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Write $x=g^{\eta }$. Applying $D_x$ and then changing the summation variable from $k$ to $\ell =k-\alpha \eta$ givesHence this state is an eigenvector of $D_x$ with eigenvalue $\boxed{{\omega }^{\alpha \beta \eta }}$.

Let a projective measurement have mutually orthogonal projections $P_i$ satisfying ${\sum }_i{P}_i=I$. On a density operator $\rho$, the Born rule and the Lüders rule giveconditional on outcome $i$. On the first part of a bipartite quantum system, replace $P_i$ by $P_i\otimes I_2$.

The subsystems are isolated when there is no interaction term coupling them. Their Hamiltonian operator has the formso their subsequent unitary time evolution factorizes as $U_1\otimes U_2$.

The reduced density matrix of subsystem 1 iswhere the partial trace is characterized byfor every local observable $M_1$. Thus ${\rho }_1$ contains exactly the statistics accessible by measurements on subsystem 1.

Suppose a projective measurement $\{Q_j\}$ is performed on subsystem 2 and its outcome is not communicated. The resulting nonselective state isFor every $M_1$, the cyclic property of the trace and ${\sum }_j{Q}_j^2=I_2$ giveHence ${\rho }_1^{\prime }={\rho }_1$. No local projective measurement on subsystem 1 can reveal whether the remote unreported measurement occurred. This is quantum no-signalling, which prevents a choice made at a spacelike separation from transmitting information faster than light and makes the measurement formalism compatible with relativistic causality.

The proposed nondisturbing device would violate no information without disturbance. In an ordinary quantum instrument, let $M_{i\alpha }$ be the Kraus operators associated with classical output $i$. If every pure state $|\psi ⟩$ remains unchanged even conditional on the displayed outcome, every nonzero $M_{i\alpha }|\psi ⟩$ must be parallel to $|\psi ⟩$. A linear operator for which every vector is an eigenvector is a scalar multiple of the identity, so $M_{i\alpha }=c_{i\alpha }I$. Its output probabilityis independent of the state. It cannot equal $⟨\psi |P_i|\psi ⟩$ for arbitrary projectors. Equivalently, repeated nondisturbing samples would permit quantum state tomography of one specimen and then quantum cloning, contradicting ordinary quantum theory.

Such devices would not necessarily enable superluminal signalling. One consistent operational extension could make every sequence of outputs depend only on the local reduced density matrix and local settings. Since an unreported remote measurement leaves that matrix unchanged, all local device statistics would remain unchanged too. Other extensions could add nonlocal outcome-dependent rules and permit signalling, but that behavior is additional to the device specification rather than forced by it.

Set $c=⟨\psi |\varphi ⟩$. The off-diagonal Hermiticity identity holds automatically, whilemust be real. Thusis necessary and sufficient. For $g\ne 0$, put $r=gc/|g|\in \mathbb{R}$. Since the two normalized spanning vectors are linearly independent, $|r|<1$.

The normalized eigenvectors and eigenvalues areDirect application of $H$ verifies the eigenvalue equations. Their unnormalized inner product isso the distinct eigenvectors are orthogonal, as required for a Hermitian operator. If $g=0$, $H=0$ and every vector is an eigenvector.

This is the unilateral left shift. It is not compact because $S{e}_{n+1}=e_n$. For every $|\lambda |<1$,is an eigenvector with eigenvalue $\lambda$. The norm bound and closedness then give

The power method starts from $q_0\ne 0$ and iteratesIf the eigenvalue of largest modulus is separated by a spectral gap and the initial vector is nonorthogonal to a corresponding eigenvector, the iterates converge in direction to that eigenvector.

Inverse iteration with shift $s$ instead solvesIt applies the power method to $(A-sI{)}^{-1}$ and therefore finds an eigenvector whose eigenvalue is closest to $s$.

Diagonalize the real Laplace-Lagrange secular matrix $\mathbf{A}$. Its eigenvalues areand choose corresponding real eigenvectors ${\mathbf{v}}_k$. If $V=({\mathbf{v}}_1{\mathbf{v}}_2)$, the initial complex eccentricity vector determines complex mode coefficientsThe matrix exponential solution isThuswhere a negative eigenvector component may equivalently be made positive by adding $\pi$ to its phase. The $g_k$ are secular precession frequencies, each eigenvector fixes the planets' eccentricity ratio and relative apsidal orientation, and ${\beta }_k$ fixes the phase selected by the initial conditions.

The multiplicative order $r$ makes the states $|a^k\mathrm{mod}N⟩$, $0\le k<r$, distinct and cyclic under $U_a$. ThereforeThus each $|{\psi }_s⟩$ is an eigenvector withThese are the Fourier eigenvectors of the cyclic modular-multiplication orbit.

