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For a reversible Markov chain with invariant law $\mu$, its Markov kernel acts as a self-adjoint operator $P$ on $L^2(\mu )$. An absolute spectral gap means that $\Vert P\Vert$ on the mean-zero functions is at most $r<1$. Geometric ergodicity implies this gap for a reversible chain. Indeed, for bounded mean-zero $g$ supported where the geometric prefactor is bounded, the total-variation estimate gives $⟨g,P^{2n}g⟩\le C_g{r}^{2n}$. The spectral theorem for normal operators puts its spectral measure in $[-r,r]$. Such test functions are dense in the mean-zero subspace.

Choose an auxiliary Riemannian metric $h$ and define the skew-adjoint bundle map $A$ by $\omega (u,v)=h(Au,v)$. Thenis a compatible almost complex structure. The positive square root is defined fiberwise by the spectral theorem for normal operators, and its smooth dependence on $A$ makes $J$ smooth.

For a unitary operator $W$, $W{W}^{†}=I$. Thus for every nonnegative integer $j$,Insert this into the convergent matrix exponential series in the finite-dimensional qubit setting:Consequently unitary conjugation commutes with the exponential:For an unbounded self-adjoint Hamiltonian, the same identity follows from the spectral theorem for normal operators, with the domain transported by $W$; no unbounded power-series manipulation is needed.

For a unitary matrix, $U{U}^{†}=1$, so $(U^{†}MU{)}^r=U^{†}{M}^{r}U$. Cyclicity of the trace then gives $\mathrm{tr}(U^{†}{M}^{r}U)=\mathrm{tr}(M^r)$ for both powers entering the action. Thus the action is invariant under unitary conjugation.

The spectral theorem for normal operators diagonalizes a finite-dimensional Hermitian matrix by a unitary matrix. Applying the invariance to that diagonal form givesOnly the eigenvalues enter, not the choice of eigenvectors.

We prove the finite-dimensional spectral theorem for normal operators by induction on dimension. A complex linear operator has an eigenvector $v$ with eigenvalue $\lambda$, by the fundamental theorem of algebra. For a normal operator $T$, the operator $T-\lambda I$ is also normal, andfor every $w$. Applying this to $v$ shows $T^{\ast }v=\overline{\lambda }v$.

Normalize $v$. Its orthogonal complement $W=v^{\perp }$ is invariant under both $T$ and $T^{\ast }$: the adjoint operator identities pair $Tw$ with $v$ through $T^{\ast }v$, and $T^{\ast }w$ with $v$ through $Tv$. The restricted operator on $W$ remains normal, and its adjoint is the restriction of $T^{\ast }$. Induction supplies an orthonormal eigenbasis of $W$, which together with $v$ gives one for $V$.

Conversely, in an orthonormal basis of eigenvectors, $T$ is diagonal and $T^{\ast }$ is the diagonal matrix of the conjugate eigenvalues. The two diagonal operators commute. Thus normality is equivalent to an orthonormal eigenbasis. The zero-dimensional case is immediate.

The Riesz-Markov-Kakutani representation theorem identifies the dual of the complex space of continuous functions on a compact space $C(K)$ with finite regular complex Borel measures on $K$: each bounded linear functional has a unique representationHere $|\mu |$ is the variation measure; positive functionals correspond exactly to positive regular measures. This is the measure representation theorem, rather than the Hilbert-space representation of a functional by a vector.

For the spectral construction use the convention that the Hilbert inner product is linear in its first argument. A unital subalgebra of $\mathcal{B}(H)$ is understood to contain $I_H$: an algebra whose abstract identity is a smaller projection could not give a resolution normalized by $P(K)=I_H$. On a compact Hausdorff space, a resolution of the identity is understood to be a normalized regular projection-valued measure: its scalar measures are regular, its values are orthogonal projections, and it is countably additive in the strong operator topology. Regularity is part of the usual convention needed for the uniqueness assertion; the statement is interpreted in this sense.

