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Use the normalized eigenfunctions and their weighted inner product ${\int }_a^{b}w\overline{{\psi }_m}{\psi }_{n}dx={\delta }_{mn}$. Orthogonality givesFor complex trials, $F$ means the corresponding Hermitian quadratic form with $|{\psi }^{\prime }{|}^2,|\psi {|}^2$; it agrees with the displayed functional on real trials. If $a_0\ne 0$, division by $|a_0{|}^2$ yieldsSince ${\lambda }_n\ge {\lambda }_0$, this is at least ${\lambda }_0$. The Rayleigh-Ritz variational principle therefore gives an upper bound from any admissible trial, improved by minimizing over trial parameters or enlarging the trial space. Equality holds when the trial is entirely in the lowest eigenspace.

The displayed ratio using $b_n$ presupposes $a_0\ne 0$. If $a_0=0$, use the undivided quotient $\sum {\lambda }_n|a_n{|}^2/\sum |a_n{|}^2$ instead; the same upper-bound argument still applies. No ground-state coefficient can be divided out in that case.

Expand the quadratic form of this Gram matrix:For $\mathbf{c}\ne 0$, linear independence of the real vectors implies ${\sum }_i{c}_i{\mathbf{x}}_i\ne 0$, so its squared Euclidean norm is strictly positive. Thusproving the positive definiteness of a Gram matrix in the independent-vector case.

Take $\gamma >0$ and set $A=\pi /2$. The Rayleigh quotient of the trial for the quantum harmonic oscillator isIts denominator is $\pi /(2\gamma )$. The kinetic numerator is ${\gamma }^2$ times that denominator. For the remaining integral, put $t=\gamma x$ and useobtained from ${\cos }^{2}t=(1+\cos 2t)/2$ and integration by parts. Thus the compact cosine trial bound for the quantum harmonic oscillator isDifferentiation gives $R^{\prime }=2\gamma -({\pi }^2-6)/(6{\gamma }^3)$, so the minimizing parameter satisfiesThe quotient tends to infinity at either end of $\gamma >0$, and the two positive terms are equal at this stationary point, proving it is the global minimum. ThereforeFor comparison, the exact ground state has value $1$. The trial is admissible for the energy quadratic form: it is continuous, has compact support, and its weak derivative is square-integrable. Its first derivative jumps at the support endpoints, so it need not belong to the smaller twice-differentiable operator domain; this does not invalidate the Rayleigh-Ritz variational principle bound.

The quadratic form has real symmetric matrixIts unit eigenvectors can be chosen as $e_u=(1,1,0)/\sqrt{2}$ with eigenvalue $1$, and $e_v=(1,-1,0)/\sqrt{2}$, $e_w=(0,0,1)$ with eigenvalue $9$. SetThe orthogonal diagonalization turns the surface equation intoIt is an ellipsoid, specifically a prolate spheroid centered at the origin. Its long principal axis is the line in direction $\boxed{(1,1,0)/\sqrt{2}}$, with semiaxis length $\boxed{3}$. Each direction in the plane perpendicular to that line has semiaxis length $\boxed{1}$; the displayed $e_v,e_w$ give one orthonormal choice of the two short principal axes. Their nonuniqueness reflects the repeated eigenvalue $9$ and rotational symmetry about the long axis.

The quadratic form is ${\mathbf{x}}^{T}Q\mathbf{x}$ for the real symmetric matrixThe $y$ axis has eigenvalue $-3$. On the $xz$ plane, the characteristic polynomial isThus the orthogonal diagonalization from part (a) yieldsFor later use, one possible orthonormal basis isThen $u=(3x+z)/\sqrt{10}$, $v=y$, $w=(x-3z)/\sqrt{10}$. This explicit choice also confirms that the Euclidean norm is preserved: $x^2+y^2+z^2=u^2+v^2+w^2$.

The disk still has only three degrees of freedom and hence three normal modes: two centre coordinates and one in-plane rotation. Five springs alter the restoring quadratic form but add no disk coordinates. The fivefold rotational symmetry makes the translational stiffness proportional to the two-dimensional identity matrix, and makes translation-rotation coupling vanish. Therefore the two translation modes are always degenerate; the rotational mode is separate, with possible accidental equality of its frequency for particular parameters. The general formulas in small oscillations of a disk attached to prestretched radial springs apply with $N=5$.