The shift $S|x⟩=|x-1\mathrm{mod}Q⟩$ acts on a Fourier state asThus ${\mathrm{QFT}}_Q|a⟩$ is an eigenvector of $S$ with eigenphase $a/Q$. Apply the unitary part of exact quantum phase estimation for $S$ to a zeroed control register and this Fourier state. It writes the eigenphase label coherently:

Over $\mathbb{C}$, the operator representing $r$ has an eigenvector $v$ with eigenvalue $\lambda \ne 0$. The relation $tr{t}^{-1}=r^{-1}$ givesTherefore $\mathrm{span}\{v,tv\}$ is a nonzero $G$-invariant subspace of dimension at most two. If $V$ is irreducible it equals this subspace, so $\dim V\le 2$.

If $P=P^T=P^2$ and $\mathrm{tr}P=1$, the spectral theorem gives eigenvalues $1,0,0$. Thus $P$ is orthogonal projection onto a line andfor a unit vector $X$; the only choices are $X$ and $-X$.

The smooth map $\pi :S^2\to S_3(\mathbb{R})$, $X\mapsto X{X}^T$, has image $G{r}_{1,3}(\mathbb{R})$ and fibers $\{X,-X\}$. Near $X_0$, restrict to the hemisphere $X\cdot X_0>0$; there $\pi$ is one-to-one and has smooth inverse obtained by choosing the positive local unit eigenvector. Combining this with a chart of $S^2$ supplies two-dimensional local parametrizations. Hence $G{r}_{1,3}(\mathbb{R})$ is a two-dimensional submanifold, diffeomorphic to $\mathbb{R}{P}^2$.

The product states $|n_1{n}_2⟩$, $n_a\in \{0,1\}$, are exact eigenvectors, withThus the energies attached to $|00⟩,|01⟩,|10⟩,|11⟩$ are respectively

For $Av=\lambda v$, $v^{\ast }Av=\lambda v^{\ast }v$ is real, so $\lambda$ is real. The spectral theorem writes positive definite $C=U\mathrm{diag}({\lambda }_i)U^{\ast }$; taking positive square roots gives $B$. For $H$, eigenvalues are $1,3$ with normalized eigenvectors $(-i,1)/\sqrt{2}$ and $(i,1)/\sqrt{2}$, so these columns form a suitable $U$. Finally expand $v$ in an orthonormal eigenbasis: the Rayleigh quotient is a weighted average of eigenvalues, bounded above by ${\lambda }_{max}$ and attaining it on a top eigenvector.

Let $A,B,C$ represent $a,b,c$. Since $c$ is central, $C$ commutes with $A$ and $B$. Over the complex number field $\mathbb{C}$, $C$ has an eigenvalue $\gamma$, and its corresponding eigenspace is invariant under all three operators. The irreducibility of $V$ therefore makes this eigenspace all of $V$, so $C=\gamma I$. Taking the trace ofgives $(\dim V)\gamma =0$ by the cyclic property of the trace; hence $C=0$.

The remaining operators $A$ and $B$ commute. Two commuting operators on a nonzero finite-dimensional complex vector space have a common eigenvector, whose span is invariant. Irreducibility therefore forces $\dim V=1$. Conversely, every pair $(\alpha ,\beta )\in {\mathbb{C}}^2$ defines a one-dimensional irreducible representation byThese are all the finite-dimensional irreducible representations.

The real spectral theorem gives real eigenvalues and orthogonal distinct eigenspaces. Constructively, a Householder reflection sends an eigenvector to $e_1$; symmetry makes the remaining block symmetric. Induction diagonalizes that block, producing orthogonal $Q$ with $Q^{T}AQ=\Lambda$. Multiplying by $Q$ gives $AQ{e}_k={\lambda }_{k}Q{e}_k$.

For a one-dimensional velocity perpendicular to a transverse magnetic field, $c_f^2=\gamma p/\rho +B^2/({\mu }_0\rho )$. The primitive-variable characteristic matrix has two advective eigenvalues and the two speeds $u\pm c_f$. Integrating the fast eigenvectors keeps $p/{\rho }^{\gamma }$ and $B/\rho$ constant. The state parameter obeys ${\rho }_t+(u\pm c_f){\rho }_x=0$, so the characteristic speed satisfies the Inviscid Burgers equation.

The strictly positive case states that a real square matrix $W$ with $W_{ij}>0$ has a positive simple eigenvalue $\lambda$ and positive left and right eigenvectors, with all other eigenvalues strictly smaller in modulus. The Brouwer fixed-point theorem applied to $v\mapsto Wv/(1^{T}Wv)$ on the nonnegative unit simplex gives $v>0$ and $Wv=\lambda v$. For another eigenvector $z$, maximize $|z_i|/v_i$; the triangle inequality gives $|\mu |\le \lambda$. Equality forces every modulus ratio and complex argument to agree because every entry is positive, so $z$ is proportional to $v$ and $\mu =\lambda$. Apply the same argument to $W^T$ for a positive left eigenvector; its positive pairing with $v$ rules out a generalized eigenvector at $\lambda$, establishing algebraic simplicity. For merely nonnegative matrices a nonnegative leading eigenvector exists. An irreducible nonnegative matrix has a positive leading eigenvector and simple Perron eigenvalue; a primitive nonnegative matrix has the strict modulus gap.