The Commutative Gelfand--Naimark theorem makes the Gelfand transform an isometric unital star-isomorphism $A\to C(K)$, where $K={\Phi }_A$ is compact Hausdorff. Denote its inverse by $\pi :C(K)\to A\subset \mathcal{B}(H)$. For $x,y\in H$, the bounded linear functional $f\mapsto ⟨\pi (f)x,y⟩$ has norm at most $\Vert x\Vert \Vert y\Vert$. Riesz-Markov-Kakutani gives a regular complex measure ${\mu }_{x,y}$ such thatUniqueness makes these measures sesquilinear in $x,y$. For $f\ge 0$, the identity $\pi (f)=\pi (\sqrt{f})\pi (\sqrt{f}{)}^{\ast }$ shows positivity, so ${\mu }_{x,x}$ is positive with total mass $\Vert x{\Vert }^2$.

For every Borel set $B$, the bounded sesquilinear form $(x,y)\mapsto {\mu }_{x,y}(B)$ determines an operator $P(B)$ by Hilbert-space Riesz representation theorem, withIn particular $0\le P(B)\le I$ and $P(B)$ is self-adjoint. We now verify the projection identity rather than assume it.

For continuous $f$, uniqueness of the representing measure applied to continuous test functions givesThese identities imply $P(B)\pi (f)=\pi (f)P(B)$. Testing once more against a continuous $g$ givesThe restriction $1_B{\mu }_{x,y}$ of a finite regular Borel measure is still regular, so uniqueness in Riesz-Markov-Kakutani yields ${\mu }_{P(B)x,y}=1_B{\mu }_{x,y}$. Consequently, for any Borel $C$,Taking $C=B$ proves that $P(B)$ is an orthogonal projection. Also $P(K)=I$ and $P(\varnothing )=0$. For disjoint $B_j$, these projections are orthogonal, and scalar countable additivity gives weak countable additivity. If $B={\bigcup }_j{B}_j$, then $R_N=P(B)-{\sum }_{j\le N}P(B_j)$ is the projection of the remaining union andThus countable additivity holds in the strong operator topology. The scalar measures are the regular measures already constructed. This completes the scalar-measure construction of a projection-valued measure.

By the integral theorem permitted in the question, continuous $f$ satisfies $⟨(\int fdP)x,y⟩=\int fd{\mu }_{x,y}=⟨\pi (f)x,y⟩$, and hence $\int fdP=\pi (f)$. For $T\in A$, this proves the spectral theorem for a commutative operator algebra:Any other regular resolution giving these integrals has the same scalar integrals on all of $C(K)$; Riesz-Markov-Kakutani uniqueness forces the same scalar measures and therefore the same projections on every Borel set.

For nonempty open $U\subset K$, compact Hausdorff normality supplies a nonzero continuous function $f$ supported in $U$. If $P(U)=0$, the stated squared-norm identity for spectral integrals gives $\pi (f)=\int fdP=0$, contradicting the isometry of $\pi$. This proves the full support of a faithful spectral measure propertyFaithfulness of the representation is essential here.

The spectral theorem for normal operators says that a bounded normal operator $T$ on a nonzero complex Hilbert space has a unique regular projection-valued measure $E$ on $\sigma (T)$ such that $T=\int zdE(z)$. Its support is all of $\sigma (T)$, and bounded Borel functions have the associated Borel functional calculus for a normal operator.

For the proof sketch, $A=C^{\ast }(I,T)$ is commutative because $T$ commutes with $T^{\ast }$; polynomials in these two operators commute, as do their norm limits. The map ${\Phi }_A\to \sigma (T)$, $\phi \mapsto \phi (T)$, is onto by the character of an algebra formula and spectral permanence for C-star algebras. It is one-to-one because a character of an algebra preserves the star operation and its values on $T,T^{\ast }$ determine it on their dense polynomial algebra. It is therefore a homeomorphism from compact ${\Phi }_A$ to the Hausdorff spectrum. Transport the resolution just constructed through this homeomorphism. It gives the formula for $T$ and full support; conversely a regular resolution for $T$ gives the same integrals for polynomials in $T,T^{\ast }$, hence by density the same continuous functional calculus and the same resolution. This argument works without separability of $H$.

