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Imposing vanishing off-diagonal entries with auxiliary integrations and an off-diagonal Grassmann Gaussian integral gives a determinant of the commutator map on the off-diagonal matrix units. Its eigenvalues are ${\lambda }_i-{\lambda }_j$, $i\ne j$. Up to a constant sign, their product is the squared Vandermonde determinant. It contributes $-2{\sum }_{i<j}\log |{\lambda }_i-{\lambda }_j|$ to the eigenvalue action, encoding repulsion between distinct eigenvalues. Diagonal zero modes and constant normalization factors must be treated separately.

If a real square matrix satisfies $M^2=-I$, split the absolutely convergent matrix exponential into even and odd powers to obtain $e^{tM}=I\cos t+M\sin t$. It is the fundamental matrix for ${\mathbf{x}}^{\prime }=M\mathbf{x}$. Consequently every solution has period $2\pi$, and the eigenvalues over the complex numbers belong to $\{i,-i\}$.

Two square matrices $A,B$ over the same field are similar if $B=S^{-1}AS$ for an invertible matrix $S$. They can describe one linear operator using two different ordered bases, with the same basis used for input and output in each description. Fix the convention $x=Sy$: $x$ is the old coordinate vector, $y$ the new coordinate vector, and the columns of $S$ are the new basis vectors in old coordinates. This is the convention of the change-of-basis matrix.^[ref]

For a concrete linear operator on the physical plane, let $T$ double horizontal components and multiply vertical components by five. In the old standard basis $\mathcal{E}=(e_1,e_2)$, its matrix of a linear map is $A=\mathrm{diag}(2,5)$. Choose the new ordered basis $\mathcal{F}=(b_1,b_2)$ with $b_1=e_1$, $b_2=e_1+e_2$. ThenThe second column of $B$ says $T(b_2)=-3b_1+5b_2=2e_1+5e_2$. Its off-diagonal entry records how the chosen basis vectors describe the output; the physical linear operator still stretches along the old horizontal and vertical directions.

Take the physical vector $p=2e_1+e_2=b_1+b_2$. Its two coordinate vectors, and the two descriptions of $T(p)$, areReconstruction gives $T(p)=4e_1+5e_2=-b_1+5b_2$. The output coordinate vectors differ, but the endpoint in the physical plane is the same.

This is why matrix similarity is useful: a convenient basis can simplify calculations without changing the linear operator. A diagonalizable matrix can be replaced by a diagonal matrix, making repeated application a calculation with powers of its eigenvalues. In the example, $A$ makes the two independent stretch factors visible, while $B$ expresses the same action in directions chosen for another purpose. Matrix similarity invariants separate properties of the linear operator from features of its address system.

Matrix similarity invariants are quantities unchanged by replacing $A$ with $S^{-1}AS$. Important examples are the characteristic polynomial, trace, determinant, matrix rank, and eigenvalues with their algebraic multiplicities. The characteristic polynomial convention here is ${\chi }_A(t)=\det (tI-A)$, where $I$ is the identity matrix.^[ref]

The short determinant proof works over any field: because $tI-B=S^{-1}(tI-A)S$, multiplicativity of the determinant givesThus its roots, the eigenvalues, retain their algebraic multiplicities in any field extension where the polynomial splits. Also $\det B=\det A$. The cyclic property of the trace gives $\mathrm{tr}(S^{-1}AS)=\mathrm{tr}(AS{S}^{-1})=\mathrm{tr}A$; equivalently, the trace is determined by the coefficient of $t^{n-1}$ in the characteristic polynomial for an $n$-by-$n$ matrix.^[ref]

For the two matrices in the opening example,Their entries differ, yet these quantities agree. Agreement is a useful necessary test for matrix similarity; it is not sufficient in general, as same eigenvalues do not imply similarity illustrates.

The relation also preserves repeated application. Adjacent factors $S{S}^{-1}$ cancel to giveFor the opening example, matrix powers therefore have the explicit formAt $k=2$, this is $\left(\begin{array}{cc}4& -21\\ 0& 25\end{array}\right)$. Acting on $y=(1,1{)}^T$ gives $(-17,25{)}^T$, which reconstructs as $S(-17,25{)}^T=(8,25{)}^T=A^{2}x$. More generally $q(B)=S^{-1}q(A)S$ for every polynomial $q$, so the minimal polynomial is also unchanged. Multiplication by invertible matrices preserves matrix rank, giving another invariant without finding eigenvalues.

