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For $X$ in the dual cone $C$, copositivity of $Q-\lambda I$ gives $⟨X,Q-\lambda I{⟩}_F\ge 0$. An upper-bounding Lagrangian for the maximization is thereforeIts supremum over unrestricted $\lambda$ is finite precisely when $\mathrm{tr}X=1$. Minimizing that upper bound gives the conic programThis is completely positive optimization, not simply semidefinite programming: $C$ is the completely positive cone.

There is also a direct equality certificate. Every feasible $X={\sum }_j{u}_j{u}_j^T$, $u_j\ge 0$, has ${\sum }_j\Vert u_j{\Vert }_2^2=1$. Its objective is a weighted average of the nonnegative-sphere Rayleigh quotients, hence is at least $\alpha$. For a minimizing unit vector $x_{\ast }$ from part (g), $X_{\ast }=x_{\ast }{x}_{\ast }^T$ is feasible and has objective $\alpha$. Thus $\boxed{X_{\ast }=x_{\ast }{x}_{\ast }^T}$ attains the dual and both values agree, without relying on unverified regularity assumptions.

Eigenvectors belonging to distinct eigenvalues are linearly independent. With $n$ distinct eigenvalues there are $n$ eigenvectors forming a basis, and the matrix is diagonalizable. In this eigenbasis, the Rayleigh quotient is a weighted average of the eigenvalues, so it is maximized by choosing $\mathbf{v}$ in the eigenspace of the largest eigenvalue.

A self-adjoint operator $T$ satisfies $⟨Tx,y⟩=⟨x,Ty⟩$. If $Tx=\lambda x$, then $\lambda ⟨x,x⟩=⟨Tx,x⟩=⟨x,Tx⟩=\overline{\lambda }⟨x,x⟩$, so $\lambda \in \mathbb{R}$. An eigenvector exists over $\mathbb{C}$; its orthogonal complement is $T$-invariant. Induction on dimension produces an orthonormal basis of eigenvectors, proving the finite-dimensional spectral theorem.

A self-adjoint $T$ is positive definite exactly when every eigenvalue is positive, since $⟨Tx,x⟩={\sum }_i{\lambda }_i|x_i{|}^2$ in an eigenbasis. The quadratic forms of two positive-definite operators add, so their sum is positive definite.

The same eigenbasis shows that $\mathrm{spec}T\subset [a,b]$ exactly when $T-\lambda I$ is positive definite for every $\lambda <a$ and negative definite for every $\lambda >b$. Finally, if $v$ is a unit eigenvector of $\alpha +\beta$ with eigenvalue $\gamma$, the Rayleigh quotient bounds giveThus every eigenvalue of $\alpha +\beta$ lies in $[a+c,b+d]$.

For $q(t)=B{q}_0$, the energy ratio is the Rayleigh quotient of the Hermitian positive-definite matrix $B^{†}B$:Expanding $q_0$ in the orthonormal right singular vectors makes this a weighted average of ${\sigma }_j^2$. HenceAny nonzero vector in the largest right-singular subspace is optimal if ${\sigma }_1$ is repeated. Its amplified state is proportional to the paired left singular vector. For a non-normal matrix $L$, this optimal initial direction need not be an eigenvector of $L$, explaining transient growth from non-normal modes.

For a normalized trial state $\psi$, expand it in an orthonormal basis of energy eigenstates, $\psi ={\sum }_n{c}_n{\psi }_n$. The Rayleigh quotient is thenThis is the quantum variational principle: minimizing the expectation of the Hamiltonian operator over any trial family gives an upper bound on the ground-state energy.

Let $\psi$ be the normalized ground-state wavefunction and form the normalized dilation ${\psi }_{\lambda }(x)={\lambda }^{1/2}\psi (\lambda x)$. Its kinetic and potential expectations areThe true ground state makes the energy stationary at $\lambda =1$, soThus the power-law quantum virial theorem gives

If $n\le -3$ and $A>0$, the two sides have opposite signs, because $T>0$ and $V>0$. If $A<0$, the scaled energy ${\lambda }^{2}T+{\lambda }^{-n}V$ tends to $-\infty$ as $\lambda \to \infty$, since $-n>2$, so the Hamiltonian is not bounded below. If $A=0$, the free particle has no normalizable ground state. Hence there is no normalizable ground state for $n\le -3$.

