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Taking the adjoint of $U=M+iN$ gives $U^{†}=M-iN$. In particular, the Hermitian parts of a unitary matrix are uniquely determined:Expanding both unitary identities givesSubtracting shows $MN=NM$, and adding gives $M^2+N^2=I$. If $Mv=\mu v$, commutation implies $M(Nv)=\mu Nv$. The assumed simple spectrum makes the eigenspace one-dimensional, so $Nv=\nu v$. Thus $M,N$ have a common orthonormal eigenbasis.

On such a vector, $Uv=(\mu +i\nu )v$. If the corresponding unitary eigenvalue is $u=e^{i\theta }$, thenThe formulas for $M,N$ also show the same eigenvalue correspondence directly from $Uv=uv$ and $U^{†}v=u^{\ast }v$. Degeneracies do not prevent existence of a common basis, although the simple-spectrum assumption makes the elementary proof immediate.

The absolutely convergent matrix exponential series givesfor Hermitian $H$. The exponents $iH$ and $-iH$ commute, so their exponentials multiply to the exponential of their sum:Hence $U$ is unitary. Alternatively, an orthonormal eigenbasis of $H$ turns the exponential eigenvalues into $e^{i{\lambda }_j}$, all on the unit circle.

The spectral theorem for real symmetric matrices supplies real eigenvalues and an orthonormal eigenbasis. Compute eigenvectors, normalize them, and within each repeated-eigenvalue eigenspace choose an orthonormal basis. Distinct eigenspaces are already orthogonal because symmetry gives $(\lambda -\mu )v^{T}w=0$ for eigenvectors with eigenvalues $\lambda ,\mu$.

Use the resulting basis vectors as columns of $O$. ThenThe order and signs of the columns can be chosen freely; rotations within a repeated-eigenvalue eigenspace are also allowed.

Let $A{v}_j=a_j{v}_j$. Commutation gives $A(B{v}_j)=B(A{v}_j)=a_{j}B{v}_j$, so $B$ preserves each eigenspace of $A$. Since the eigenvalues of $A$ are nondegenerate, each eigenspace is one-dimensional and $B{v}_j=b_j{v}_j$. The spectral theorem for a Hermitian matrix lets us choose these vectors as an orthonormal eigenbasis. With $U=(v_1,\dots ,v_n)$,This is simple spectrum gives simultaneous diagonalization. The common eigenvectors can be chosen up to their phases and ordering. In fact degeneracy does not prevent a common eigenbasis for commuting Hermitian matrices, but the one-dimensional argument already suffices here.

The convergent matrix exponential series respects similarity, so $e^A=U{e}^{{\Lambda }_A}{U}^{†}$ and likewise for $B$. Scalar exponentials multiply on each diagonal entry, givingThe essential hypothesis is commutation; Hermiticity supplies this convenient diagonalization proof.

The spectral theorem for a real symmetric matrix gives real eigenvalues and an orthonormal eigenbasis. Compute the roots of $\det (A-\lambda I)=0$, find a basis for each eigenspace, and orthonormalize within any repeated-eigenvalue eigenspace. Eigenvectors belonging to distinct eigenvalues are already orthogonal: if $Av=\lambda v$ and $Aw=\mu w$, symmetry gives $(\lambda -\mu )v^{T}w=0$.

Putting the normalized basis vectors in the columns of an orthogonal matrix $Q$ yieldsFor repeated eigenvalues, the choice of axes inside their eigenspace is not unique.

Assume a complete orthonormal eigenbasis and a self-adjoint operator bounded below, with the trial state in its operator domain or associated energy-form domain. The expansion and orthogonality giveConsequentlyIt is a weighted average of eigenvalues, all at least $E_0$. This is the Rayleigh-Ritz variational principle; equality holds precisely when the trial state lies in the ground-state subspace. A continuous part of the spectrum would require the corresponding spectral integral in place of part of the sum.

