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The null space consists of concentration patterns on which transport produces no change. Here it is the familyEvery box has the same concentration. Ocean circulation may continue, but inflow and outflow of the chemical tracer balance in each box. With $\mathbf{b}=0$ these are exactly the steady states; physically admissible members have $\alpha \ge 0$. With nonzero $\mathbf{b}$, adding $\alpha 1$ to any particular steady state leaves its transport balance unchanged, but a null space vector alone need not balance the sources and sinks.

To identify the tracer inventory, apply conserved inventory from a left null vector. Directly summing each column givesConsequently, for the supplied ordinary differential equations,For $\mathbf{b}=0$, $M$ is a conserved quantity. The connected box model equalizes concentrations, approaching the unique member of the null space with the initial inventory:The arithmetic mean is appropriate here because the supplied matrix has zero column sums. In a general box model the conserved quantity is $w^T\mathbf{c}$, where $w$ is a positive left null space vector; for physical concentrations expressed per unit volume, these weights represent the box volumes up to a common factor. One must establish the weights from the equations, rather than infer equal volumes from a schematic drawing.

Nonnegative initial concentrations remain nonnegative under transport alone. Indeed $A+2I$ is a nonnegative matrix, sohas nonnegative entries for $t\ge 0$: it is a positive matrix semigroup. Its column sums are one by $1^{T}A=0$. To see why equalization occurs, set $d=\mathbf{c}-(M/4)1$ for $\mathbf{b}=0$; then $1^{T}d=0$ andThe right side vanishes only for constant $d$, which in the zero-sum vector subspace means $d=0$. Its strict negative bound on the unit sphere in that vector subspace gives decay to zero. Thus the matrix exponential redistributes inventory without creating negative concentrations; arbitrary prescribed sinks can still exhaust the tracer if they are allowed to act indefinitely.

Differentiate the planar rotation matrix at angle zero:The even and odd powers in the matrix exponential therefore giveThis exact formula gives the whole one-parameter subgroup. Taking $X=J$ means unit angular velocity: at $t=\pi /2$, $\exp (tJ)=J$ sends $(1,0{)}^T$ to $(0,1{)}^T$, a counterclockwise quarter-turn. At $t=\pi$, it equals $-I$; at $t=2\pi$, it returns to $I$. Choosing $X=2J$ reaches the quarter-turn at $t=\pi /4$. Changing the generator changes the speed, while preserving the circular path.

For a moving vector $v(t)=R_{t}v(0)$, the ordinary differential equation $v^{\prime }(t)=Jv(t)$ says its velocity is perpendicular to its position. The beautiful compression is that one constant matrix specifies every angle of this endless family of rotations.

A matrix exponential defines a positive matrix semigroup if $e^{tA}$ maps nonnegative vectors to nonnegative vectors for all $t\ge 0$. Here the inequality means entrywise nonnegativity, with zero entries allowed. In finite dimensions this holds exactly when the real matrix $A$ has nonnegative off-diagonal entries: if $A+\gamma I\ge 0$, expand $e^{tA}=e^{-\gamma t}{\sum }_{k\ge 0}{t}^k(A+\gamma I{)}^k/k!$; conversely, the off-diagonal entries of $e^{tA}=I+tA+O(t^2)$ force $a_{ij}\ge 0$ for $i\ne j$. Zero column sums additionally preserve $1^{T}c$ under $c^{\prime }=Ac$.

Multiplying the two matrix exponentials with the Pauli matrix multiplication law givesOne has $C^2+|\mathbf{v}{|}^2=1$. If $|\mathbf{v}|\ne 0$, put $\theta =2\arccos C$ and $\mathbf{n}=\mathbf{v}/|\mathbf{v}|$. The exponential of a Pauli vector gives $U_F^N=I\cos (N\theta /2)-i\mathbf{n}\cdot \mathbf{\sigma }\sin (N\theta /2)$. Its action on a Bloch vector is a rotation through $N\theta$ about $\mathbf{n}$, and its quasienergies are $\pm \hslash \theta /(2T)$ modulo $2\pi \hslash /T$. When $U_F=\pm I$, its axis is arbitrary and the Bloch vector is unchanged.

Reachability in control theory asks about a particular initial state $x_0$ and target $x_1$: does some admissible control input make $x(T)=x_1$ at the declared time $T$, or at some allowed duration? Controllability requires this for every pair in the declared state space. This is a property of state evolution, distinct from graph reachability.