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.

The maximally mixed density operator on an $N$-dimensional Hilbert space assigns equal weights to every vector of any orthonormal basis. It is invariant under every unitary operator. Its purity is $\mathrm{Tr}({\rho }_{\ast }^2)=1/N$, the minimum possible because the nonnegative eigenvalues sum to one and obey the Cauchy-Schwarz inequality. For $N>1$ it is a mixed state; for $N=1$ it is the sole pure state. A rank-$k$ projector has probability $k/N$ in this state.

Under the energy-closed boundary assumptions of Backus' necessary condition for dynamo action, the magnetic energy equation gives ${\dot{E}}_B\le 2s(t)E_B$ after discarding nonnegative resistive dissipation. Integrating gives an upper bound by the time average of the largest spatial eigenvalue of the rate-of-strain tensor. Thus the field-amplitude exponent is at most its space-time supremum. The exponent of squared energy is twice the field-amplitude exponent.

A POVM $\{E_i\}$ gives outcome probability distributions $p_i=\mathrm{Tr}(\rho E_i)$ and $q_i=\mathrm{Tr}(\sigma E_i)$ by the Born rule. Their classical overlap, the Bhattacharyya coefficient, is $B(p,q)={\sum }_i\sqrt{p_i{q}_i}$. Fuchs and Caves showed that minimizing this overlap over all POVMs gives the unsquared quantum fidelity:The same POVM is applied to either possible input quantum state. Each measurement in quantum mechanics has $B(p,q)\ge F$; some measurement attains equality, including when the density operators have zero eigenvalues. Thus quantum fidelity quantifies the smallest classical overlap that observation can reveal. For fidelity of commuting mixed states, measurement in a common eigenbasis attains the minimum. This operational interpretation compares statistics without assigning a probability that two mixed states are equal. ^{[ref] [ref]}

Here $Z$ has the standard normal distribution. Applied to the generating function $e^{sx-s^2/2}$, the Gaussian moment-generating function replaces $s$ by $e^{-t}s$, proving the Hermite eigenvalue identity. Positivity and preservation of Gaussian measure then give the $L^2$ contraction and identify the full Ornstein-Uhlenbeck semigroup by polynomial density.

For $u_t=(a{u}_x{)}_x$, this conservative finite difference method uses coefficient samples at edge midpoints. For smooth $a,u$, its spatial residual is $h^2(a{u}_{xxxx}/12+a^{\prime }{u}_{xxx}/6+a^{\prime \prime }{u}_{xx}/8+a^{\prime \prime \prime }{u}_x/24)+O(h^4)$. With zero Dirichlet boundary conditions, $-v^T{L}_{h}v=h^{-2}{\sum }_j{a}_{j+1/2}(v_{j+1}-v_j{)}^2$. If $0<a\le a_{+}$, its eigenvalues lie in $[-4a_{+}/h^2,0]$, giving the mesh-independent Forward Euler method bound $\Delta t/h^2\le 1/(2a_{+})$. Constant $a=a_{+}$ proves sharpness as the grid is refined.

A mixed state is a density operator that is not a rank-one orthogonal projection. In finite dimension it is characterized by $\mathrm{Tr}({\rho }^2)<1$, so its purity of a density operator distinguishes it from a pure state.

Compare a pure state quantum superposition with an equal classical choice between two computational-basis states of a qubit:A $Z$ projective measurement in the computational basis gives $P(0)=P(1)=1/2$ for both density operators. An $X$ projective measurement in the Hadamard basis $|\pm ⟩=(|0⟩\pm |1⟩)/\sqrt{2}$ distinguishes them: the Born rule gives $P(+)=1$ for ${\rho }_{+}$ and $P(+)=1/2$ for ${\rho }_{\mathrm{mix}}$. These comparisons use separate copies of each preparation; measuring $Z$ first changes the state tested by a subsequent $X$ measurement.

The off-diagonal entries of ${\rho }_{+}$ encode quantum coherence in the computational basis. Such quantum coherence depends on the basis: ${\rho }_{+}$ is diagonal in the Hadamard basis but remains a pure state. Mixedness is independent of the basis, because a unitary change of basis preserves the eigenvalues and the purity of a density operator.

