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For increasing sites and an ordered nonnegative normalized B-spline basis, the matrix $A_{ij}=N_j(x_i)$ is a totally nonnegative matrix. One way to see the minors' signs is knot insertion. Each inserted knot changes coefficients by $b_j={\alpha }_j{a}_j+(1-{\alpha }_j)a_{j-1}$ with $0\le {\alpha }_j\le 1$, using the standard constant endpoint pieces. This is a nonnegative rectangular bidiagonal map, whose minors are nonnegative. Products preserve that property by the Cauchy–Binet formula. Insert each site to full knot multiplicity; a suitable refined coefficient is then the value $s(x_i)$. Thus collocation is an ordered row submatrix of the refinement map, and has nonnegative minors. The strict support conditions $t_i<x_i<t_{i+k}$ give invertibility by the Schoenberg–Whitney theorem; its inverse has the checkerboard inverse of a totally nonnegative matrix sign pattern.

The directed vertex-edge incidence matrix has one $+1$ and one $-1$ per nonloop column and a zero column for a loop. Every square subdeterminant belongs to $\{0,1,-1\}$. Induct on its size: a column with no nonzero entries makes the determinant zero; a column with one nonzero entry permits a cofactor expansion to a smaller determinant; and if every column has two nonzero entries, their opposite signs make the column sums zero, giving linearly dependent rows. Negating rows preserves absolute subdeterminants. Consequently the unsigned vertex-edge incidence matrix of a bipartite graph is totally unimodular: negate all rows in one part. The assignment problem is an instance, with row vertices and column vertices forming the two parts.

For the real two-by-two matrix space, the Frobenius inner product is $\mathrm{tr}(A^{t}B)={\sum }_{i,j}{a}_{ij}{b}_{ij}$. The displayed four matrices form an orthonormal basis. The first spans the identity direction, the next two span the traceless symmetric matrices, and the last spans the antisymmetric matrices. Symmetric and antisymmetric matrices are orthogonal because transposition changes the sign of their trace pairing; pairing with the identity gives the trace. This basis therefore displays the orthogonal direct sum into these three natural pieces.

For any fixed real $n$-by-$n$ matrix $A$, as the real parameter $t$ tends to zero,The big O notation means the remainder is bounded in absolute value by $C{t}^2$ for some constant $C$ near zero. The determinant stays positive for sufficiently small $t$, so its oriented volume factor is also its ordinary volume factor. Thus the trace is the first-order fractional volume change, measured from the identity matrix.

A proof follows directly from multilinearity of the determinant. Each column of $I+tA$ is $e_j+tA{e}_j$. The constant term chooses all the standard basis columns and equals one. A term containing exactly one factor $t$ replaces just column $j$ by $A{e}_j$; expanding that vector in the standard basis leaves only its $e_j$ component, $a_{jj}$, because every other choice repeats a column. Summing gives $\mathrm{tr}A$. Terms with two or more replacements have order $t^2$ or higher. This also proves the derivative of the determinant at the identity matrix.

This coefficient has a direct visual meaning. At the point $e_j$, the linear map $x\mapsto Ax$ has component $a_{jj}$ along direction $e_j$. In a small step $x\mapsto x+tAx$, these are the first-order changes of the square's or box's sides in their own directions. Summing them gives its fractional area or volume change; the off-diagonal entries initially change shape rather than contributing to this first-order total.

For the constant-coefficient linear system of differential equations $x^{\prime }=Ax$, the exact linear flow is $x(t)=e^{tA}x(0)$, using the matrix exponential. Its volume factor isThis is the matrix exponential determinant identity. One can derive it without diagonalization: set $E(t)=e^{tA}$, whose matrix inverse is $E(-t)$, and use $E^{\prime }=AE$ in Jacobi's formula to get $V^{\prime }=\det E\mathrm{tr}(E^{-1}AE)=(\mathrm{tr}A)V$. The scalar differential equation with initial value one gives the displayed exponential function. Thus positive trace causes volume expansion, negative trace causes contraction, and zero trace preserves volume, even when individual directions behave differently.

In two dimensions, compare four generators at $t=0.4$:The first and third diagonal matrices expand or contract both directions. The second expands horizontally and contracts vertically by reciprocal factors, making a rectangle of unchanged area. For the last nilpotent matrix, $A^2=0$, so its matrix exponential is exactly $I+tA$; it sends $(x,y)$ to $(x+ty,y)$, a shear mapping. Zero trace allows substantial shape change and does not mean the flow map is stationary.