An $n\times n$ matrix is diagonalizable over a chosen field exactly when it has a basis of $n$ linearly independent eigenvectors. Equivalently, $P^{-1}AP$ is diagonal for some invertible $P$. Distinct eigenvalues suffice but are not necessary. For example,is not diagonalizable: its only eigenvalue is $1$ and its eigenspace is the one-dimensional span of $(1,0{)}^T$. It is a nontrivial Jordan block.

A Hermitian matrix satisfies $A^{†}=A$, where the dagger denotes conjugate transpose. If $Av=\lambda v$ with $v\ne 0$, thenThe denominator is positive real, soFor $Av=\lambda v$ and $Aw=\mu w$, Hermiticity also givesThusThis is the necessary qualification to the PDF's orthogonality wording: arbitrary eigenvectors belonging to a repeated eigenvalue need not be orthogonal. Within each repeated eigenspace, Gram-Schmidt orthogonalization produces an orthonormal basis; together the eigenspaces give an orthonormal eigenbasis by the finite-dimensional spectral theorem. For example, $(1,0{)}^T$ and $(1,1{)}^T$ are nonorthogonal eigenvectors of the identity matrix.

The $xy$ block has characteristic polynomial $(2-\lambda {)}^2-a^2$. A convenient orthonormal eigenbasis for the full real symmetric matrix iswith eigenvalues $2-a$, $2+a$, and $c$, respectively. Hence its orthogonal diagonalization isFor repeated eigenvalues, every nonzero combination within the corresponding eigenspace is also an eigenvector; the displayed basis remains valid. Introduce principal axes coordinatesThe quadratic form becomesThe principal-axis reduction of a quadric therefore has three regimes.

For $0<a<2$, all coefficients are positive: the surface is an ellipsoid centred at the origin, withFor $a=2$, the $X$ term disappears. The surface is an elliptic cylinder, unbounded along $e_{+}$, whose cross-section is $4Y^2+c{Z}^2=1$:For $a>2$, there is one negative coefficient, so the surface is a one-sheet hyperboloid:Its waist at $X=0$ is an ellipse with semiaxes $(a+2{)}^{-1/2}$ and $c^{-1/2}$. At other $X$, both waist radii are multiplied by $\sqrt{1+(a-2)X^2}$. There is no intercept on its longitudinal axis. Its asymptotic cone is $(a+2)Y^2+c{Z}^2=(a-2)X^2$. If positive eigenvalues coincide, the corresponding choice of principal axes is not unique, reflecting a rotational symmetry of the surface.

Choose the common orthonormal eigenbasis established in part (ii), and let $U$ be the unitary matrix of its columns. ThenConsequentlyThus $N$ is unitarily diagonalizable, even though it need not be Hermitian. As a consistency check, commutation also gives $N{N}^{†}=N^{†}N=A^2+B^2$, so $N$ is a normal matrix, as also follows from unitary diagonalization of a normal matrix.

Suppose first that $AB=BA$, and take an eigenvector $v_j$ of $A$, with $A{v}_j=a_j{v}_j$. Commutation givesSince the eigenvalue $a_j$ is simple, its eigenspace is one-dimensional, so $B{v}_j=b_j{v}_j$ for some scalar $b_j$. Thus every vector in an eigenbasis of $A$ is also an eigenvector of $B$. The simple spectrum of $B$ ensures these account for its eigenvector directions as well. Eigenvectors are understood up to nonzero scaling and permutation, rather than as literally identical numerical vectors.

Conversely, if $v_1,\dots ,v_n$ are a common eigenbasis with $A{v}_j=a_j{v}_j$ and $B{v}_j=b_j{v}_j$, thenfor every basis vector. Hence $AB-BA=0$. ThereforeThis is simple spectrum gives simultaneous diagonalization. Without simple spectra, commuting Hermitian matrices still possess a common orthonormal eigenbasis, but not every arbitrary eigenvector of one matrix must be an eigenvector of the other.