Even the quantum state ensemble is not unique:The off-diagonal entries of the two Hadamard basis projectors cancel. Thus the density operator retains precisely what measurements on the system can reveal, while an accessible preparation label carries additional information. A mixed state can also be the reduced density matrix of an entangled state, even when the whole system is in a pure state; classical ignorance about a selected component is not its only physical origin. ^[ref]

For $\gcd (x,n)=1$ and order $r$, multiplication by $x$ modulo $n$ is a unitary operator. On the orbit of one it cyclically permutes $r$ basis states. The displayed orthonormal eigenstates have eigenvalues $e^{2\pi is/r}$, and $|1⟩=r^{-1/2}{\sum }_s|u_s⟩$. Thus quantum phase estimation on this efficiently prepared state samples a uniform $s$ without knowing $r$. Controlled powers are implemented by quantum modular exponentiation. Sufficient phase resolution allows continued-fraction recovery in quantum order finding; the reduced denominator equals $r$ only when $s$ is coprime to $r$.

For a dimensionless chemostat with Monod equation kinetics,the total nutrient equivalent $z=c+n/\alpha$ satisfies $z^{\prime }=\beta -z$. The positive equilibrium is $c_{\ast }=1/(\alpha -1)$, $n_{\ast }=\alpha (\beta -c_{\ast })$, provided $\alpha >1$ and $\beta >c_{\ast }$. The Jacobian matrix has eigenvalues $-1$ and $-n_{\ast }/(1+c_{\ast }{)}^2$, so the positive equilibrium is locally asymptotically stable.

For a linear ordinary differential equation with holomorphic coefficients on a punctured disc, analytic continuation of a basis gives an invertible monodromy matrix. A nonzero eigenvector with eigenvalue $\lambda =e^{2\pi i\sigma }$ selects a solution $w$ for which $z^{-\sigma }w$ is single-valued and holomorphic. Its Laurent series gives $w=z^{\sigma }{\sum }_{n\in \mathbb{Z}}{c}_n{z}^n$. No regular-singular hypothesis is needed; infinitely many negative powers may occur.

For the homogeneous linear differential equation $\dot{Y}=K(t)Y$ with continuous coefficient matrix of period $P>0$, the monodromy matrix of a periodic linear system is the one-period map from $Y(0)$ to $Y(P)$. If $X$ is any fundamental matrix,Its eigenvalues are the Floquet multipliers. Multipliers are nonzero, since a fundamental matrix is invertible. Changing the starting phase conjugates $N$, so the Floquet multipliers and Jordan blocks are independent of that choice.^[ref]

This turns a continuous-time problem into repeated matrix multiplication:If $Nv=mv$, the associated solution of a differential equation obeys $Y(t+P)=mY(t)$ at every time, not merely at the sampled times. In particular, $Nv=v$ gives a nonzero $P$-periodic solution, and every nonzero $P$-periodic solution arises this way. If $Nv=-v$, the state changes sign after one period and repeats after two. A unit-modulus Floquet multiplier different from $1$ need not give repetition after one period, and a generalized eigenvector in a nontrivial Jordan block need not even give bounded motion.

For the scalar example under periodic solution, $N$ is the one-by-one matrix $(e^{2\pi a})$. Numerically solving a matrix problem uses the same idea: integrate one initial value problem for each column of $X(0)=I$ through one period, then compute the eigenvalues of $X(P)$. This is useful when an explicit solution of a differential equation is unavailable: one period determines the Floquet growth rate over arbitrarily many periods.

For piecewise constant coefficients, which are also covered by the one-period construction, $N$ is the time-ordered product of the segment matrix exponentials, with the earliest segment on the right. For example, take period $P=2$ and alternateSince $K_1^2=K_2^2=0$,Replacing the coefficient by its mean would instead give $e^{K_1+K_2}=\left(\begin{array}{cc}\cosh 1& \sinh 1\\ \sinh 1& \cosh 1\end{array}\right)$ and the incorrect rate $0.5$. The matrices do not commute. The scalar mean-coefficient formula therefore cannot simply be transferred to matrix systems. Nevertheless the Liouville formula for a fundamental matrix gives the correct determinant even in the noncommuting case:In the example, this is $1=m_{+}{m}_{-}$; it fixes the product of the Floquet multipliers, rather than either multiplier separately.

In Sur les équations différentielles linéaires à coefficients périodiques (1883), Gaston Floquet calls a function satisfying $f(t+P)=mf(t)$ periodic of the second kind, and distinguishes the genuinely periodic function obtained when $m=1$. This historical terminology captures precisely the difference between repetition of the coefficient and repetition of the state.^[ref]