\mathsf{T}}(D^{2}f)Q$. The cyclic property of the trace and $Q{Q}^{\mathsf{T}}=I$ preserve its matrix trace. This makes the Laplacian useful when the behavior depends on spatial differences equally in every direction, rather than on the arbitrary choice of axes.

For a vector field, the Euclidean Laplacian acts separately on components expressed in a fixed Cartesian coordinate system. Components in a position-dependent basis require differentiating that basis too; applying a scalar formula separately to polar components omits those terms.^[ref] The scalar formula in Laplacian in polar coordinates is valid away from $r=0$; smoothness at the origin must be handled in coordinates that remain regular there.

For a constant real square matrix $A$, the autonomous differential equation $\dot{x}=Ax$ has the linear flow ${\phi }_t(x)=e^{tA}x$, where $e^{tA}$ is the matrix exponential. The identity $e^{(t+s)A}=e^{tA}{e}^{sA}$ gives the flow map composition law, and $e^{-tA}$ reverses the evolution. Its determinant is $e^{t\mathrm{tr}A}$, so the matrix trace gives the instantaneous fractional rate of volume change, as illustrated in trace and local volume change. A zero trace preserves volume even when the flow stretches one direction and contracts another.

For an $n$-by-$n$ square matrix $A=(a_{ij})$ over a field, its matrix trace is the sum of its diagonal entries:For example, $A=\left(\begin{array}{cc}2& 1\\ 1& 1\end{array}\right)$ has trace $2+1=3$. Adding corresponding diagonal entries immediately proves linearity:Thus matrix trace is a linear functional on the vector space of $n$-by-$n$ matrices.^[ref]

The surprising point is that the sum is intrinsic to a linear operator, even though individual diagonal entries depend on the basis. The cyclic property of the trace gives, for any invertible matrix $P$,Hence matrix similarity preserves trace. For the example above, choose the change-of-basis matrix $P=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$. ThenThe diagonal entries have changed but their sum has not. This makes trace useful for extracting a feature of the underlying linear operator without finding a preferred basis.

Over the complex numbers, or over the real numbers after allowing complex eigenvalues, the trace equals their sum with algebraic multiplicity:No diagonalizable matrix hypothesis is needed.^[ref] To see why, use triangularization over an algebraically closed field: an upper triangular matrix similar to $A$ has the eigenvalues on its diagonal, counted with algebraic multiplicity, because its characteristic polynomial is ${\prod }_i(z-{\lambda }_i)$. Similarity preserves trace, so summing that diagonal proves the formula.

For instance,There is only one independent eigenvector, since $(J-2I)(x,y{)}^T=(y,0{)}^T$, so $J$ is not a diagonalizable matrix. Nevertheless its repeated eigenvalue contributes twice. For our earlier $A$, the characteristic polynomial is $z^2-3z+1$, its eigenvalues are $(3+\sqrt{5})/2$ and $(3-\sqrt{5})/2$, and their sum is $3$. An inexpensive diagonal sum therefore gives spectral information before solving the characteristic polynomial.

The trace of a real matrix does not give that linear map's volume factor: that factor is the absolute value of its determinant. For example, $2I_2$ has trace $4$ and area factor $4$, but $2I_3$ has trace $6$ and volume factor $8$. Their occasional numerical agreement in two dimensions is a coincidence. The connection between trace and local volume change gives a geometric interpretation near the identity matrix, and describes the volume rate of a flow map generated by a linear system of differential equations.