The first-order statement and the exact flow map should be distinguished. For $A=\mathrm{diag}(1,-1)$, $\det (I+tA)=1-t^2$, whereas $\det (e^{tA})=1$ for all real $t$. Zero trace eliminates the first-order change in the approximate step, and it preserves volume exactly in the matrix exponential flow. The examples use an abstract parameter $t$ and coordinate vectors; no physical units are assumed.

For a symmetric nonnegative integer matrix, the two neighboring prefix shapes along the diagonal are equal. In the RSK growth-diagram local rule, set $\mu =\nu$ and take alternating sums of row lengths. The two occurrences of the alternating sum of $\mu$ cancel, leaving $o(\lambda )=o(\rho )+a_{ii}$. Induction along the diagonal proves the formula. Each column contributes one to this alternating sum exactly when its length is odd.

For $P=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$, the off-diagonal equations in $P^2=P$ are $b(a+d-1)=c(a+d-1)=0$. Thus either $b=c=0$ or the trace is one. The condition alone is not sufficient: $\left(\begin{array}{cc}1& 1\\ 1& 0\end{array}\right)$ has trace one but is not an idempotent matrix. For a two-by-two matrix of trace one, idempotence is equivalent to zero determinant, by the Cayley-Hamilton theorem.

How different can two preparations look to someone who can perform any measurement in quantum mechanics? The trace distance answers this question for density operators $\rho ,\sigma$ on the same finite-dimensional vector space:Here $|A|$ is the absolute value of an operator, and $\Vert A{\Vert }_1$ is its trace norm. Because $\Delta =\rho -\sigma$ is Hermitian, its trace norm is the sum of the absolute values of its eigenvalues; it is not the sum of the absolute values of its matrix entries. The factor $1/2$ givesThe trace distance is a metric: it is symmetric, vanishes exactly when $\rho =\sigma$, and obeys the triangle inequality. Identical preparations have trace distance zero; preparations with orthogonal occupied eigenspaces have trace distance one and can be identified perfectly in a single measurement in quantum mechanics.

Operationally, the trace distance is the largest total variation distance between the outcome probability distributions obtainable by a POVM. Equivalently, the variational characterization of trace distance says it is the largest possible probability gap for one event in a binary measurement in quantum mechanics. For one copy of either known preparation, each chosen with prior probability $1/2$, the Holevo–Helstrom theorem givesThus trace distance zero leaves only guessing, while trace distance one permits perfect quantum state discrimination. The optimum ranges over measurements; different measurement bases can expose different amounts of distinguishability. ^[ref]

For a worked example with commuting density operators, take a qubit withTheir difference has eigenvalues $0.5,-0.5$, so $D=(|0.5|+|-0.5|)/2=0.5$. Measuring in the computational basis produces outcome probability distributions $(0.8,0.2)$ and $(0.3,0.7)$, whose total variation distance is also $0.5$. Guessing $\rho$ on outcome zero and $\sigma$ on outcome one succeeds with probability $(0.8+0.7)/2=0.75$. A projective measurement in the Hadamard basis instead produces $(0.5,0.5)$ for both preparations and gives no discrimination advantage.

The trace-distance contraction under quantum channels ensures that a common quantum channel, including the partial trace, cannot increase the trace distance. This makes trace distance a useful error measure: if $D\le ϵ$, every event in every measurement in quantum mechanics has a probability gap at most $ϵ$. Losing access to a subsystem can hide distinguishability, but cannot create more of it.

For positive distinct diagonal entries $c_1<\cdots <c_n$, the critical points of this function are diagonal sign matrices $D=\mathrm{diag}({\epsilon }_1,\dots ,{\epsilon }_n)$ with $\prod {\epsilon }_i=1$. Criticality forces $CX$ symmetric; its square is $C^2$, so it commutes with the distinct-spectrum diagonal matrix $C^2$ and is diagonal. In the skew coordinates $a_{ij}$ its Hessian is $-{\sum }_{i<j}(c_i{\epsilon }_i+c_j{\epsilon }_j)a_{ij}^2$. The sign of each parenthesis is that of ${\epsilon }_j$, giving Morse index ${\sum }_{j:{\epsilon }_j=1}(j-1)$.

