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Conic optimization using the completely positive cone. For a real symmetric matrix $Q$, the trace-normalized program $\min \{⟨Q,X{⟩}_F:X\in {\mathrm{CP}}_n,\mathrm{tr}X=1\}$ minimizes a weighted average of nonnegative-unit-vector Rayleigh quotients. The weights are the squared norms of the factors in $X={\sum }_j{x}_j{x}_j^T$. Hence a minimizing rank-one factor attains the same value as the original orthant minimum.

Conic optimization using the copositive cone. The constraint $Q-\lambda I\in {\mathrm{COP}}_n$ supplies an exact reformulation of a Rayleigh quotient minimum on the nonnegative orthant. Exact conic formulation does not itself provide an efficient membership algorithm.

With vanishing surface pressure/mass density and regular admissible displacements, integration by parts makes the axisymmetric adiabatic displacement operator symmetric in the mass density-weighted inner product. Completing the pressure square gives the displayed energy form. Nonnegative form for every admissible displacement excludes exponentially growing axisymmetric adiabatic modes in the Cowling approximation. A negative trial Rayleigh quotient proves negative spectrum by the Rayleigh-Ritz variational principle. Nonnegative stratification and epicyclic coefficients give a simple sufficient stability condition.

Let a real symmetric matrix $M$ have ordered eigenvalues ${\lambda }_1\ge {\lambda }_2\ge \cdots \ge {\lambda }_n$, with $n\ge 2$. If $A$ is obtained by deleting one matching row and column, its largest eigenvalue ${\mu }_1$ satisfiesThe Rayleigh quotient of $A$ is that of $M$ restricted to vectors whose deleted coordinate is zero. This gives the upper bound. Some nonzero linear combination of the top two orthonormal eigenvectors has that coordinate zero; normalizing it gives a quotient at least ${\lambda }_2$. The argument does not require distinct eigenvalues.

Let a real symmetric matrix have simple eigenpairs $({\lambda }_i,w_i)$ and letThen its Rayleigh quotient isAfter one shifted inverse solve and normalization, the relative coefficients areThus the eigenvector direction error is cubed at each sufficiently close iterate.

For the magnetized isothermal atmosphere, set $a=1/(2H)$, $s=v_s^2>0$, $b=v_a^2>0$ and $\Delta =i{k}_x{\xi }_x+i{k}_y{\xi }_y+(i{k}_z+a){\xi }_z$. After weighting the displacement by $e^{z/(2H)}$, the restoring Hermitian matrix $\mathsf{K}$ satisfiesThus $gH\le s$ makes the form nonnegative for every wavevector. If $gH>s$, choose $k_z=0$, $k_x\ne 0$, ${\xi }_z=1$, ${\xi }_x=-a/(i{k}_x)$ and ${\xi }_y=g/(si{k}_y)$. The squares vanish, and sufficiently small nonzero $k_y$ gives a negative form. The Rayleigh quotient proves instability when these wavelengths are admissible. Since $s=\gamma c_s^2$ and $\beta =2c_s^2/b$, the condition is precisely $\beta (\gamma -1)<1$.

For a static smooth ideal magnetohydrodynamics equilibrium with conservative external forces and surface terms eliminated by the boundary conditions, the linear displacement equation is $\rho \ddot{\mathbf{\xi }}=\mathcal{F}\mathbf{\xi }$, where the force operator is a self-adjoint operator in the appropriate integral pairing. Its restoring quadratic form $\delta W=-\frac{1}{2}\int {\mathbf{\xi }}^{\ast }\cdot \mathcal{F}\mathbf{\xi }dV$ is real. Multiplying the normal mode equation by the complex conjugate of the displacement and integrating gives ${\omega }^2\int \rho |\mathbf{\xi }{|}^{2}dV=2\delta W$. Hence squared frequencies are real, and a negative restoring form gives an unstable mode through the variational Rayleigh quotient, subject to the operator domain and spectral hypotheses. Background fluid flow adds terms to this argument; the static hypothesis is essential.

An odd trial state is orthogonal to the even oscillator ground state. Its Rayleigh quotient therefore bounds the first excited energy from above, unlike an arbitrary trial state. The kinetic quadratic form permits continuous compactly supported trials with derivative jumps.

For the matrix exponential $A(t)=e^{Lt}$, maximizing the energy ratio over nonzero initial conditions givesThe Rayleigh quotient achieves its maximum at an eigenvector of $A^{\ast }A$ with largest eigenvalue, equivalently a right singular vector of $A$. Such an optimal disturbance is generally not an eigenvector of $L$.