For $n=1$ take the normalized Gaussian functionThe Gaussian integral givesand thereforeIts unique minimum satisfiesAt this point $⟨V⟩=2⟨T⟩$, consistently with the virial theorem, and the resulting variational estimate is

PutThe Rayleigh quotient shift is exactlyExpanding the denominator givesThe question prints this as $s_k={\lambda }_1-K{ϵ}^2+O({ϵ}^4)$, so its sign convention requiresunless every $a_i$ vanishes. Equivalently, $-K$ is the positive quadratic Rayleigh-quotient error.

The shifted inverse solve has the spectral decompositionOnly relative coefficients matter after normalization, so factor out $({\lambda }_1-s_k{)}^{-1}$:From the shift expansion,whereas, for $i\ge 2$,ConsequentlyAfter the final normalization,Thus an $O(ϵ)$ eigenvector-direction error becomes $O({ϵ}^3)$ in one step, proving the Local cubic convergence of Rayleigh quotient iteration for this symmetric matrix.

The kernel ridge regression estimator minimizesBy the representer theorem, its fitted-value vector isWriting $f$ for the true value vector, the variance contribution to $\mathbb{E}\Vert f-\hat{f}{\Vert }_2^2$ isThe squared bias is $\Vert \lambda (K+\lambda I{)}^{-1}f{\Vert }_2^2$. In an orthonormal eigenbasis of $K$, the scalar inequalitywhich is equivalent to $4d\lambda \le (d+\lambda {)}^2$, yieldsFor $\Vert f{\Vert }_2=1$, only the last term varies. The Rayleigh quotient of $K^{-1}$ is maximized by a unit eigenvector associated with its largest eigenvalue $1/d_n$, equivalently an eigenvector of $K$ associated with $d_n$.

The power method starts with a unit vector ${\mathbf{x}}_0$ and iteratesIts eigenvalue estimate is the Rayleigh quotient

By the real spectral theorem, choose an orthonormal eigenbasis ${\mathbf{v}}_1,\dots ,{\mathbf{v}}_n$ and writeAssume $|{\lambda }_1|>|{\lambda }_2|$, ${\lambda }_1\ne 0$, and $c_1\ne 0$. Before normalization,Thus the direction error is $O(|{\lambda }_2/{\lambda }_1{|}^k)$. Orthogonality makes the linear terms vanish from the Rayleigh quotient:ConsequentlyThis is the quadratic Rayleigh-quotient improvement for the power method.

For a decision $x$ in the unit ball, conditional expectation under the posterior givesThe symmetric positive semidefinite matrix $C+m{m}^T$ has a Rayleigh quotient maximized over $\Vert x\Vert \le 1$ by any unit eigenvector corresponding to its largest eigenvalue. Bayes decision rule therefore chooses any such eigenvector for the stated utility.

Write $\Phi \alpha ={\sum }_i{\alpha }_i\varphi (X_i)$. By the definition of the kernel matrix,andThe irrelevant positive factor $1/n$ gives exactly the stated constrained optimization.

Let $K{v}_1={\lambda }_1{v}_1$, where ${\lambda }_1>0$ is the largest eigenvalue and $\Vert v_1{\Vert }_2=1$. The generalized Rayleigh quotient is maximized byup to sign and addition of a vector in $\ker K$, which does not change $\hat{u}=\Phi \hat{\alpha }$. Repeated leading eigenvectors give further principal directions.

For $\varphi (x)=x$ and centered data, the objective is the Rayleigh quotientThis is the first-direction optimization in principal component analysis. Therefore $\hat{u}$ is any unit eigenvector of the sample covariance matrix $S$ corresponding to its largest eigenvalue.

For any nonzero state, the Rayleigh quotient isExpand $|\psi ⟩={\sum }_n{c}_n|{\psi }_n⟩$ in normalized energy eigenstates. ThenBecause the ground state is unique, equality holds exactly when all coefficients except $c_0$ vanish. Thus the Rayleigh-Ritz variational principle gives its minimum at the ray of $|{\psi }_0⟩$.

For the proposed exact state, write $\tilde{\psi }=e^{-\beta x^n}$. Its logarithmic derivatives giveThe stationary Schrödinger equation, after multiplication by $2m/{\hslash }^2$, isMatching the highest power requires $2n-2=6$, so $n=4$, and cancellation of $x^6$ requires $16{\beta }^2=1$. Normalizability selects $\beta =1/4$. Then $12\beta =3$ also cancels the quadratic term, leaving $\epsilon =2$. ThereforeThis is the exact ground state of a solvable sextic potential candidate $e^{-x^4/4}$.