By the spectral theorem for real symmetric matrices, choose an orthonormal eigenbasis ${\mathbf{u}}_1,{\mathbf{u}}_2,{\mathbf{u}}_3$ corresponding to $6,\sqrt{3},-\sqrt{3}$, with ${\mathbf{u}}_1=(1,1,1{)}^T/\sqrt{3}$. Write $P=[{\mathbf{u}}_1{\mathbf{u}}_2{\mathbf{u}}_3]$ and $\mathbf{x}=P\mathbf{\xi }$. This orthogonal change of coordinates givesThe plane constraint is therefore ${\xi }_1=0$. On that plane the quadratic form vanishes exactly whenConsequently the entire set isThe two direction vectors are nonzero and not parallel, since ${\mathbf{u}}_2,{\mathbf{u}}_3$ are linearly independent. An invertible orthogonal transformation maps each full line to a full line. Thus the set consists of precisely two infinite lines through the origin.

We use the spectral theorem to construct a sufficiently accurate approximate eigenpair. Complete $x$ to an orthonormal eigenbasis $x,x_2,\dots ,x_n$ of the symmetric matrix $B$, with corresponding eigenvalues $\lambda ,{\lambda }_2,\dots ,{\lambda }_n$. Simplicity of $\lambda$ means that every ${\lambda }_j-\lambda$ below is nonzero. PutThe orthogonal decomposition of $Ex$ shows that $x^{T}v=0$ andWith $u_{\epsilon }=x+\epsilon v$ and ${\mu }_{\epsilon }=\lambda +\epsilon a$, direct expansion cancels every first-order term:Also $\Vert u_{\epsilon }{\Vert }_2^2=1+{\epsilon }^2\Vert v{\Vert }_2^2\ge 1$. Thus its eigenpair residual has Euclidean norm $O({\epsilon }^2)$ in Big O notation.

For completeness, a real symmetric matrix $M$ always has an eigenvalue within $\Vert (M-\mu I)u{\Vert }_2/\Vert u{\Vert }_2$ of any real $\mu$ and nonzero $u$. Indeed, writing $u={\sum }_j{c}_j{q}_j$ in an orthonormal eigenbasis with eigenvalues ${\theta }_j$ givesApply this symmetric eigenvalue residual bound to $M=B+\epsilon E$. There is an eigenvalue ${\lambda }_{\epsilon }$ such thatwhich provesThis establishes the first-order perturbation of a simple symmetric eigenvalue without presupposing differentiability of the perturbed eigenvector. If $n=1$, the sum defining $v$ is empty and the displayed eigenvalue formula is exact.

An observable is represented by a self-adjoint operator. For an ideal projective measurement, let $P_a$ be its spectral projector for outcome $a$. The Born rule gives $p(a)=⟨\psi |P_a|\psi ⟩$, or $p(a)=\mathrm{tr}(\rho P_a)$ for a density matrix $\rho$. Conditional on obtaining $a$, the Lüders rule givesFor a degenerate outcome this projects onto the whole associated eigenspace, rather than arbitrarily selecting a vector within it. An immediate repeated ideal measurement returns the same outcome, provided there is no intervening evolution or disturbance.

Commuting finite-dimensional observables admit a common orthonormal eigenbasis; their projectors commute and ideal joint measurements are compatible. Noncommuting observables generally have no complete common eigenbasis, and measurement order can affect the subsequent statistics. Particular common eigenvectors can nevertheless exist, so a nonzero commutator does not mean that every state has a nonzero uncertainty in both quantities.

For a quantitative constraint, apply the Cauchy-Schwarz inequality to $(A-⟨A⟩)|\psi ⟩$ and $(B-⟨B⟩)|\psi ⟩$. The imaginary part of their inner product is $⟨[A,B]⟩/(2i)$, giving the Robertson uncertainty principleThe canonical commutation relation $[x,p]=i\hslash$ gives $\Delta x\Delta p\ge \hslash /2$. Retaining the real part of the same inner product gives the stronger Robertson-Schrödinger uncertainty relation, including a covariance term. These are constraints on the preparation of a quantum state, not simply statements about a poorly calibrated instrument.

