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For a diagonalizable matrix, the active spectrum consists of eigenvalues whose eigenspaces have nonzero components of the starting vector in its spectral decomposition. These components are orthogonal projections for a self-adjoint operator, but need not be orthogonal projections for a general diagonalizable matrix. With a unique largest modulus in this active spectrum, the power method selects its eigendirection; an exactly absent dominant component is never created in exact arithmetic.

A Hermitian matrix has a unitary spectral decomposition $A=Q\mathrm{diag}({\lambda }_j)Q^{†}$ with real eigenvalues ${\lambda }_j$. The matrix exponential respects similarity, giving $e^A=Q\mathrm{diag}(e^{{\lambda }_j})Q^{†}$. Its eigenvalues are positive and its conjugate transpose equals itself. In particular $v^{†}{e}^{A}v={\sum }_j{e}^{{\lambda }_j}|(Q^{†}v{)}_j{|}^2>0$ for $v\ne 0$, so it is a Hermitian positive-definite matrix.

The density operators on a finite-dimensional Hilbert space form a convex set. Its extreme points are exactly the rank-one projectors, hence exactly the pure states. To prove this, suppose $|\psi ⟩⟨\psi |={\sum }_i{a}_i{\rho }_i$ with $a_i>0$ and ${\rho }_i$ positive semidefinite. For $w\perp \psi$, $0={\sum }_i{a}_i⟨w|{\rho }_i|w⟩$ is a sum of nonnegative numbers, so every term vanishes. Diagonalizing ${\rho }_i$ shows that $⟨w|{\rho }_i|w⟩=0$ implies ${\rho }_{i}w=0$. Thus each ${\rho }_i$ is supported on the line spanned by $\psi$, and its unit trace forces ${\rho }_i=|\psi ⟩⟨\psi |$.

Conversely, a density operator of rank at least two has a spectral decomposition with at least two positive eigenvalues. If $\lambda$ is one of them, $0<\lambda <1$ and $\rho =\lambda |e⟩⟨e|+(1-\lambda )\sigma$, where $\sigma$ is the normalized remainder. These are distinct density operators, proving that $\rho$ is not an extreme point.

For a positive-definite Hermitian matrix with spectral decomposition $A=U\mathrm{diag}({\lambda }_j)U^{\ast }$, the matrix logarithm is $\log A=U\mathrm{diag}(\log {\lambda }_j)U^{\ast }$. More generally it is a branch of the inverse of the matrix exponential when such a branch exists.

A density matrix has unit trace. The Pauli matrices are traceless, so $\mathrm{Tr}\hat{\rho }=C$ and hence $C=1$. Using $\mathrm{Tr}({\sigma }_i{\sigma }_j)=2{\delta }_{ij}$ in the expectation value givesConsequently the required Bloch vector and density matrix areThe two eigenvalues are $3/4$ and $1/4$, both positive, so this is a valid density operator. Its purity of a density operator isTherefore it is a mixed state. Equivalently, its spectral decomposition is $\rho =(3/4)|+x⟩⟨+x|+(1/4)|-x⟩⟨-x|$, an ensemble of the two distinct spin one-half along an axis states, rather than a single pure state.

For an open chain there are $L-1$ bonds, with both endpoint Ising spins independently summed. Retain the same spin-chain transfer matrix, but replace the trace by endpoint vectors. With $1=(1,1{)}^{\mathsf{T}}$,This follows directly by expanding the product of $L-1$ Boltzmann factors. If ${\mathbf{v}}_{\pm }$ are orthonormal eigenvectors of the symmetric matrix $W$, the equivalent spectral decomposition isFixed endpoint Ising spins replace the corresponding $1$ by a basis vector; extra endpoint energies replace its entries by the corresponding Boltzmann factors. The leading bulk Helmholtz free energy per spin agrees with the periodic result, because the boundary changes only subextensive terms. In particular, $L=1$ gives $Z_1^{\mathrm{open}}=1^{\mathsf{T}}1=2$, as required for one free Ising spin.