Finally choose disjoint nonempty relatively open sets $U,V\subset \sigma (T)$ around two distinct spectral points, and put $Q=E(U)$. Full support gives $Q\ne 0$ and $E(V)\ne 0$, while $QE(V)=0$, so $Q\ne I$. The spectral integral, or multiplicativity of its Borel calculus with $1_U$, gives $QT=TQ$ and also $Q{T}^{\ast }=T^{\ast }Q$. Thus $Y=QH$ is closed, nonzero and proper, and $T(Qx)=Q(Tx)\in Y$. The spectral projection gives a reducing subspace conclusion isIn fact it is a reducing subspace, since it is also invariant under $T^{\ast }$.

Put $B=\Vert f{\Vert }_{\infty }$, $v={\mathrm{Var}}_{\mu }(f)$, and $N_n=n-m_n$. We first obtain a uniform-in-$f$ stationary bound, then compare the burned-in trajectory with a stationary one.

For a reversible Markov chain, geometric ergodicity gives an absolute L2 spectral gap of a reversible Markov chain. Thus, for some $r<1$, the Markov kernel viewed as an operator satisfiesOne can use the rate in the total-variation estimate. To see why, reversibility makes $P$ a self-adjoint operator. For bounded mean-zero $h$ supported on a set where $M$ is bounded,The spectral theorem for normal operators writes the left side as $\int t^{2k}d{\nu }_h(t)$. Any positive spectral mass where $|t|>r$ would contradict the inequality as $k\to \infty$. These test functions are dense in $L_0^2(\mu )$: truncate a function and restrict it to $\{M\le j\}$, then correct its mean using a bounded function supported on a fixed positive-measure set where $M$ is bounded. Hence the spectral restriction extends to the entire mean-zero space.

Let $(W_i)$ be the stationary chain. For $h=f-\mu f$, the stationary covariance bound for reversible Markov chains givesConsequently

The initial law of the burned-in trajectory is $P^{m_n}(x,\cdot )$. Its total variation distance from $\mu$ is at most ${\epsilon }_n=M(x)r^{m_n}$. Attaching the same conditional future evolution cannot increase that distance, so the entire length-$N_n$ path laws differ by at most ${\epsilon }_n$. For either path the sum $S$ has $|S|\le B{N}_n$. The dual bound for total variation distance gives a difference of at most $2B^2{N}_n^2{\epsilon }_n$ in the second moments, and at most $4B^2{N}_n^2{\epsilon }_n$ in the squares of the means. This burn-in total variation comparison yieldsSince $m_n=\lfloor n^{1/3}\rfloor$, $N_n{r}^{m_n}\to 0$. Dividing by $N_n$ givesThe constant depends on the chain, not on $f$.

For the common eigenvector $v$, its orthogonal complement $v^{\perp }$ is invariant under both $\alpha$ and ${\alpha }^{\ast }$. Indeed, if $w\perp v$, thenand the analogous calculation applies to ${\alpha }^{\ast }w$. The restriction of $\alpha$ to $v^{\perp }$ is therefore again normal. Induction on $\dim V$, beginning with the normalized vector $v/\Vert v\Vert$, produces an orthonormal eigenbasis. This proves the finite-dimensional spectral theorem for normal operators.

Proceed by induction on $\dim V$. Over $\mathbb{C}$, $\alpha$ has an eigenvector $v$, say $\alpha v=\lambda v$. The normal operator $\alpha -\lambda I$ satisfies the norm equality from part (c), soThus ${\alpha }^{\ast }v=\overline{\lambda }v$. Part (a) now shows that $v^{\perp }$ is invariant under both $\alpha$ and ${\alpha }^{\ast }$. The restriction of $\alpha$ to $v^{\perp }$ is normal. By induction it has an orthonormal eigenbasis, and adjoining the normalized vector $v$ proves the finite-dimensional spectral theorem for normal operators.

Hence

After spatial discretization, a linear time-dependent PDE produces an updateStability over $0\le n\Delta t\le T$ requires a bound $\Vert Q_h^n\Vert \le C_T$ independent of the mesh. If $Q_h$ is normal, the spectral theorem for normal operators givesso eigenvalue analysis is decisive. More generally, if $Q_h=X_h{\Lambda }_h{X}_h^{-1}$, thenThus eigenvalues suffice only when the eigenvector condition numbers are uniformly bounded and unit-circle eigenvalues are semisimple. Defective or increasingly nonnormal matrices can have large powers even though every eigenvalue lies in the unit disk.