A sequential spin one-half example makes the change in statistics explicit. Prepare $|+z⟩$. An immediate $S_z$ measurement is certainly positive. If an $S_x$ measurement is inserted first, its outcomes each have probability $1/2$ and leave states $|\pm x⟩$. Each then yields either $S_z$ result with probability $1/2$, so the final positive fraction is $1/2$. The change reflects projection onto incompatible spin eigenstates, even with ideal measuring apparatus. The theory of measurement in quantum mechanics allows more general state-update rules, but the operator description and the Born rule still constrain their probabilities.

Solving the time-independent Schrödinger equation with $\psi \propto e^{ik\varphi }$ gives $E={\hslash }^2{k}^2/(2m{R}^2)$. The periodic boundary condition requires $e^{2\pi ik}=1$, hence $k=n\in \mathbb{Z}$. With the angular integration measure $d\varphi$, wavefunction normalization givesso a convenient orthonormal eigenbasis isThe $n=0$ ground state is nondegenerate. For each $|n|>0$, $u_n$ and $u_{-n}$ span a two-dimensional energy eigenspace; real sine and cosine combinations are an equally valid basis. These opposite-circulation eigenstates have angular momentum $n\hslash$ and $-n\hslash$.

If instead the wavefunction is normalized with respect to physical arc length $ds=Rd\varphi$, its expression is $e^{in\varphi }/\sqrt{2\pi R}$. The two conventions describe the same probability, since $|u_n{|}^{2}d\varphi =|u_n/\sqrt{R}{|}^{2}ds$. The remaining parts use $d\varphi$.

In the ordered product basis $(|\uparrow \uparrow ⟩,|\uparrow \downarrow ⟩,|\downarrow \uparrow ⟩,|\downarrow \downarrow ⟩)$, the two-spin XY Hamiltonian isThe transverse term exchanges opposite spins and leaves parallel spins unchanged. Diagonalizing its two-dimensional block gives the complete orthonormal eigenbasisThe antisymmetric state is the singlet state; the other three are triplet states. Accidental equality of energies does not invalidate this eigenbasis.

For a finite irreducible Markov chain with at least two states satisfying detailed balance, its Dirichlet form of a Markov chain obeysExpand $f-\pi (f)$ in an orthonormal eigenbasis. The denominator is ${\sum }_{j\ge 2}{a}_j^2$, and the numerator is ${\sum }_{j\ge 2}(1-{\lambda }_j)a_j^2$. Their ratio is at least $1-{\lambda }_2$ and equality holds for the second eigenfunction. Thus the optimal Poincare inequality for a reversible Markov chain constant is $1/\gamma$, for the ordinary rather than absolute spectral gap.

Vary the coupling in the opening example. The eigenvectors $e_{+}=(1,1{)}^T/\sqrt{2}$ and $e_{-}=(1,-1{)}^T/\sqrt{2}$ form a fixed orthonormal eigenbasis, with eigenvalues $3+t$ and $3-t$. Writing a vector as $u{e}_{+}+v{e}_{-}$ givesFor $|t|<3$, both directions cost energy and $A_t$ is a positive-definite matrix. At $t=3$, the quadratic form is $6u^2$: the entire line spanned by $e_{-}$ has zero energy, and $A_t$ is a singular positive semidefinite matrix. At $t=-3$, the flat line is instead spanned by $e_{+}$. For $|t|>3$, the quadratic form is an indefinite quadratic form; one direction lowers the energy without bound as its displacement grows. At $t=4$, for instance, $q_t(e_{+})=7$ while $q_t(e_{-})=-1$.

The geometric transition is precise. For $|t|<3$, $q_t=1$ is an ellipse. At $t=3$, it consists of the two parallel lines $u=\pm 1/\sqrt{6}$, extending freely in the zero-energy direction. For $t>3$, it is a hyperbola. The boundary case retains nonnegative energy but loses both the bounded positive level sets and the unique zero-energy vector.

Let $B$ be a finite-dimensional Hermitian matrix and $a<b$ real. The spectral theorem gives $v={\sum }_j{c}_j{x}_j$ in an orthonormal eigenbasis, soIf the spectrum avoids the closed interval, every factor product is positive. Consequently the quadratic form is positive for $v\ne 0$ and zero for $v=0$. More precisely it is bounded below by ${\min }_j[({\lambda }_j-a)({\lambda }_j-b)]\Vert v{\Vert }^2$. This is a finite-dimensional instance of testing a spectral gap with a polynomial in the operator.

