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Use the shallow-water approximation, a hydrostatic vertically mixed layer, a deep motionless ambient, and no ambient fluid entrainment. Particle loss changes reduced gravity but changes total layer volume only at the discarded dilute-particle order. Likewise thermal expansion changes the equation of state while the leading Boussinesq approximation retains volume conservation. DefineDepth integration of mass conservation, horizontal momentum balance, particle transport, and the heating law givesHere the integrated excess hydrostatic pressure is $b{h}^2/2$. There is no particle source from the bed. Bed drag and the particle contribution to inertia are neglected at this order. The equivalent variable-buoyancy shallow water equations areThe term $h{b}_x/2$ is necessary when heating or particle loss makes the reduced gravity vary horizontally.

For $v=(h,u,\varphi ,\theta {)}^T$, the quasilinear system $v_t+A(v)v_x=S(v)$ hasIts characteristic polynomial factors asThus the characteristic curves have slopesOn the two repeated contact characteristic curves, the particle and heating laws are $D\varphi =-W_s\varphi /h$ and $D\theta =\beta \varphi$. Writing $c_g=\sqrt{bh}$ and $D_{\pm }={\partial }_t+(u\pm c_g){\partial }_x$, the gravity-wave compatibility relations areFor $h>0$ and $b>0$, gravity-wave right eigenvectors can be taken as $(1,\pm c_g/h,0,0{)}^T$. The contact eigenspace has dimension two: $\delta u=0$ and $b\delta h+(h/2)\delta b=0$, with independent $\delta \varphi ,\delta \theta$. Therefore $A$ has a full set of real eigenvectors.

The system is hyperbolic throughout strict static stability, but not strictly hyperbolic because the contact speed is repeated. At $b=0$ the eigenvalues coalesce and this diagonalizability is lost; when $b<0$ the gravity-wave speeds become complex, reflecting the loss of a statically stable lower layer.

An eigenvalue of a matrix $A$ is a scalar $\lambda$ for which $Av=\lambda v$ for some nonzero eigenvector $v$. Its corresponding eigenspace is

Write $v=(a,b,c,d{)}^T$. The displayed matrix is the outer product $A=v{v}^T$, soIts column space is contained in $\mathrm{span}\{v\}$ and is nonzero because $Av=\Vert v{\Vert }^{2}v\ne 0$. Hence $A$ is a rank-one matrix.

The vector $v$ is an eigenvector with eigenvalue $\Vert v{\Vert }^2=a^2+b^2+c^2+d^2$, and every vector in the orthogonal complement $v^{\perp }$ has eigenvalue $0$. ThusSince ${\mathbb{R}}^4=\mathrm{span}\{v\}\oplus v^{\perp }$ is a direct sum of these eigenspaces, $A$ has an eigenbasis and is therefore a diagonalizable.

Put $a={\lambda }_P$ and $b={\lambda }_Q$. The Pauli operators are Hermitian and satisfy $P^2=Q^2=I$ and $PQ+QP=0$, soThus $V$ is a unitary operator. For every Pauli operator $R$, according as $R$ commutes or anticommutes with $P$ and $Q$, expansion of $VRV$ gives one of $\pm R$ or $\pm RPQ$, up to the phase that makes it Hermitian. It is therefore another Pauli operator, so $V$ normalizes the Pauli group and is a Clifford operation. Finally,which is the normalized projection onto the eigenvalue-$b$ eigenspace of $Q$. Hence $V$ maps the eigenvalue-${\lambda }_P$ eigenspace of $P$ onto the eigenvalue-${\lambda }_Q$ eigenspace of $Q$.

Put $P_1={\sigma }_1\otimes {\sigma }_2$ and $|\psi ⟩=|A{⟩}^{\otimes 2}$. After the first Hadamard gate and the two controlled Pauli gates, the joint state isThe final Hadamard gate changes this toConditioned on ancilla outcome $s\in \{0,1\}$, the normalized data state is therefore

Write the input in the two eigenspaces of $P_1$ aswhere the displayed eigenstates are normalized. A direct PBC measurement giveswith probabilities $|\alpha {|}^2$ and $|\beta {|}^2$. Hence the ancilla circuit and the Pauli measurement have identical outcome distributions and conditional data states after identifying the Pauli outcome with $(-1{)}^s$.

With ${\chi }_A(t)=\det (tI-A)$, direct expansion givesThus the eigenvalues are $3,1,-1,-2$. Kernel calculation gives eigenspace dimensions $1,2,1,1$, respectively.

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.

For $\dot{A}=D^{2}f(t)B$ and $\dot{B}=iA$, with $f$ alternating between $+1$ and $-1$ for intervals of length $T$, the monodromy matrix of a periodic linear system has determinant one and trace $\cosh (\sqrt{2}DT)+\cos (\sqrt{2}DT)>2$. Its reciprocal Floquet multipliers give positive dominant Floquet growth rate for $D,T>0$, despite the zero mean of $f$. An exceptional initial state in the stable eigenspace decays.

For closed manifolds with the product Riemannian metric, the positive Laplace-Beltrami operator is the sum of the two factor operators. Products of their orthonormal eigenbases form a complete basis, and eigenvalues add. Hence the product eigenspace at $\lambda$ is the displayed finite orthogonal sum of tensor products. The discrete assertion requires a discrete spectral setting.

Suppose $D$ has $n$ distinct eigenvalues over a field and $C$ commutes with $D$. Each eigenspace of $D$ is one-dimensional. Since commuting maps preserve eigenspaces, $C$ acts as a scalar on each of them. The eigenvectors of $D$ give a common eigenbasis and hence simultaneous diagonalization. For real matrices, the result gives a real eigenbasis only if the eigenvalues of $D$ are real; otherwise it applies over $\mathbb{C}$.

A spectral decomposition expresses a linear operator through its eigenvalues and projections onto its eigenspaces.

For $0\le A<I$ and normalized $|b⟩={\sum }_j{\beta }_j|u_j⟩$, filtering by $h(\lambda )=\lambda$ succeeds withThe inequality follows by replacing each nonnegative ${\lambda }_j^2$ by its minimum and using ${\sum }_j|{\beta }_j{|}^2=1$. Equality holds exactly when all occupied eigenvectors belong to the minimum-eigenvalue eigenspace. If ${\lambda }_{\min }=0$, success is positive exactly for inputs outside the kernel of $A$.

A weight-space decomposition expresses a Lie algebra representation as a direct sum of simultaneous eigenspaces for a Cartan subalgebra. Each summand is the weight space for its weight. In complex finite-dimensional representations of a semisimple Lie algebra, the simple coroots act as commuting diagonalizable operators, giving this decomposition.