For a constant-coefficient scheme on the whole line or a periodic grid, the discrete Fourier transform diagonalizes translation-invariant difference operators. This is Von Neumann stability analysis: insert $u_m^n=G(\theta {)}^n{e}^{im\theta }$ and require $|G(\theta )|\le 1$. Its advantages are simplicity, sharp mesh restrictions, and direct identification of unstable wavelengths. Its limitations are boundaries, variable coefficients, nonlinearities, and nonnormality; frozen-coefficient Fourier analysis then gives at most local evidence.

As a successful implicit example, the Backward Euler diffusion scheme haswhere the periodic or homogeneous-Dirichlet discrete Laplacian $D_h$ is symmetric negative semidefinite. Its eigenvalues are $(1-\Delta t{\lambda }_j(D_h){)}^{-1}\in (0,1]$, proving unconditional discrete-$2$-norm stability.

For a failure, consider explicit upwinding for $u_t+a{u}_x=0$ on a finite inflow grid:Its lower-triangular nonnormal upwind amplification matrix has only the eigenvalue $1-\mu$, so eigenvalues alone incorrectly suggest stability for $0\le \mu \le 2$. For $1<\mu <2$, however, interior alternating data are amplified by $|1-2\mu |>1$ before the boundary is felt, and the matrix powers have no mesh-uniform bound. The correct range is $0\le \mu \le 1$. This example isolates the missing hypothesis: spectral radius does not control powers of a nonnormal matrix family.

Let $\Phi :M_d(\mathbb{C})\to M_D(\mathbb{C})$ be a linear map. It is positive when $X\ge 0$ implies $\Phi (X)\ge 0$, and completely positive whenis positive for every ancillary dimension $r$. In finite dimensions it is enough to check $r=d$.

A finite family of Kraus operators $A^{\alpha }:{\mathbb{C}}^d\to {\mathbb{C}}^D$ defines the Kraus representationThis map is completely positive because, for every positive semidefinite matrix $Y$ on the enlarged space,

Conversely, use the unnormalized maximally entangled vector $|\Omega ⟩={\sum }_{j=1}^d|j⟩|j⟩$. Complete positivity makes the Choi matrixpositive semidefinite. By the spectral theorem for normal operators, $C_{\Phi }={\sum }_{\alpha =1}^R|v_{\alpha }⟩⟨v_{\alpha }|$, where $R=\mathrm{rank}{C}_{\Phi }\le dD$. Reshape each $v_{\alpha }\in {\mathbb{C}}^D\otimes {\mathbb{C}}^d$ into a matrix $A^{\alpha }$ by $|v_{\alpha }⟩={\sum }_{\mu j}{A}_{\mu j}^{\alpha }|\mu ⟩|j⟩$. The Choi matrix inversion formulathen gives $\Phi (X)={\sum }_{\alpha }{A}^{\alpha }X(A^{\alpha }{)}^{†}$. Thus finite Kraus representations characterize finite-dimensional completely positive maps.

The additional normalizationmakes $\Phi$ trace preserving and hence a quantum channel; ${\sum }_{\alpha }{A}^{\alpha }(A^{\alpha }{)}^{†}=I_D$ instead makes it unital.

A positive contraction is a positive operator whose operator norm is at most one. Its spectrum lies in $[0,1]$, so the spectral theorem for normal operators gives $M^2\le M$. Positive contractions are the effects in a POVM and occur in the pure-state gentle measurement bound.

The spectral theorem for normal operators gives a unique decompositionwhere $A_{+},A_{-}\ge 0$ and their supports are orthogonal. They are called the positive and negative parts of $A$. If $\mathrm{Tr}A=0$, then $\mathrm{Tr}{A}_{+}=\mathrm{Tr}{A}_{-}=\Vert A{\Vert }_1/2$.

For a bounded linear operator between Hilbert spaces, the spectral regularization method for quadratic Tikhonov regularization has scalar coefficient $s/(s^2+\alpha )$. Since $(s-\sqrt{\alpha }{)}^2\ge 0$, this coefficient is at most $1/(2\sqrt{\alpha })$. The spectral theorem for normal operators transfers this scalar bound to the operator norm, providing the noise term in the noise-bias decomposition for linear regularization.