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A quadratic phase on ${\mathbb{F}}_p^n$ is a function $x\mapsto e_p(q(x))$, where $q$ is a quadratic form and $e_p(t)=e^{2\pi it/p}$. Its third multiplicative derivative is one, so its $U^3$ norm is one.

Suppose the Bayesian deviance is locally $D(\theta )=D(\hat{\theta })+(\theta -\hat{\theta }{)}^{T}J(\theta -\hat{\theta })$ with positive-definite $J$, and the posterior is normal with mean $m$ and covariance $\Sigma$. Writing $b=m-\hat{\theta }$, expectation and variance of the quadratic form giveConsequently the usual effective parameter count in DIC is $p_D=\mathrm{tr}(J\Sigma )$, whereas half the deviance variance is $\mathrm{tr}\{(J\Sigma {)}^2\}+2b^{T}J\Sigma Jb$. A locally flat prior gives $m=\hat{\theta }$ and $\Sigma =J^{-1}$, so both counts equal the parameter dimension. Informative priors generally destroy their equality. The expectation formula requires only the first two moments; normality is used for the variance formula.

Let $J$ be the plane rotation matrix through $\pi /2$, generating a cyclic group of order four. A quadratic form $a{x}^2+b{y}^2$ has its two coefficients exchanged by a quarter-turn and unchanged by a half-turn. Thus the pullback action on this two-dimensional space gives the displayed group representation: $D(J^k)=D(J{)}^k$, with $D(J{)}^2=I$. It is not a faithful representation, because its kernel of a group homomorphism is $\{I,J^2\}$. Its two invariant coefficient lines correspond to $x^2+y^2$ and $x^2-y^2$, on which a quarter-turn acts by $+1$ and $-1$, respectively. Restricting to this diagonal space is valid for these signed coordinate permutations; a general linear transformation can introduce an $xy$ term.

Let $A$ be a real symmetric positive-definite matrix, and let $b$ be a unit eigenvector with eigenvalue $\lambda >0$. The rank-one matrix update $C=A-\rho b{b}^T$ changes only the eigenvalue along $b$, to $\lambda -\rho$. Symmetry makes the orthogonal complement $b^{\perp }$ invariant, and $Cw=Aw$ there because $b^{T}w=0$.

In detail, write $x=tb+w$ with $w\perp b$. ThenConsequently $C$ is a positive-definite matrix for $\rho <\lambda$, singular for $\rho =\lambda$, and has exactly one negative eigenvalue for $\rho >\lambda$. In dimension at least two the last case is indefinite. In dimension one it is negative definite instead: there is no nonzero orthogonal-complement direction. The distinction matters when classifying a stationary point of a quadratic form.

The Rayleigh quotient measures the quadratic form of a self-adjoint operator per unit squared norm. For a nonzero vector $x$ and a finite-dimensional Hermitian matrix $A$,For a real symmetric matrix, replace the conjugate transpose $x^{\ast }$ by $x^T$. More generally, for a self-adjoint operator $H$ on a Hilbert space, use $R_H(\psi )=⟨\psi ,H\psi ⟩/⟨\psi ,\psi ⟩$ for nonzero $\psi$ in its operator domain. On an eigenvector it returns the exact eigenvalue. Multiplying the vector by any nonzero scalar leaves the quotient unchanged, so it measures a direction rather than a chosen length.

For the symmetric matrix $A=\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$ from the finite-dimensional spectral theorem example,The first value is not an eigenvalue: it is an average of the two directional responses. To see every direction, write the unit vector $q(\theta )=\cos \theta e_{+}+\sin \theta e_{-}$ in the orthonormal eigenbasis. ThenThe largest value is $3$ along $e_{+}$; the smallest is $1$ along $e_{-}$. A direction halfway between them has quotient $2$. In the left panel, the color on the unit circle represents the quotient; in the right panel, the same values are plotted against the angle measured from $e_{+}$.

The Hermitian Rayleigh quotient range explains why this is useful: evaluating one trial vector gives an upper bound on the smallest eigenvalue and a lower bound on the largest one, without solving a characteristic polynomial. At an extreme eigenvector, a small angular error changes the quotient only quadratically. Near $e_{+}$ in this example, $3-R_A(q(\theta ))=2{\sin }^2\theta =2{\theta }^2+O({\theta }^4)$. This is the geometric reason for quadratic Rayleigh-quotient improvement for the power method and motivates the Rayleigh-Ritz method.