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

For linear maps $A,B$ on a finite-dimensional vector space, the cyclic property of the trace gives $\mathrm{tr}(AB-BA)=\mathrm{tr}(AB)-\mathrm{tr}(BA)=0$. In matrix entries both traces are ${\sum }_{i,j}{A}_{ij}{B}_{ji}$ after exchanging indices. Thus the trace of a finite-dimensional Lie algebra representation vanishes on the derived algebra. In particular, for a perfect Lie algebra, every representing matrix has zero trace.

The trace of the curvature form of a connection is a global closed differential form of degree two. Under a frame change $e^{\prime }=eg$, the connection matrix and curvature change by $A^{\prime }=g^{-1}Ag+g^{-1}dg$ and ${\Theta }^{\prime }=g^{-1}\Theta g$. Thus trace is independent of frame. The Bianchi identity gives $d\Theta =\Theta \wedge A-A\wedge \Theta$. Taking the trace cancels the two terms: one-forms commute with two-forms in the graded sense, and renaming the matrix indices gives equality of their traces. Locally $\mathrm{tr}\Theta =d\mathrm{tr}A$, since the trace of $A\wedge A$ vanishes by pairing its off-diagonal terms. It is the curvature of the determinant connection.

Let $S$ be a real symmetric matrix and $A$ a real antisymmetric matrix of the same size. Invariance of the trace under transpose and the cyclic trace identity giveHence $\mathrm{tr}(SA)=0$. Equivalently these two matrices are orthogonal for the Frobenius inner product, since $⟨S,A{⟩}_F=\mathrm{tr}(S^{T}A)=\mathrm{tr}(SA)$. Together with the unique symmetric and antisymmetric parts of a matrix, this is an orthogonal decomposition of the real square matrices.

With row vectors, external arrival rates $\nu$ and transient routing matrix $P$, the total arrival rates satisfy $\alpha =\nu +\alpha P=\nu (I-P{)}^{-1}$. These include all internal migrations and do not depend on colony service speeds.

For a nearest-neighbour classical chain with finitely many spin states and dimensionless bond energy $E(s,t)$, its transfer matrix is $W_{st}=e^{-E(s,t)}$. On-site energies are divided between adjacent bonds. Periodic boundary conditions give partition function $Z_N=\mathrm{tr}{W}^N$ by direct multiplication and summation over spin labels. For a strictly positive symmetric matrix, the Perron–Frobenius theorem gives ${\lim }_{N\to \infty }{N}^{-1}\log Z_N=\log {\lambda }_{\max }$. A diagonal local observable $O$ has expectation $\mathrm{tr}(O{W}^N)/Z_N$, tending to $v_{\max }^{T}O{v}_{\max }$ for a normalized dominant eigenvector.

A method for counting or weighting configurations by processing successive slices and recording only the information that crosses the current boundary. Compatible boundary states define a matrix of transitions, so repeated slices can be handled by matrix powers. It is especially useful when the boundary is much smaller than the whole configuration.

For a time-homogeneous continuous-time Markov chain, $P_{ij}(t)={\mathbb{P}}_i(X_t=j)$ and the matrices satisfy $P(s+t)=P(s)P(t)$, the Chapman-Kolmogorov equation. On a finite state space with Q-matrix $Q$, the matrix exponential gives $P(t)=e^{tQ}$.

The distinction from a discrete-time transition matrix is the elapsed-time parameter: $P(t)$ describes where the process is after $t$ time units, allowing any number of intervening jumps. Each $P(t)$ is a stochastic matrix, with $P(0)=I$. A Q-matrix instead records instantaneous rates and has row sums zero. Sampling at a fixed spacing $\Delta$ yields a Discrete-time Markov chain with transition matrix $P(\Delta )$. See Hao Wu's MIT lecture 21, transition semigroups and lecture 22, infinitesimal rates.

For a linear map $T:V\to W$, the dual map $T^{\ast }:W^{\ast }\to V^{\ast }$ sends a linear functional $\phi$ on the output space to the input measurement $T^{\ast }(\phi )=\phi \circ T$. It reverses direction because a measurement of the output can be evaluated on an input by first applying $T$:For $T(x,y)=(x+y,2y)$ and $\phi (u,v)=2u-v$, the input measurement is $T^{\ast }\phi (x,y)=2(x+y)-2y=2x$. The output measurement ignores the part of this transformation contributed by $y$.

If $T$ has matrix $A$ in specified bases, a row representing $\phi$ changes to $\ell A$. Storing the coefficients of a linear functional as a column in the corresponding dual basis gives $A^T{\ell }^T$, explaining why the transpose of a linear map is represented by the matrix transpose. This algebraic operation uses no inner product; it agrees with a matrix transpose even over the complex numbers, rather than automatically introducing complex conjugation. The adjoint operator is defined using an inner product and is a different construction. ^[ref].