The addition formulas for the hyperbolic functions implySince $M=B(y)/2$, its fourth matrix power isAlternatively, the orthonormal eigenvectors $(1,\pm 1)/\sqrt{2}$ have eigenvalues $e^{\pm y}/2$. Raising these eigenvalues to the fourth power gives the same result by spectral decomposition.

Classical multidimensional scaling begins with a symmetric dissimilarity matrix $D$, rather than necessarily with measured coordinates. Its aim is to find a low-dimensional Euclidean configuration representing those dissimilarities. Square the entries, not the matrix product, and defineFor centered coordinates $X$, squared Euclidean distances satisfy $D_{ij}^2=B_{ii}+B_{jj}-2B_{ij}$ with $B=X{X}^T$. Multiplication on both sides by $J$ annihilates the two single-index terms, proving the double-centering formula.

If $B$ is positive semidefinite, use its spectral decomposition $B=U\Lambda U^T$ to constructKeeping all positive eigenvalues reproduces the original Euclidean distances exactly; their number is the minimal embedding dimension. Keeping fewer gives a low-rank approximation to the Gram matrix. It minimizes squared Gram-matrix error, often called strain, rather than necessarily the sum of squared errors in raw distances. Negative eigenvalues signal that the dissimilarities cannot be represented exactly by Euclidean coordinates; discarding them produces an approximation whose adequacy should be inspected.

In the upper-right sketch, the supplied distances come from four points in a rectangle, and the two-dimensional reconstruction recovers that shape. The overall translation, rotation and reflection of the configuration are not identifiable from distances. Classical scaling reconstructs centered geometry from pairwise distances. If the input distances are computed from centered multivariate observations, classical scaling and principal component analysis give the same score configuration up to an orthogonal transformation: their matrices $Z{Z}^T$ and $Z^{T}Z$ share the nonzero squared singular values.

Let a density operator on a Hilbert space ${\mathcal{H}}_S$ have the spectral decompositionIntroduce an auxiliary Hilbert space ${\mathcal{H}}_R$ with an orthonormal basis containing $r$ vectors $|j{⟩}_R$, and setThis vector has norm one. Its partial trace isThis explicitly constructs a purification of a density operator. The same construction works for an arbitrary trace-class density operator: its nonzero eigenvalues form a finite or countable summable family, and the displayed vector converges in the Hilbert space norm. The entropy inequality below is understood with finite Von Neumann entropies, avoiding undefined differences of infinities.

Apply this construction with $S=AB$ to obtain a joint pure state on $ABC$. The Schmidt decomposition across $AB:C$, $A:BC$ and $B:AC$ respectively givesNow apply Subadditivity of Von Neumann entropy to the reduced density matrices on $BC$ and $AC$:These inequalities say $S(A)-S(B)\le S(C)$ and $S(B)-S(A)\le S(C)$. Replacing $S(C)$ by $S(AB)$ proves the Araki–Lieb inequality:The subadditivity used here concerns the actual reduced density operators, rather than statistical independence. In particular it follows from nonnegativity of quantum relative entropy, sinceand similarly for $AC$. Thus no independence assumption on the purification of a density operator is needed.

The printed normalization has its factors reversed. For the displayed map $\Phi (X)={\sum }_k{A}_{k}X{A}_k^{†}$, the condition ${\sum }_k{A}_k{A}_k^{†}=I$ says $\Phi (I)=I$, which is unitality. Trace preservation instead requiresThese are distinct trace-preserving and unital Kraus conditions. To disprove the printed equivalence, take $0<\gamma <1$ andThey obey ${\sum }_j{B}_j{B}_j^{†}=I$, butThus their completely positive map is not trace preserving. They are the adjoints of the operators of an amplitude damping channel. That channel is trace preserving but sends $I$ to $\mathrm{diag}(1+\gamma ,1-\gamma )$, so it cannot satisfy the printed normalization in any Kraus representation either. The corrected characterization is proved next.

Work in finite-dimensional input and output Hilbert spaces, with a linear map $\Phi$ on operators. Suppose first that $\Phi (X)={\sum }_k{A}_{k}X{A}_k^{†}$ and ${\sum }_k{A}_k^{†}{A}_k=I$. For every auxiliary dimension $m$ and every positive operator $W$ on the extended space,Each summand is positive because conjugation preserves positivity. This is exactly the definition of a completely positive map. Cyclicity of the trace givesso the map is also trace preserving.