There is a concrete historical link with normal modes. In The Theory of Sound (1877), Volume I, Sections 88–89, Lord Rayleigh studies the period of a constrained trial vibration. He shows that the period has a stationary point at a normal mode and that a trial type bounds the longest natural period. These energy ratios are a mechanical setting for the modern Rayleigh quotient and generalized Rayleigh quotient; the book describes the result through vibration periods, rather than today's matrix notation.^[ref]

For $-u^{\prime \prime }+u=f$ with homogeneous Dirichlet boundary conditions, the conforming piecewise-linear hat functions on a uniform grid give a diffusion stiffness matrix $S$ and a mass matrix $M$. Their nonzero entries are $S_{ii}=2/h$, $S_{i,i+1}=-1/h$, $M_{ii}=2h/3$, and $M_{i,i+1}=h/6$, with symmetric counterparts. The sum is a positive-definite matrix because its coefficient quadratic form is $\int ((u_h^{\prime }{)}^2+u_h^2)$. For forcing $f(x)=x$, the load at $x_i$ is $h{x}_i$.

The real case of the finite-dimensional spectral theorem says that a self-adjoint operator on a finite-dimensional real inner product space has a real orthonormal eigenbasis. In orthonormal coordinates this is the spectral theorem for real symmetric matrices, $A=Q\Lambda Q^T$. The worked ellipse under finite-dimensional spectral theorem shows why both orthogonal diagonalization and orthogonality matter: the new coordinates separate the quadratic form into squares without distorting lengths or angles.

A real normal matrix need not have real eigenvalues. For example, the quarter-turn $\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$ is an orthogonal matrix, but its characteristic polynomial is $z^2+1$ and its eigenvalues are $i,-i$. Over the complex numbers it has an orthonormal eigenbasis; over the real numbers it has no eigenvector. Self-adjointness supplies the reality conclusion in the real spectral theorem.

For continuous $P\ge 0,Q$, a scalar second variation $q=\int (P{\phi }^{\prime 2}+Q{\phi }^2)$ is strictly positive on nonzero fixed-endpoint Lipschitz variations if and only if every regularized form $q_{\eta }$ for $\eta >0$ has no conjugate point in $(a,b]$, and the original weak Jacobi boundary problem has no nonzero Lipschitz solution. The Jacobi positivity criterion applies to each regularized coefficient $P+\eta >0$. Letting $\eta$ decrease to zero gives $q\ge 0$. For a nonnegative quadratic form, $q(\phi )=0$ forces its mixed bilinear form with every test function to vanish, by considering $q(\phi +t\psi )$ for both signs of $t$. This is the weak Jacobi equation, so the kernel condition gives strict positivity. Necessity follows by the same argument in reverse. It distinguishes strict positivity from a uniform coercive estimate when the derivative coefficient degenerates.

Expanding the squared velocity $v_G+\omega \times r$ and integrating over mass removes the mixed term because the centre-of-mass first moment is zero. The remaining rotational integral is half the scalar product of angular velocity and angular momentum about the centre of mass. Principal axes diagonalize that rotational quadratic form.

For distinct ordered observations $x_i$, let $s_u$ be their natural cubic spline interpolant with values $u_i$. Its roughness is a quadratic form, $\int (s_u^{\prime \prime }{)}^2=u^{T}Ku$, where $K$ is positive semidefinite and its nullspace consists of vectors sampled from affine functions. Minimizing penalized least squares $\Vert y-u{\Vert }^2+\lambda u^{T}Ku$ gives $\hat{u}=(I+\lambda K{)}^{-1}y$. Thus the smoothing matrix is $A_{\lambda }=(I+\lambda K{)}^{-1}$ and its effective degrees of freedom are $\mathrm{tr}(A_{\lambda })$. As $\lambda \downarrow 0$ the fit approaches interpolation and the trace approaches $n$; as $\lambda \to \infty$ it approaches the least-squares affine line and the trace approaches $2$.

For stage states $Y_i=y_n+h{\sum }_j{a}_{ij}{k}_j$ and a vector field with $Y_i^T{k}_i=0$, expansion of the new squared norm gives $\Vert y_{n+1}{\Vert }^2-\Vert y_n{\Vert }^2=-h^2{\sum }_{i,j}{M}_{ij}{k}_i^T{k}_j$, where $M_{ij}=b_i{a}_{ij}+b_j{a}_{ji}-b_i{b}_j$. Hence $M=0$ preserves the quadratic form exactly for every well-defined step. The same argument with a fixed symmetric matrix in the inner product preserves any quadratic invariant of the vector field. In particular it preserves the Euclidean norm for a skew-symmetric matrix differential equation.