In finite dimension, the kernel and image of the dual map are described by annihilators of vector subspaces:

For compatible finite matrices, the transpose reverses the factor order. Entrywise, $[(AB{)}^T{]}_{ij}={\sum }_k{A}_{jk}{B}_{ki}={\sum }_k(B^T{)}_{ik}(A^T{)}_{kj}$, proving the identity. The analogous formula for conjugate transpose is $(AB{)}^{†}=B^{†}{A}^{†}$. Commutativity of scalar entries permits the last rearrangement; the matrix products themselves are not assumed to commute.

Consider an original ideal calculation with no heating, decoherence, detuning or spectator modes. In the ordered basis $(|g,n⟩,|e,n-1⟩)$ for $n\ge 1$, the red motional sideband has $H_r=\hslash \eta \Omega \sqrt{n}{\sigma }_x/2$. Here ${\sigma }_x$ is the Pauli matrix exchanging the two basis states. Since ${\sigma }_x^2=I$, its unitary time evolution isThus $U_n^{†}{U}_n=I$, and excitation from $|g,n⟩$ has probability $P_r(n,t)={\sin }^2(\theta \sqrt{n}/2)$. For $n=0$ it vanishes. The blue motional sideband similarly gives $P_b(n,t)={\sin }^2(\theta \sqrt{n+1}/2)$. A red Rabi pi pulse calibrated for $n=1$ has $\theta =\pi$; it is not a Rabi pi pulse for every occupied $n$.

Choose $\eta =0.08$, $\Omega /(2\pi )=100\mathrm{kHz}$ and $\nu /(2\pi )=2\mathrm{MHz}$. The $n=1$ red Rabi pi pulse lasts $t_{\pi }=\pi /(\eta \Omega )=62.5\mu \mathrm{s}$. Let motion have the Boltzmann distribution of a quantum harmonic oscillator, with chosen mean occupation $\bar{n}=0.25$:Here $\sum p_n=1$, $\sum n{p}_n=\bar{n}$ and $p_0=0.8$. Reindex the red sum using $n=m+1$ and $p_{m+1}=r{p}_m$: $P_r(t)=r{P}_b(t)$ for equal resonant pulse parameters, even outside the short-pulse limit. At times with $P_b>0$, the ratio $R=P_r/P_b$ therefore gives $\bar{n}=R/(1-R)$. A nonthermal distribution, unequal drive strengths, detuning or state-detection bias can invalidate that thermometer. Ground-state population cannot generally be deduced from the mean alone without specifying the distribution.

The chosen pulse gives $P_r\simeq 0.181393$ and $P_b\simeq 0.906967$. These are internal excitation probabilities, not quantum gate fidelities or cooling-cycle efficiencies. An independent matrix exponentiation of each two-state block agrees with the sine formula and preserves normalization. Summing $n=0,\dots ,79$ omits probability $r^{80}<1.21\times 10^{-56}$; because every sine squared lies between zero and one, this also bounds each omitted response. The thermal relation supplies a separate dimensional check:Here $k_B$ is the Boltzmann constant. Substitution recovers $\bar{n}=0.25$; the thermal root-mean-square expansion parameter is $\eta \sqrt{2\bar{n}+1}\simeq 0.098$. This checks the populated part of the Lamb-Dicke regime, while exponentially rare large $n$ still exceed any fixed small-parameter criterion. The pulse is a leading-order model; it does not bound corrections from the full coupling or experimental noise.

A linear map on a finite-dimensional vector space over an algebraically closed field has both an upper triangular matrix and a lower triangular matrix in suitable bases. One proof uses invariant subspaces and annihilators to make the induction step entirely in the dual space.

Choose an eigenvector $v$ and let $U=\mathrm{span}\{v\}$. The dual map preserves the annihilator of a vector subspace $U^0$, of dimension one less. By induction, choose a basis of $U^0$ making the restriction upper triangular. Extend this to a basis of the whole dual space; its matrix $C$ remains upper triangular because $U^0$ is an invariant subspace. In the corresponding primal basis, the matrix is $C^T$, which is lower triangular. Reversing the primal basis makes it upper triangular. Dimension zero and one start the induction. This also allows the original eigenvector to be the first vector of the resulting upper triangular basis.