Applying the implicit midpoint rule to the simple harmonic oscillator system givesFactor the change in the quadratic form $y^2+z^2$ and substitute these equations:Thus every step preserves the first integralFor completeness, the implicit equations are solvable for every real $h$: putting $s=h/2$, their coefficient matrix has determinant $1+s^2>0$ and the update isThis is an orthogonal matrix, as follows from $(1-s^2{)}^2+4s^2=(1+s^2{)}^2$. It therefore preserves the Euclidean norm, explaining geometrically the exact energy conservation. This is a concrete instance of Runge-Kutta conservation of quadratic invariants. Preservation of the circle does not assert that the numerical and exact solutions have the same phase at a fixed nonzero step size.

Orthogonality means zero inner product, not merely that two objects look different. For complex vectors, $⟨u,v⟩={\sum }_i{u}_i^{\ast }{v}_i$; for functions a common choice is $⟨u,v⟩=\int w(x)u(x{)}^{\ast }v(x)dx$ with positive weight $w$. An orthonormal set has $⟨u_m,u_n⟩={\delta }_{mn}$.

For an orthogonal set, the component of $f$ along $u_n$ has coefficientA complete orthonormal set gives $f={\sum }_n{c}_n{u}_n$ in the appropriate norm and Parseval's identity $\Vert f{\Vert }^2={\sum }_n|c_n{|}^2$. Without completeness there can be a nonzero residual orthogonal to the entire chosen set. Orthogonal nonzero vectors are linearly independent, but independence alone does not imply orthogonality.

Fourier coefficients are projections becauseLikewise, for a self-adjoint operator, $A{u}_n={\lambda }_n{u}_n$ gives $({\lambda }_n-{\lambda }_m)⟨u_m,u_n⟩=0$, so distinct eigenvalues have orthogonal eigenvectors. Degeneracy does not itself force a chosen pair to be orthogonal, but an orthonormal basis can be selected within each eigenspace. The differential version uses the correct Sturm-Liouville weight and boundary conditions.

For mechanical normal modes with unequal masses the relevant inner product is $⟨u,v{⟩}_{\mathsf{M}}=u^{†}\mathsf{M}v$. This makes the modes in D14 orthogonal even when an ordinary unweighted dot product does not. Expanding in such coordinates diagonalizes the kinetic and potential quadratic forms, explaining why modal motions evolve independently and their energies add.

A projective quadric is the zero locus of a nonzero homogeneous quadratic polynomial $q(v)=v^{T}Bv$, with $B$ symmetric over a field of characteristic different from two. A projective point on it is singular when all first partial derivatives vanish. Since $\nabla q(v)=2Bv$, the quadric is nonsingular exactly when $B$ is invertible.

Over the complex numbers, a nonzero quadratic form in three variables can be diagonalized by an invertible change of basis. Its rank classifies a projective conic:

The zero polynomial is excluded from the definition of a conic. Thus these last two cases are precisely the singular conics.

Expand the increment of the quadratic form, using $A^T=A$:The last term obeys $|h^{T}Ah|\le \Vert A\Vert \Vert h{\Vert }^2=o(\Vert h\Vert )$. Thus $q$ is differentiable andThis is the quadratic form gradient. The derivative is a linear functional of the increment $h$, whereas the gradient is its representing vector under the Euclidean inner product.

For a nonzero everywhere attractive weak potential, $\gamma <0$, the pole has $k=i\kappa$ with $\kappa =m|\gamma |/{\hslash }^2>0$. The exterior waves decay as $e^{-\kappa |x|}$, giving one shallow bound state to leading weak-coupling order:Its size ${\kappa }^{-1}$ is much larger than the potential range because $\kappa a=O(\lambda )\ll 1$. For an everywhere repulsive potential there is no negative-energy normalizable bound state: the kinetic and potential quadratic forms are both nonnegative. Its lower-half-plane pole is a virtual state, not a stable bound state. More complicated sign-changing potentials with a vanishing integral require higher-order analysis and are not covered by this simple leading result.

The wavefunction is real and vanishes continuously at both endpoints. HenceThe momentum second moment is most safely evaluated from its quadratic form:ThusThe derivative jumps where the wavefunction joins zero. On the whole real line the strong second-derivative operator therefore produces endpoint delta distributions, so the state is not in the usual operator domain of $p^2$. The finite momentum second moment, equivalently the kinetic-energy quadratic form or the integral of $p^2$ against the momentum probability density, remains well defined and gives the value above.

Each state has exactly $k(n-k)$ neighbours: choose one occupied coordinate and one unoccupied coordinate to swap. The normalized proposal distribution is thereforeThe swap relation is symmetric and the number of neighbours is constant, so $q(v\mid u)=q(u\mid v)$. The discrete uniform distribution used as the prior distribution is also constant on $S_k$, with mass ${\left(\genfrac{}{}{0ex}{}{n}{k}\right)}^{-1}$. Both factors cancel in the Metropolis–Hastings acceptance probability.