For a linear time-invariant system, solving the ordinary differential equation by a matrix exponential givesThe first term is the unforced endpoint. Reachability in control theory asks whether the control input can supply exactly the displacement $x_1-e^{AT}{x}_0$. The timing matters because the same command is carried by different amounts of subsequent dynamics. ^[ref]

Here is a contrasting mathematical example with one shared control input:The controllability matrix has rank of a matrix one. Since $\frac{d}{dt}(z_1-z_2)=0$, the difference is invariant under every allowed control input. From $(0,0)$ the reachable set is the diagonal $z_1=z_2$ when input amplitudes are unrestricted: $(1,1)$ is reachable at $T=2$ using $u=1/2$, while $(1,0)$ is impossible at every duration. Having a control input appear in both state equations did not give independent control of both states. This reachable set is pictured alongside the bounded double integrator below.

For a real angle $\alpha$ and a unit vector $\mathbf{n}$, the rotation about $\mathbf{n}$ sendsThe vector triple product identity gives the equivalent expression $\mathbf{x}+\sin \alpha \mathbf{n}\times \mathbf{x}+(1-\cos \alpha )\mathbf{n}\times (\mathbf{n}\times \mathbf{x})$. To see why it is a rotation, decompose $\mathbf{x}$ into its parallel and perpendicular components relative to $\mathbf{n}$. The parallel component is fixed, while the perpendicular component becomes $\cos \alpha {\mathbf{x}}_{\perp }+\sin \alpha \mathbf{n}\times {\mathbf{x}}_{\perp }$, the usual plane rotation. It is also the matrix exponential of the linear operator $H\mathbf{x}=\alpha \mathbf{n}\times \mathbf{x}$: the identity $H^3=-{\alpha }^{2}H$ reduces its exponential series to these three terms.

In the three-dimensional vector representation, the Hermitian matrices $(J_k{)}_{ij}=-i{ϵ}_{kij}$ generate active rotations by the matrix exponential $R(\theta ,\mathbf{n})=\exp (-i\theta n^k{J}_k)$. They satisfy $[J_i,J_j]=i{ϵ}_{ijk}{J}_k$. The real antisymmetric matrices $A_k=-i{J}_k$ are the corresponding Lie algebra generators of $SO(3)$. Their commutator gives the leading change produced by interchanging two infinitesimal rotations.

A complex matrix $A$ is skew-Hermitian when $A^{†}=-A$. Its matrix exponential $e^{tA}$ is unitary for every real $t$.

For a nonzero real three-dimensional skew-symmetric matrix, choose real orthogonal vectors $u,v$ of equal length with $Au=\theta v$, $Av=-\theta u$ and $\theta >0$, and a real unit vector $e_3$ in its kernel. The matrix exponential acts by $e^{A}u=u\cos \theta +v\sin \theta$, $e^{A}v=v\cos \theta -u\sin \theta$ and $e^A{e}_3=e_3$. Its matrix in this orthonormal basis is a planar rotation matrix with an unchanged axis. A purely imaginary nonzero eigenvalue and its complex eigenvector give $u,v$ by taking real and imaginary parts; antisymmetry proves their equal norms and orthogonality.

Use the cross-product Lie algebra convention $X=\widehat{e_1}$, $Y=\widehat{e_2}$, $Z=\widehat{e_3}$, so $[X,Y]=Z$. Put $R_x(t)=\exp (tX)$ and $R_y(t)=\exp (tY)$, using the matrix exponential. The finite group commutator isHere $\epsilon \to 0$ and big O notation refers to any fixed matrix norm. Active rotation matrices act on column vectors from right to left: first rotate about $y$ by $-\epsilon$, then about $x$ by $-\epsilon$, then about $y$ by $+\epsilon$, and finally about $x$ by $+\epsilon$. Swapping $x$ and $y$ replaces the product by its inverse and changes the leading Lie bracket to $-Z$.

Writing $c=\cos \epsilon$ and $s=\sin \epsilon$, direct matrix multiplication givesFor $\epsilon =0.1$ radians, $C_{\epsilon }{e}_1\approx (0.999950208,0.009966462,0.000501230)$, whereas $e_1+{\epsilon }^{2}Z{e}_1=(1,0.01,0)$. Thus the residual is a positive rotation about $z$ to leading order; at finite $\epsilon$ it also has smaller components. The Lie bracket captures the leading effect of the four moves rather than their entire finite motion. See Etingof, Section 7.2, Proposition 7.11.

The left panel projects the actual four arcs onto the $yz$-plane for $\epsilon =0.25$; the start and endpoint are close but distinct. The right panel divides the residual coordinates by ${\epsilon }^2$, making the limit $(\Delta y/{\epsilon }^2,\Delta z/{\epsilon }^2)\to (1,0)$ visible. This is a numerical picture of $[X,Y]=Z$ in the SO(3) Lie algebra.

The spectral abscissa is the largest real part of a matrix's eigenvalues. It describes asymptotic exponential rates of its matrix exponential, but a non-normal matrix can also have transient amplification. It is generally distinct from the Euclidean logarithmic norm, which gives an immediate Euclidean energy-growth bound.