Conversely, suppose $\Phi$ is completely positive and trace preserving, and let $|i⟩$ be an orthonormal input basis. Use the unnormalized entangled vector $|\Omega ⟩={\sum }_i|i⟩\otimes |i⟩$, with the reference factor first. Complete positivity makes the corresponding unnormalized Choi matrix positive:Its spectral decomposition can be written $J_{\Phi }={\sum }_k|v_k⟩⟨v_k|$, absorbing the nonnegative eigenvalues into the vectors. Define linear operators $A_k$ uniquely byExpanding the outer products and comparing their $(i,j)$ reference blocks yieldsThe matrix units span all operators, so linearity proves the Kraus representation $\Phi (X)={\sum }_k{A}_{k}X{A}_k^{†}$ for every $X$. Finally, trace preservation givesfor all $X$. Taking matrix units as test operators shows that the matrix in parentheses is zero. This proves both directions of the corrected CPT characterization, including the normalization rather than assuming it.

For an input quantum state ensemble $\{p_x,{\rho }_x\}$, write ${\sigma }_x=\Phi ({\rho }_x)$ and $\bar{\sigma }={\sum }_x{p}_x{\sigma }_x$. Its output Holevo quantity isThe Holevo capacity is the optimized one-use quantitywhere the supremum ranges over finite input ensembles and $S$ is the Von Neumann entropy in bits.

For $n$ uses of a memoryless quantum channel, the map is ${\Phi }^{\otimes n}$. An $n$-use code consists of a finite message set ${\mathcal{M}}_n$ of size $M_n$, an encoding state ${\rho }_m^{(n)}$ on the $n$ inputs for each message, and a decoding POVM $\{D_m^{(n)}:m\in {\mathcal{M}}_n\}$ on the output space, with $D_m^{(n)}\ge 0$ and ${\sum }_m{D}_m^{(n)}=I$. The encoding states may be entangled across uses. The measurement outcome is the receiver's decoded message, with conditional probabilityFor uniformly distributed messages the average error and rate areA rate $R$ is achievable if there is a sequence of such codes with $P_e^{(n)}\to 0$ and ${\mathrm{liminf}}_n{R}_n\ge R$. The unassisted classical capacity of a quantum channel is the supremum of achievable rates; the sender and receiver do not share prior entanglement.

The operational content of the Holevo-Schumacher-Westmoreland theorem is that ${\chi }^{\ast }(\Phi )$ is the capacity for product-state input encodings with collective output measurements. Allowing arbitrary entangled block inputs gives the regularized formulaThe limit equals the supremum because superadditivity of Holevo capacity follows by taking product ensembles for two blocks and using additivity of the Von Neumann entropy on product states. The coding theorem makes every rate below the one-block Holevo quantity achievable through many independently encoded blocks, with collective decoding.

The converse can be seen explicitly from the Holevo bound and Fano's inequality. For a uniform transmitted message $M$ and decoded message $\hat{M}$,The first inequality is obtained by subtracting the Fano bound on $H(M\mid \hat{M})$ from $H(M)={\log }_2{M}_n$. Dividing by $n$ and letting the average error vanish bounds every achievable rate by the regularized expression. The block-coding achievability just described supplies the reverse bound.

Under the assumed additivity of Holevo capacity across tensor powers, ${\chi }^{\ast }({\Phi }^{\otimes n})=n{\chi }^{\ast }(\Phi )$ for every $n$. ThusThe stated conditional assertion is true: under this additivity assumption, entangled input states cannot increase the asymptotic unassisted classical rate beyond the rate already achievable with product inputs. The decoder may still use collective measurements across channel outputs.

Let $D=\dim {\mathcal{H}}_A$ and introduce an isomorphic reference system $R$. Choose a basis $\{|j⟩{\}}_{j=1}^D$ and the maximally entangled stateComplete positivity makesa positive density operator. This is the normalized Choi matrix, with the reference placed first. Its spectral decomposition can be writtenThe number $r$ is at most $D^2$.