Let $Q={\Sigma }^{-1}$ and $d=v-u$. The displayed multivariate normal distribution presupposes that the covariance matrix $\Sigma$ is a positive-definite matrix, so $Q$ is a symmetric matrix. Expanding the difference of quadratic forms givesConsequentlyThe truncation at zero is essential: a log acceptance ratio can be positive, but the logarithm of an acceptance probability cannot.

For the particular proposed swap, let $a=i_1$ be the coordinate changing from zero to one and $b=i_2$ the coordinate changing from one to zero. Then $d=e_a-e_b$, and the untruncated scalar simplifies further toOnly two rows and four entries of $Q$ are needed for this evaluation. This is Metropolis sampling on fixed-cardinality subsets.

A second-order nondivergence-form elliptic operator on a domain $D\subseteq {\mathbb{R}}^d$ has the formOnly the symmetric part of $a$ affects the Hessian matrix term, so one may take $a(x)$ symmetric. The operator is a uniformly elliptic operator if there is a constant $\kappa >0$, independent of $x$, such thatA common bounded-coefficient convention additionally requires an upper bound $\sum a_{jk}(x){\xi }_j{\xi }_k\le \Lambda |\xi {|}^2$ with $\Lambda <\infty$. The lower bound says that the principal quadratic form cannot degenerate in any direction or at any location.

For an explicit stochastic example, standard $d$-dimensional Brownian motion has infinitesimal generatorIts principal quadratic form is $|\xi {|}^2/2$, so it is uniformly elliptic with $\kappa =\Lambda =1/2$. More generally, a sufficiently regular stochastic differential equation $d{X}_t=b(X_t)dt+\Sigma (X_t)d{B}_t$ has diffusion generator $b\cdot \nabla +\frac{1}{2}(\Sigma {\Sigma }^{\mathsf{T}}):D^2$; it is uniformly elliptic when $\Sigma {\Sigma }^{\mathsf{T}}$ has a uniformly positive lower bound. A conservative diffusion process has no zeroth-order term, since its infinitesimal generator annihilates constants.

The eigenvector identity $Ab=\frac{1}{2}b$ gives $\nabla f(-b)=-2Ab+b=0$. Since $H_f=2A$ is a positive-definite matrix, this is a strict local minimum. A global certificate follows by completing the quadratic form:Positive definiteness makes the first term nonnegative, with equality only at $x=-b$. Thus $-b$ is the unique global minimum, with value $-1/2$.

Write $(x^{T}b{)}^2=x^{T}b{b}^{T}x$ and put $C=A-b{b}^T$, which is symmetric. Differentiating this quadratic form givesUnlike $A$, $C$ need not be a positive-definite matrix. Its behaviour must be checked before classifying the stationary point.

If $A$ is a skew-symmetric matrix, the scalar $x^{\top }Ax$ equals its own transpose, soand it is zero over the real numbers.

Conversely, suppose the quadratic form $x^{\top }Ax$ vanishes for every real vector. Substituting $x=e_i$ gives $a_{ii}=0$. Substituting $x=e_i+e_j$ gives $a_{ij}+a_{ji}=0$ for $i\ne j$. Therefore $A^{\top }=-A$. This proves the quadratic-form criterion for a skew-symmetric matrix:

A convex cone containing no nonzero line through the origin, equivalently $K\cap (-K)=\{0\}$. The copositive cone is pointed: a matrix in both it and its negative has zero quadratic form on the nonnegative orthant; testing $e_i$ and $e_i+e_j$ kills all entries.

The displayed bilinear form is the polar form of a quadratic form. In odd characteristic of a field, $Q(x)=B_Q(x,x)/2$ recovers the quadratic form; in characteristic of a field two, different forms can have the same alternating bilinear form. Over ${\mathbb{F}}_2$, two such forms differ by a linear map to the field, giving the affine parametrization of binary quadratic refinements.

For the Dirichlet boundary conditions on $(-a,a)$, take the two-dimensional even trial space spanned by $(a^2-x^2)$ and $x^2(a^2-x^2)$. Using the kinetic-energy quadratic form gives, for $b=\lambda a^2$,The minimum occurs at $b=(-49+4\sqrt{133})/13$ and equals $7-\sqrt{133}/2\simeq 1.2337187027$. It is slightly above the exact infinite square well coefficient ${\pi }^2/8\simeq 1.2337005501$. The zero extension of a trial function has derivative jumps at the walls, so the appropriate admissibility statement uses the Dirichlet kinetic-energy form; it need not belong to the unrestricted full-line second-derivative operator domain.