For a finite-dimensional Hermitian matrix $A$, its spectral projection $P_{\lambda }$ is the orthogonal projection onto the eigenspace for $\lambda$. In an orthonormal eigenbasis, the displayed sum is over every basis vector with that eigenvalue. The projection is independent of the choice of basis within a repeated eigenspace. For distinct eigenvalues $\lambda ,\mu$,This is the spectral decomposition grouped by distinct eigenvalues: decompose a vector into orthogonal pieces, then scale each piece. For the matrix $A=\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$,Thus $P_3(1,0{)}^T=(1/2,1/2{)}^T$ and $P_1(1,0{)}^T=(1/2,-1/2{)}^T$: the two pieces add back to $(1,0{)}^T$. Squaring the matrix simply squares each scalar, $A^2=9P_3+P_1$, and similarly the matrix exponential is $e^{tA}=e^{3t}{P}_3+e^t{P}_1$. Spectral projections of a self-adjoint matrix turn repeated matrix calculations into scalar calculations on fixed vector subspaces.

In the ordered spin-one basis $(m=1,0,-1)$, the second angular momentum operator satisfies $J_2^3=J_2$. Therefore its matrix exponential reduces to $I-i\sin \theta J_2+(\cos \theta -1)J_2^2$. A half-turn exchanges the two outer weights and negates the middle weight. Applied to the pion isospin triplet, this fixes the rotation factor in G parity.

For a Dirac spinor a spatial rotation acts in each chiral block by a matrix exponential $e^{\pm i\theta \mathbf{n}\cdot \mathbf{\sigma }/2}$, depending on angle orientation. Since $(\mathbf{n}\cdot \mathbf{\sigma }{)}^2=I_2$ for a unit axis, the Pauli matrix multiplication law gives this exponential as $\cos (\theta /2)I_2\pm i\sin (\theta /2)\mathbf{n}\cdot \mathbf{\sigma }$. At $\theta =2\pi$ it is $-I_2$, and at $4\pi$ it is $I_2$. A vector returns already at $2\pi$. This describes the lift of the rotation loop to the Spin group, rather than a single-valued representation of the ordinary rotation group on spinors.

A matrix with the same number of rows and columns represents an endomorphism of a finite-dimensional vector space after choosing a basis. Square matrices support trace, determinant, powers and the matrix exponential.

Consider $q$ identical length-$n$ arms joined to one hub, with XY exchange interactions of strength $J_0/\sqrt{q}$ from the hub to each first site and strength $J_k$ between sites $k,k+1$ on each arm. Let $|s_0⟩$ be the hub excitation and $|s_k⟩=q^{-1/2}{\sum }_{a=1}^q|a,k⟩$. This orthonormal symmetric single-excitation subspace is invariant, and its restricted Hamiltonian operator iswith endpoint terms omitted. At the first link the $q$ hub contributions add to $J_0$; all other links act identically on the arms. Thus the linear isometry $F|k⟩=|s_k⟩$ intertwines $H$ with the path Hamiltonian operator $H_T={\sum }_{k=0}^{n-1}{J}_k(|k⟩⟨k+1|+|k+1⟩⟨k|)$: $HF=F{H}_T$. Expanding the matrix exponential gives $e^{-iHt}F=F{e}^{-i{H}_{T}t}$. Perfect quantum state transfer from zero to $n$ on the path therefore produces an equal excitation superposition at the $q$ arm tips. Since all other qubits are zero, the tip state is a pure W state, up to the transfer's common phase.

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 finite-state continuous-time Markov chain, a transition intensity matrix has nonnegative off-diagonal rates $q_{rs}$ and $q_{rr}=-{\sum }_{s\ne r}{q}_{rs}$. Every row sums to zero. In the time-homogeneous case its transition probability matrix is the matrix exponential $P(t)=e^{tQ}$. An absorbing state has a zero row.

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 $L=\left(\begin{array}{cc}{\lambda }_1& 0\\ 1& {\lambda }_2\end{array}\right)$ with distinct real negative eigenvalues, the matrix exponential is $A=\left(\begin{array}{cc}a& 0\\ b& d\end{array}\right)$, where $a=e^{{\lambda }_{1}t}$, $d=e^{{\lambda }_{2}t}$ and $b=(a-d)/({\lambda }_1-{\lambda }_2)$. The optimal energy amplification of a linear system isIts eigenvectors are nonorthogonal, and immediate energy growth occurs exactly when ${\lambda }_1{\lambda }_2<1/4$. If ${\lambda }_2>{\lambda }_1$, putting $\alpha =({\lambda }_2-{\lambda }_1{)}^{-1}$ gives $G(t)\sim (1+{\alpha }^2)e^{2{\lambda }_{2}t}$. The optimal asymptotic initial direction $(\alpha ,1)$ is an adjoint eigenvector, while the eventual state aligns with the right eigenvector $(0,1)$.