Here an index state is a chosen basis vector $|j{⟩}_R$ in the reference. The corresponding relative state of $|v_{\ell }⟩$ is the generally unnormalized vectorThus $|v_{\ell }⟩={\sum }_j|j{⟩}_R|{\eta }_{\ell j}{⟩}_A$. Define a linear map $A_{\ell }$ by its columns:Then $(I_R\otimes A_{\ell })|\Omega ⟩=\sqrt{{\lambda }_{\ell }}|v_{\ell }⟩$, soExpand both sides in reference matrix units. The block with reference indices $i,j$ on the left is $D^{-1}\Phi (|i⟩⟨j|)$; on the right it is $D^{-1}{\sum }_{\ell }{A}_{\ell }|i⟩⟨j|A_{\ell }^{†}$. Since matrix units span all operators and $\Phi$ is linear,This constructs the Kraus representation explicitly from the relative states, rather than assuming it.

Finally, trace preservation implies for every density operator $\rho$ thatTaking every rank-one orthogonal projection as $\rho$ forces the Hermitian operator in brackets to have the same quadratic form as the identity. ThereforeThe argument is finite-dimensional, as appropriate to the quantum-information setting; a channel with a different output dimension has the same construction with rectangular Kraus operators.

The stated condition is the definition of an extreme point in the convex set of density matrices. Suppose first that $\rho =|\psi ⟩⟨\psi |$ with $\Vert \psi \Vert =1$, and consider any convex combination $\rho ={\sum }_i{a}_i{\rho }_i$ with all $a_i>0$. For every $|w⟩$ orthogonal to $|\psi ⟩$,Every term is nonnegative, so $⟨w|{\rho }_i|w⟩=0$ for each $i$. To see that this implies ${\rho }_{i}w=0$, expand ${\rho }_i={\sum }_j{\mu }_j|u_j⟩⟨u_j|$, ${\mu }_j\ge 0$. The equality ${\sum }_j{\mu }_j|⟨u_j|w⟩{|}^2=0$ implies ${\mu }_j⟨u_j|w⟩=0$ for every $j$, which kills ${\rho }_{i}w$. Thus each density matrix ${\rho }_i$ vanishes on ${\psi }^{\perp }$. By self-adjointness its range lies in the line $\mathbb{C}\psi$, so ${\rho }_i=c_i|\psi ⟩⟨\psi |$. Its unit trace forces $c_i=1$, establishing the required purity.

Conversely, suppose $\rho$ is not rank one. In its spectral decomposition, at least two eigenvalues are positive. Select a normalized eigenvector $|e_1⟩$ with positive eigenvalue ${\lambda }_1$; then $0<{\lambda }_1<1$. DefineThe remaining eigenvalues show that $\sigma$ is a density matrix. The decomposition $\rho ={\lambda }_1|e_1⟩⟨e_1|+(1-{\lambda }_1)\sigma$ is a convex combination of two different density matrices, neither equal to $\rho$. This violates the given purity condition. Hence the extreme points of the density-operator state space obey

Set $X=\rho -\sigma$. The trace distance is $D(\rho ,\sigma )=\frac{1}{2}\Vert X{\Vert }_1$, where $\Vert X{\Vert }_1=\mathrm{Tr}\sqrt{X^{†}X}$. Since $X$ is a Hermitian operator, its spectral decomposition is $X={\sum }_j{x}_j|j⟩⟨j|$. Define its positive part of a Hermitian operator and negative part of a Hermitian operator byThey are positive semidefinite operators, satisfy $X=Q-R$ and $QR=0$, and obey $|X|=Q+R$. ThereforeThe states have equal trace, so $\mathrm{Tr}X=0$ and $\mathrm{Tr}Q=\mathrm{Tr}R=D(\rho ,\sigma )$. This is the spectral-parts formula for trace distance.

Let $q_y=⟨y|\rho |y⟩$. Positivity and normalization of the density operator imply $q_y\ge 0$ and ${\sum }_y{q}_y=1$. The completely dephasing channel produceswhich is already a spectral decomposition. Its Von Neumann entropy is thereforeThe quantum output entropy is exactly the Shannon entropy of the basis-measurement probabilities, including possible zero eigenvalues via $0{\log }_{2}0=0$.

