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Let $A$ be a companion matrix of a monic irreducible polynomial $f\in F[x]$ of degree $n$. Form the $N$-by-$N$ block matrix $B$ with $A$ on its diagonal, $I_n$ immediately above the diagonal, and zero elsewhere. If the field characteristic is $p>0$, write $f(x)=h(x^q)$ with $q=p^r$ maximal, so that $h^{\prime }\ne 0$; in characteristic zero take $q=1$ and $h=f$. ThenThus a separable polynomial gives $m_B=f^N$, but an irreducible polynomial with zero formal derivative can give a smaller exponent. For example, over the rational function field ${\mathbb{F}}_2(t)$, the irreducible polynomial $f(x)=x^2-t$ has $q=2$, so a two-block chain still has minimal polynomial $f$.

To prove the formula, write $B=D+K$, where $D$ has diagonal blocks $A$ and $K$ is the block shift. These are commuting operators, and $K^j$ vanishes exactly when $j\ge N$. The binomial theorem in positive field characteristic gives $B^q=D^q+K^q$. Expanding $h(D^q+K^q)$ gives $f(B)=K^{q}H$, with $H$ commuting with $K$. The diagonal blocks of $H$ are $h^{\prime }(A^q)$, which are invertible by Bézout's identity for polynomials, since $h$ is irreducible with nonzero formal derivative and $h(A^q)=0$. Therefore $f(B{)}^s=K^{qs}{H}^s$ vanishes exactly when $qs\ge N$. Finally, the block upper triangular matrix determinant gives the displayed characteristic polynomial, while the Cayley-Hamilton theorem and irreducibility force the minimal polynomial to be a power of $f$.

Let $p_T=n{k}_{B}T$, $C=⟨\mathbf{R}{\mathbf{R}}^T⟩$, and $M=I+\alpha \sigma$, so the single-bead mobility is $M/\zeta$. Assume an incompressible solvent, constant $n,H,T,\zeta$, negligible bead inertia, and a symmetric positive-definite $M$. The last requirement makes the resistance invertible and Brownian diffusion physically admissible. At a given position and time, $M$ depends on the moment $C$ but not on the individual configuration $\mathbf{R}$.

Use the source's gradient convention $K_{ij}={\partial }_i{u}_j$, and write $G=K^T$ for the component-first gradient. Then $\mathbf{u}({\mathbf{r}}_2)-\mathbf{u}({\mathbf{r}}_1)\simeq G\mathbf{R}$. The spring forces on the two beads are ${\mathbf{F}}_1=H\mathbf{R}$ and ${\mathbf{F}}_2=-H\mathbf{R}$. Their overdamped probability currents areHere the Brownian term written as $-k_{B}T\nabla \psi$ is a force weighted by probability density. Its corresponding conditional osmotic force is $-k_{B}T\nabla \log \psi$; treating the given term as an unweighted deterministic force per bead would not produce the stated Fokker–Planck equation.

With midpoint $\mathbf{x}=({\mathbf{r}}_1+{\mathbf{r}}_2)/2$, the derivatives are ${\nabla }_{r_1}=\frac{1}{2}{\nabla }_x-{\nabla }_R$ and ${\nabla }_{r_2}=\frac{1}{2}{\nabla }_x+{\nabla }_R$. Subtracting the bead currents gives the configuration currentThe midpoint current is $\mathbf{u}\psi$ after neglecting the specified spatial Brownian diffusion. Configuration probability conservation therefore givesSince $G\mathbf{R}$ has components $R_j{\partial }_j{u}_i$, this agrees with the source's $\mathbf{R}\cdot \nabla \mathbf{u}$. The factor two in both spring relaxation and configuration diffusivity comes from relative motion of both beads.

The second moment $C$ is the unnormalized conformation tensor. Now multiply by $R_i{R}_j$ and integrate over configuration space. Assume sufficiently decaying tails so every boundary term at infinite separation vanishes. The affine current gives $GC+C{G}^T$; spring drift gives $-(2H/\zeta )(MC+CM)$; two integrations by parts of diffusion give $(4k_{B}T/\zeta )M$. HenceBecause $\sigma =nHC-p_{T}I$, $M$ is a polynomial in $C$ and commutes with it. ThusThis is the Hookean dumbbell conformation equation with anisotropic mobility, an exact second-moment closure within the stated model; no Gaussian approximation to $\psi$ was needed.

The upper-convected derivative of the identity is $I^{\nabla }=-(G+G^T)$. Applying it to the polymer stress therefore givesConsequently the coefficients and the closed stress equation arewhere ${\sigma }^{\nabla }=D_t\sigma -\sigma K-K^T\sigma$ in the printed convention. Here $\lambda$ is a polymer relaxation time, unrelated to the inverse-length parameter of question c; $\mu$ is the polymer viscosity scale. The coefficient $\alpha$ has inverse-stress units. In the dimensional mobility parameter in the Giesekus model, the dimensionless factor is $b=\alpha p_T$, not the dimensional $\alpha$ printed in this problem.

The upper-convected rate removes the affine stretching and rotation of material connectors. If $\dot{F}=GF$, then a tensor transported purely as $C=F{C}_0{F}^T$ has $C^{\nabla }=0$. Under a time-dependent rigid rotation $Q$, the rate transforms as $Q{C}^{\nabla }{Q}^T$: this is material frame indifference, not an ordinary componentwise time derivative. The resulting equation is the Giesekus model in these dimensional conventions.

