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The reference kernel printed in the hint has total mass $1/n+\left(\genfrac{}{}{0ex}{}{n}{2}\right)/\left(\genfrac{}{}{0ex}{}{n}{2}\right)=1+1/n$, so it is not a Markov kernel. We explicitly replace it by the normalized random transposition shuffle: choose two positions independently and uniformly and swap them. Its probabilities areThese sum to one, and its ordinary spectral gap is $\tilde{\gamma }=2/n$, the value intended in the hint. This value can also be certified by the Aldous spectral gap theorem: the single-label generator has rate $2/n^2$ between distinct positions, so on centered functions it acts as multiplication by $-2/n$. Both shuffles have the same uniform distribution on a finite set as stationary distribution.

We use the Canonical paths comparison theorem in the following precise form. For two finite reversible Markov chains with the same positive stationary distribution $\pi$, route each directed transition $(x,y)$ of $\tilde{P}$ along a positive-capacity path ${\eta }_{xy}$ of $P$. Ifthen $\tilde{\mathcal{E}}(f,f)\le A\mathcal{E}(f,f)$ and $\gamma \ge \tilde{\gamma }/A$. Indeed, telescope each difference along its path, apply the Cauchy-Schwarz inequality, and sum with weights $\pi (x)\tilde{P}(x,y)/2$. The load definition gives the Dirichlet form of a Markov chain inequality; the common variance denominator then gives the spectral gap inequality. Identity transitions use empty paths.

Put $s_k=(k,k+1)$. A transposition $(i,j)$ is routed using the wordof length $L_{ij}=2(j-i)-1\le 2n$. Each generator occurs at most twice; write its occurrence count as $N_k(i,j)\le 2$. For a fixed directed adjacent edge $(\sigma ,\sigma s_k)$ and each occurrence of $s_k$ in a word, there is exactly one starting permutation whose translated word crosses this edge at that occurrence. This follows because right multiplication by the prefix is a bijection of the symmetric group. Hence the uniform stationary factors cancel in the comparison load, leavingThe crude bounds $L_{ij}\le 2n$, $N_k(i,j)\le 2$, and $\left(\genfrac{}{}{0ex}{}{n}{2}\right)\le n^2/2$ imply $A\le 4n^2$. ThereforeOnly pairs with $i\le k<j$ contribute, and there are $k(n-k)\le n^2/4$ of these. Keeping this restriction gives the stronger $A\le 2n^2$ and $\gamma \ge 1/n^3$. Neither comparison bound needs the exact adjacent-shuffle spectral gap; the normalization repair of the reference kernel is essential.

False under the hypotheses as printed. Nonzero coefficients do not ensure the two-column dominant-subspace condition.

For a counterexample, take $n=3$, eigenvaluesand an orthonormal eigenbasis $w_1,w_2,w_3$ for whichThe two displayed coefficient vectors are orthonormal and can be completed to an orthogonal matrix, so such a real symmetric matrix $A$ exists. Every $b_i$ and $c_i$ is nonzero, butIndeed, with $t=1/\sqrt{2}$,is a nonzero multiple of $w_1$, while the remaining direction in
$\mathrm{span}(A^k{e}_1,A^k{e}_2)$ isThusBy the simultaneous iteration interpretation of the QR algorithm, the leading two columns of the accumulated $Q$ factor span this same space. The leading block $B_k$ is therefore an orthogonal-coordinate representation of the compression of $A$ to this space. Sincewe havefor every relevant $k$, rather than $\{-2,2\}$. Hence the requested Hausdorff convergence does not hold.

For completeness, the intended statement becomes true if one addsLet $Z_k$ consist of the first two columns of ${\overline{Q}}_{k-1}$. The Accumulated QR factorization identity givesAfter division by $|{\lambda }_n{|}^k$, every component along
$w_1,\dots ,w_{n-2}$ tends to zero because of the strict spectral gap, while $\Delta \ne 0$ makes the two surviving dominant components independent. Therefore these two-dimensional subspaces converge to the dominant invariant subspaceThere are two-by-two orthogonal matrices $U_k$ such thatBecauseup to the harmless index convention at $k=0$, we obtainin the matrix 2-norm. Orthogonal similarity preserves the spectrum, and the stated spectral perturbation bound now gives

For a reversible transition matrix $P$ on $L^2(\pi )$, the spectral gap isEquivalently for a lazy chain,

Markov chain Monte Carlo asymptotic variance for a stationary Markov chain and $\psi \in L^2(\pi )$ iswhenever the limit and series exist. If a reversible Markov chain has positive $L_0^2(\pi )$ spectral gap $\gamma$, then the spectral theorem for normal operators on a separable Hilbert space gives

Part c gives the integral representationThe integrand vanishes on the diagonal $y=x$. Thus the assumed off-diagonal Peskun ordering impliesfor every $f\in L^2(\pi )$.