By spectral decomposition, the Von Neumann entropy is the Shannon entropy of the eigenvalues:Throughout, logarithms have base two, so the Von Neumann entropy and Shannon entropy are measured in bits. The continuous convention $0{\log }_{2}0=0$ includes zero eigenvalues.

The constraints $-I\le T\le I$ are in the order of Hermitian operators. They imply $-1\le ⟨{\alpha }_i|T|{\alpha }_i⟩\le 1$. Using the spectral decomposition of $X$,Choose $T=\mathrm{sgn}(X)$, with eigenvalues $+1$ on the positive spectral subspace, $-1$ on the negative spectral subspace, and zero on the kernel. It satisfies the constraints and attains equality. The trace-norm variational principle for Hermitian operators is therefore

To prove the Holevo–Helstrom theorem, write a binary POVM as $\{E,I-E\}$, where $0\le E\le I$, and associate the first outcome with $\rho$. Any larger outcome set followed by a binary decision can be grouped into this form. Put $\Delta =p\rho -(1-p)\sigma$. Its success probability isThe substitution $T=2E-I$ bijects the allowed effects with the interval $-I\le T\le I$. Since $\mathrm{Tr}\Delta =2p-1$,Choosing $E$ as the positive spectral projection of $\Delta$ attains this value, with an arbitrary decision on its kernel. This proves the optimal success probability and constructs an optimal measurement.

The positive part of a Hermitian operator and negative part of a Hermitian operator are the positive operatorsThus $X=X_{+}-X_{-}$ and $X_{+}{X}_{-}=0$. In this convention the negative part itself is nonnegative. This is the positive-negative decomposition of the Hermitian operator.

The operator absolute value is defined by the unique positive operator square rootThe scalar identity $|\lambda |=\max (\lambda ,0)+\max (-\lambda ,0)$ applied in the spectral decomposition proves

Take the real inner productThe stationary distribution is strictly positive for an irreducible Markov chain. Detailed balance gives $⟨Pf,g{⟩}_{\pi }=⟨f,Pg{⟩}_{\pi }$, so $P$ is a self-adjoint operator. Choose a real orthonormal eigenbasis $f_1,\dots ,f_m$, where $m=|S|$, $P{f}_j={\lambda }_j{f}_j$, $f_1\equiv 1$, ${\lambda }_1=1$, and $\pi (f_j)=0$ for $j\ge 2$.

For fixed $y$, apply the spectral decomposition to $g_y(z)=1_{\{y\}}(z)/\pi (y)$. Its coefficient against $f_j$ is $f_j(y)$, whereas $(P^t{g}_y)(x)=P^t(x,y)/\pi (y)$. Thus the spectral kernel expansion of a reversible chain isThis holds for all integers $t\ge 0$; at $t=0$ each multiplier is one, including when ${\lambda }_j=0$. Negative eigenvalues are allowed, and retain their signs for odd $t$.

Memorylessness means that the ensemble's average density operator is ${\rho }^{(n)}={\pi }^{\otimes n}$. Its spectral decomposition isThe product eigenvalues giveZero eigenvalues contribute zero to the Von Neumann entropy and are excluded from logarithmic typicality tests.

The quantum typical subspace is the span of eigenvectors whose eigenvalues satisfyIts orthogonal projection $P_{\epsilon }^{(n)}$ selects precisely the classical typical set for the eigenvalue distribution $q$. The typical-set cardinality bounds and the weak law of large numbers therefore givefor any fixed $\delta >0$ and all sufficiently large $n$.

Since $A|x⟩=(-1{)}^{f(x)}|x⟩$, the spectral decomposition givesReplace the middle gate of the previous compute-phase-uncompute construction bya one-qubit rotation about the z-axis. ThenThe circuit uses the gates of $C$, their inverses, and one fixed rotation; it returns the quantum ancilla to $|0⟩$. The same construction implements $e^{itA}$ with middle gate $e^{itZ}$, an instance of simulation of a computable diagonal Hamiltonian.