Suppose $A\in {\mathbb{C}}^{n\times n}$ is a diagonalizable matrix, $n\ge 2$, with an eigenbasis $v_1,\dots ,v_n$ and spectral decompositionAssume the dominant eigenvalue ${\lambda }_1$ is a simple eigenvalue andThen the normalized power iteration approaches the one-dimensional eigenspace of ${\lambda }_1$. Its asymptotic direction-error bound has factor $\gamma =|{\lambda }_2/{\lambda }_1|<1$. This spectral gap is a gap in modulus: if $g=|{\lambda }_1|-|{\lambda }_2|$, then $\gamma =1-g/|{\lambda }_1|$. A small relative gap gives slow convergence.^[ref]

The short proof is the exact eigenbasis identityThe remaining sum has norm bounded by a constant times ${\gamma }^k$. This proves convergence of direction, allowing a changing sign or complex factor. A tiny $c_1$ or a poorly conditioned, nonorthogonal eigenbasis can delay visible convergence: the condition number of $V$ affects constants, while $\gamma$ controls the eventual rate. The bound need not be an equality; if the initial coefficient of the second-largest eigenvalue vanishes, the active spectrum of the power method gives a faster rate. When $\gamma =0$, all unwanted components disappear after one multiplication.^[ref]

The hypothesis is $c_1=(V^{-1}{x}_0{)}_1\ne 0$, not an ordinary nonzero inner product with the right eigenvector. Let $w_1^{\ast }$ be the first row of $V^{-1}$, a left eigenvector normalized so $w_1^{\ast }{v}_j$ is $1$ for $j=1$ and $0$ otherwise; $\ast$ denotes conjugate transpose. Then $c_1=w_1^{\ast }{x}_0$. For a Hermitian matrix, choose an orthonormal eigenbasis and the criterion becomes $v_1^{\ast }{x}_0\ne 0$: the starting vector must be nonorthogonal to the dominant eigenvector. For a general diagonalizable matrix, the nonorthogonal eigenbasis makes these two tests different, as illustrated under power iteration failure cases.

For each unit vector iterate compute the Rayleigh quotient ${\rho }_k=x_k^{\ast }A{x}_k$ and the eigenpair residual $r_k=A{x}_k-{\rho }_k{x}_k$. Use the same multiplication $A{x}_k$ for both diagnostics and the next normalized iterate. For any complex matrix, this Rayleigh quotient minimizes $\Vert A{x}_k-\mu x_k{\Vert }_2$ over complex scalars $\mu$, because $x_k^{\ast }{r}_k=0$. Under power iteration convergence for a diagonalizable matrix, ${\rho }_k\to {\lambda }_1$ and $\Vert r_k{\Vert }_2\to 0$, even when the vectors alternate signs or phases.^[ref]

A dimensionless stopping quantity iswhen the denominator is positive; $\Vert A{\Vert }_2$ is the operator norm, and a known upper bound may replace it if computing it is costly. A small eigenpair residual certifies a small backward error of an approximate eigenpair, but does not establish that the eigenvalue found is dominant.

For a Hermitian matrix with orthonormal eigenbasis $u_j$, write $x_k={\sum }_j{a}_j{u}_j$. Then the independently useful identityshows that some eigenvalue lies within $\Vert r_k{\Vert }_2$ of ${\rho }_k$. If a specific target eigenvalue ${\lambda }_1$ is separated from all the others by ${\delta }_k={\min }_{j>1}|{\lambda }_j-{\rho }_k|>0$, it also gives ${\epsilon }_k\le \Vert r_k{\Vert }_2/{\delta }_k$. This separation is measured from the computed Rayleigh quotient; it differs from the dominant-modulus spectral gap. For a nonnormal matrix, the same orthogonal expansion is unavailable and a small eigenpair residual alone need not imply a comparably small distance to the spectrum.