For the Gaussian trial state ${\psi }_{\alpha }=e^{-\alpha x^2/2}$, normalization cancels from the quotient. With respect to the probability density proportional to $e^{-\alpha x^2}$,HenceThe stationary equation isWriting $z={\alpha }^2$, the quadratic has exactly one positive root,Since the quotient tends to infinity as $\alpha \downarrow 0$ or $\alpha \to \infty$, this is the unique global minimizer. ThusUsing the stationary equation to replace $15/(8{\alpha }_{\ast }^3)$ by ${\alpha }_{\ast }/6+1/(2{\alpha }_{\ast })$ gives the best estimateThis is the Gaussian variational estimate for a solvable sextic potential.

The exact eigenfunction $e^{-x^4/4}$ is positive and has no nodes. The nodeless theorem for a one-dimensional ground state therefore identifies it as the true ground state, with dimensionless energy $2$. The variational value is consistent: since ${\alpha }_{\ast }^2>3/2$,so $E_0^{\ast }>{\hslash }^2/m=E_0$, as every trial-state upper bound must satisfy.

For a nonzero real vector $x$, the Rayleigh quotient of the matrix $A$ isThe Rayleigh quotient iteration starts from a unit vector $v_0$. Given $v_k$, compute the shiftsolve the linear systemand normalize in the Euclidean norm:The next eigenvalue estimate is ${\mu }_{k+1}=R_A(v_{k+1})$. The process stops when the eigenpair residual is sufficiently small.

Multiplyby $\hat{w}$ and integrate from the rigid bottom to an upper endpoint $z$ at which $\hat{w}{\hat{w}}_z=0$. The bottom term also vanishes because $\hat{w}(0)=0$. Integration by parts givesTherefore the horizontal wavenumber has the Rayleigh quotientFor a trapped wave the upper endpoint may be taken to infinity because the eigenfunction decays.

LetBecause the Rayleigh quotient is stationary with respect to first-order changes of its eigenfunction, only the explicit dependence of $l^2$ on $c=\omega /k$ contributes when the quotient is differentiated. ThuswhereIt follows thatSince $\omega =kc$, the horizontal group velocity is

For a stationary ridge wave, $c=0$. Assume $\overline{u}>0$, so downstream is the positive $x$ direction. The stated inequality isThe denominator in the group-velocity formula is then positive, andThe atmospheric internal gravity waves generated by the ridge consequently carry their wave packet downstream.

For a real symmetric matrix with $|{\lambda }_1|>|{\lambda }_2|$ and a starting vector having nonzero component in the leading eigendirection, power-method direction errors arewhereas the Rayleigh quotient error isThe improvement occurs because orthogonality cancels terms linear in the direction error.

For a radial stellar oscillation, put ${\xi }_r=r\eta$. The self-gravitating adiabatic equation is $({\Gamma }_{1}P{r}^4{\eta }^{\prime }{)}^{\prime }+r^3[(3{\Gamma }_1-4)P{]}^{\prime }\eta +\rho r^4{\omega }^2\eta =0$. It is a Sturm-Liouville problem whose Rayleigh quotient tests radial stability. The familiar constant-exponent threshold is ${\Gamma }_1=4/3$; a variable exponent requires the full integral test.

Let $\hat{P}$ project onto a leading unit eigenvector of $H+W$, with $H$ as in the Davis-Kahan curvature lemma. Its maximizing Rayleigh quotient gives $\mathrm{tr}(H(P-\hat{P}))\le \mathrm{tr}(W(\hat{P}-P))$. Duality of the operator norm and trace norm, together with the rank-two bound $\Vert \hat{P}-P{\Vert }_{\ast }\le \sqrt{2}\Vert \hat{P}-P{\Vert }_F$, proves the estimate. No gap condition on the perturbed matrix is needed.

The weighted operator $L=W^{-1}[-(P\xi \prime )\prime +Q\xi ]$, with real coefficients and $P,W>0$ in the interior, defines a Sturm-Liouville problem. Its realization as a self-adjoint differential operator requires a domain whose endpoint conditions annihilate the boundary form $[P(\eta {\prime }^{\ast }\xi -{\eta }^{\ast }\xi \prime )]$. Its Rayleigh quotient uses the weight $W$ in the denominator.

For quantum witness embedding $J$, verifier unitary $U$ and accepting projector $\Pi$, the witness acceptance operator is the positive contraction $F=J^{†}{U}^{†}\Pi UJ$. The maximum acceptance over normalized quantum witnesses is ${\lambda }_{\max }(F)$ by the Rayleigh quotient. Restricting to basis quantum witnesses only tests its diagonal entries and need not find that maximum.