The isotropic subtraction is fixed by zero polymer extra stress at thermal equilibrium. With no flow and positive mobility, the stationary distribution isso $\boxed{{\sigma }_{\mathrm{eq}}=0}$. The constitutive equation correspondingly has a zero-stress quiescent state and a Newtonian low-rate limit $\sigma \simeq \mu (K+K^T)$. Without subtracting exactly $p_{T}I$, the spring contribution would retain an equilibrium isotropic stress and the equation would acquire corresponding reference terms. Because mobility itself depends on $\sigma$, that reference cannot simply be shifted while retaining the same model. The separate fluid pressure enforcing incompressibility is not included in this polymer extra stress.

The squared angular momentum operator and one selected component are commuting operators, so a complete set of common eigenfunctions can be chosen: for the $z$ component these are the usual $|j,m⟩$ angular momentum eigenstates. Distinct components do not have a common complete eigenbasis, because their commutator is another nonzero component. A $j=0$ state is an exception in which all components vanish; noncommutation does not forbid every individual common eigenfunction.

The Cayley-Hamilton theorem statesLet $d=\dim V$. Extend the field to an algebraically closed field and choose a basis in which the matrix of $T$ is an upper triangular matrix, as permitted. Write its diagonal entries as ${\lambda }_1,\dots ,{\lambda }_d$ and put $V_i=\mathrm{span}(e_1,\dots ,e_i)$. Each $V_i$ is an invariant subspace, andApply $T-{\lambda }_{d}I$ first, then $T-{\lambda }_{d-1}I$, and continue. The resulting product sends $V_d$ into $V_0=\{0\}$. These factors are commuting operators, soThe determinant of an upper triangular matrix gives ${\chi }_T(x)=(-1{)}^d{\prod }_i(x-{\lambda }_i)$. Thus ${\chi }_T(T)=0$ after the field extension. Its entries were in the original field, whose embedding is injective, so they were already zero there. The zero-dimensional case is immediate as well.

Suppose $C=S{\Lambda }_C{S}^{-1}$ and $D=S{\Lambda }_D{S}^{-1}$ for an invertible $S$ and diagonal matrices ${\Lambda }_C,{\Lambda }_D$. Since diagonal matrices commute,Therefore simultaneous diagonalization implies commutation. This argument works over either $\mathbb{R}$ or $\mathbb{C}$ and explains why simultaneous diagonalization produces commuting operators.

The operators $X_i$ act on different qubits and therefore commute, soSince $X=HZH$,The scalar $e^{-it}$ is a global phase. Thus each factor uses two Hadamard gates and one phase gate, and applying the factors in parallel or sequentially gives an exact quantum circuit of $\boxed{3n}$ elementary gates.

Each summand of $J_X$ acts on one qubit, while each summand $Z_i{Z}_j$ of $J_Z$ acts on two. Therefore $J=J_X+J_Z$ is a 2-local Hamiltonian. We haveSplit $t$ into $r$ steps of length $\delta =t/r$ and use the second-order product formulaFor one step, $\Lambda =O(\delta n^2)$ in the stated estimate, so the spectral-norm error is $O({\delta }^3{n}^6)$. The error bound for a product of unitary operators makes the total errorIt is therefore enough to choosewith $r$ also large enough that the small-step estimate applies.

All $Z_i{Z}_j$ terms commute. A factor $e^{-i\delta Z_i{Z}_j}$ uses a constant-size circuit of two controlled-NOT gates and one phase gate, up to a global phase, so one product-formula step costs $O(n^2)$ gates. The complete Hamiltonian simulation consequently has size

Matrix similarity transports commuting operators: with $M=PD{P}^{-1}$,By ii, all such matrices are thereforeFor completeness, matrix multiplication givesThe parameterization is injective, since matrix multiplication by $P^{-1}$ and $P$ recovers the parameter diagonal matrix. Thus there are infinitely many choices. Already the distinct scalar matrices $t{I}_3$, $t\in \mathbb{R}$, give an infinite family.

Use dimensionless physical helicity $h=\mathbf{S}\cdot \hat{\mathbf{p}}$, with spin eigenvalues $\pm 1/2$ in natural units. In the Dirac representation,The entries $1$ in the supplied transformation denote two-by-two identities. Since its matrix acts only on the upper/lower spinor blocks, it commutes with the repeated Pauli block. Therefore $u^{-1}=u$ andThe eigenvalues remain the physical spin projection along the direction of motion, $+1/2$ or $-1/2$. A convention omitting the factor one half instead labels them $\pm 1$; this changes no measurement probabilities. A change of gamma-matrix representation cannot change the spin observable.

With the two diagonal matrices from the preceding part, write $D_A=PA{P}^{-1}$ and $D_B=PB{P}^{-1}$. Their diagonal products agree entry by entry, so $D_A{D}_B=D_B{D}_A$. ThereforeHence simultaneous diagonalization implies that $A,B$ are commuting operators.

For $n=2$, takeThey are commuting operators because every matrix commutes with the identity matrix. However $A$ has only the eigenvalue zero, and $\ker A=\mathrm{span}\{(1,0{)}^T\}$ has dimension one. It has no eigenbasis, so it cannot appear in any simultaneous diagonalization. Equivalently, a diagonal matrix similar to this nonzero nilpotent matrix would have to be zero, an impossibility. For any prescribed $n\ge 2$, use $A=E_{12}$ and $B=I_n$; its zero eigenspace has dimension $n-1$. For $n=1$ no such example exists.