On the mean-zero subspace, the variational characterization of the spectral gap of a positive reversible kernel isIt follows immediately thatEquivalently, the energy inequality says $K⪯Q$ in the Löwner order on $L_0^2(\pi )$. Positivity permits the operator monotonicity of the square root and hence $K^{1/2}⪯Q^{1/2}$; the spectral representations in the question identify the top spectral values and give the same gap inequality.

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 $s$.

Let $E_2=(e_1,e_2)$ andBecause $Q_k{R}_k$ is the QR factorization of $H^{k+1}$, $S_k$ is the span of the first two columns $q_1^{(k)},q_2^{(k)}$ of $Q_k$.

The first row of the tridiagonal eigenvalue equation isIn the generic case ${\lambda }_{n-1}\ne {\lambda }_n$, the two dominant coefficient vectors are independent, sinceHere $h_{12}\ne 0$, since otherwise the displayed eigenvalue equation and the nonzero $b_j$ would force all eigenvalues to equal $h_{11}$. The strict spectral gap below the dominant pair now impliesThis is two-dimensional subspace iteration. If the dominant eigenvalue is repeated, $H$ acts as a scalar on $W$; the same QR argument either captures all of $W$ or separates its missing direction from the strictly smaller eigenspaces, and the conclusion below is unchanged.

Since $q_2^{(k)}\in S_k$ and $q_3^{(k)}\perp S_k$,The space $W$ is invariant under $H$, soShifting the index does not affect the limit, and thereforeThus the unshifted QR algorithm asymptotically deflates a $2\times 2$ block associated with the equal-modulus dominant pair.

For a finite reversible lazy chain, the relaxation time is $t_{\mathrm{rel}}=1/\gamma$, where $\gamma =1-{\lambda }_2$ is the spectral gap. Its spectral profile isThe variational characterization of the spectral gap isEvery class $c_0^{+}(A)$ is contained in the class over which this last infimum is taken, so $\lambda (A)\ge \gamma$ and $\Lambda (r)\ge \gamma$. Therefore

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.

Iterate $q^{(k+1)}=A{q}^{(k)}/\Vert A{q}^{(k)}\Vert$. With a unique dominant eigenvalue separated by a spectral gap and a starting vector that is nonorthogonal to its eigenvector, the directions converge at the spectral-ratio rate.

Choose two positions independently and uniformly from $\{1,\dots ,n\}$ and swap them. This Markov chain on the symmetric group holds with probability $1/n$ and assigns probability $2/n^2$ to each unordered transposition. Its stationary distribution is uniform, and its ordinary spectral gap is $2/n$ for $n\ge 2$. The latter follows from the Aldous spectral gap theorem: the single-label continuous-time Markov chain has rate $2/n^2$ between every pair of positions, whose mean-zero eigenvalues are all $-2/n$.

The relaxation time of a finite reversible lazy Markov chain is the reciprocal of its spectral gap: $t_{\mathrm{rel}}=(1-{\lambda }_2{)}^{-1}$.

A real nonnegative normalized filter $w$ defines $g_Z=\int w(t)e^{itH}{h}_Z{e}^{-itH}dt$. In an energy eigenbasis, it multiplies each matrix element by $\hat{w}(E_i-E_j)$. A Fourier cutoff at the spectral gap removes couplings between a unique ground state and excitations. The filtered terms still sum to $H$, because the total Hamiltonian is unchanged by its own unitary time evolution. Each term commutes with the ground projector, but need not be minimized there.

For the random adjacent transposition shuffle with $n\ge 2$, the generator $P-I$ is the interchange process on a path graph with each edge rate $1/n$. The Aldous spectral gap theorem reduces its spectral gap to that of the single-label random walk. The eigenfunctions $k\mapsto \cos (\ell \pi (k-1/2)/n)$, for $0\le \ell <n$, have generator eigenvalues $-2(1-\cos (\ell \pi /n))/n$; direct substitution verifies both the interior and endpoint equations. Consequently $\gamma \sim {\pi }^2/n^3$.

For a path with $T+1$ vertices and edge weight $1/2$, the eigenvalues are $1-\cos [\pi k/(T+1)]$, $k=0,\dots ,T$. The corresponding coefficients are proportional to $\cos [\pi k(t+1/2)/(T+1)]$. The zero mode is uniform, and the spectral gap is at least $2/(T+1{)}^2$. The endpoint difference equations use reflecting ghost values, not periodic ones.

For random walk on $\{-1,1{\}}^n$ that chooses a coordinate uniformly and flips it with probability $1/2$, the Walsh characters $f_J(x)={\prod }_{j\in J}{x}_j$ form an orthonormal eigenbasis under the uniform distribution on a finite set. Flipping a coordinate in $J$ negates $f_J$, so $P{f}_J=(1-|J|/n)f_J$. Hence $1-k/n$ has multiplicity $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$, and the ordinary and absolute spectral gaps both equal $1/n$.

For a probability density function $w$ on a domain, a weighted Poincare inequality has the formThe optimal $\lambda$ is the spectral gap of the associated weighted diffusion operator.