For a block matrix $M=\left(\begin{array}{cc}A& B\\ C& D\end{array}\right)$ with $A$ an invertible matrix, the Schur complement of $A$ in $M$ is $S=D-C{A}^{-1}B$. It is the effective coefficient matrix for the remaining variables after block elimination. The definition requires invertibility of $A$, but not of $S$ or $M$. Its size is that of $D$, even when $B$ and $C$ are rectangular matrices.

In the original example under block matrix inverse, $D=5$ becomes $S=19/6$ because changing the remaining variable also changes the two variables solved through $A$. This is what makes a Schur complement more informative than simply deleting rows and columns. It retains their mediated effect, and its matrix inverse is the corresponding block of $M^{-1}$ when the block matrix inverse exists.

The same elimination appears when minimizing a quadratic form. The Schur complement as a minimized quadratic form gives a geometric example; the Schur complement determinant identity gives an exact test for an invertible matrix.

Let $M=\left(\begin{array}{cc}A& B\\ B^T& D\end{array}\right)$ be a real symmetric matrix with $A$ a positive-definite matrix. Its quadratic form satisfies the completing the square identityFor each fixed vector $y$, the unique minimizing vector is $x=-A^{-1}By$, so the minimized value is $y^{T}Sy$. Also, $M$ is a positive-definite matrix exactly when $S$ is a positive-definite matrix, under the stated assumption on $A$.^[ref]

An original scalar example isHere $A=2$, $B=1$, $D=3$, and the Schur complement is $5/2$. Holding $x=0$ would give $3y^2$, but letting $x$ relax to $-y/2$ reduces the quadratic form to $(5/2)y^2$. At $y=2$ the respective values are $12$ and $10$. Eliminating a variable therefore changes the effective cost of the retained variable.

This interpretation supports quadratic optimization and the positive-definite symmetric matrix tests used in Cholesky decomposition. In a multivariate normal distribution, a covariance matrix also gives a conditional covariance through a Schur complement; the required probabilistic hypotheses are supplied by conditional multivariate normal distribution, rather than by the algebra alone.

Half the length of one principal diameter of an ellipsoid. For a centred quadratic form $x^{T}Sx=C$ with $S$ symmetric positive definite and $C>0$, its semi-axis lengths are $\sqrt{C/{\lambda }_i}$, where ${\lambda }_i$ are the eigenvalues of $S$. The shortest centre-to-surface distance corresponds to the largest eigenvalue.

The spatial Galerkin method for the heat equation gives a mass matrix $M$ and a diffusion stiffness matrix $S$. With zero forcing, $\frac{d}{dt}(c^{T}Mc)=-2c^{T}Sc\le 0$. The quadratic form $c^{T}Mc$ is the squared L2 norm of the represented finite-element function, proving an energy contraction. For nonzero forcing, the corresponding identity has the additional term $2c^{T}F$, which can be estimated using the Cauchy-Schwarz inequality.

Simple kriging assumes a known mean function. If $m=(m(s_1),\dots ,m(s_n){)}^T$ and $m_0=m(s_0)$ are known, center the observations, predict the residual, and add back the target mean:For known zero mean, these reduce to $c^T{\Sigma }^{-1}z$ and $v_0-c^T{\Sigma }^{-1}c$. Completing the covariance quadratic form giveswhich proves minimum error variance among predictors of the form $m_0+w^T(Z-m)$. No sum-to-one constraint is needed: the known mean already supplies the offset. In the exponential example from ordinary kriging, imposing known zero mean gives weights $(0,2/5,2/5{)}^T$, prediction $2$, and error variance $3/5$. The smaller error variance uses additional knowledge of the mean; it is not a free improvement when the mean is unknown.

If $(Z_0,Z)$ has a joint multivariate normal distribution, the displayed prediction is exactly $\mathbb{E}[Z_0\mid Z=z]$, and $s_{\mathrm{SK}}^2$ is its conditional variance. A conditional expectation minimizes mean squared error among all square-integrable functions of the observations. Without a joint multivariate normal distribution, simple kriging remains the best predictor in the specified affine class, but the conditional expectation may be nonlinear and its conditional variance may depend on $z$. Thus Gaussian process regression and the second-moment kriging construction agree under these assumptions, while having different starting points.^[ref]

Every real square matrix decomposes as $A=S+K$, where $S=(A+A^T)/2$ is its symmetric part and $K=(A-A^T)/2$ is a skew-symmetric matrix. For a real column vector $x$, $x^{T}Kx$ equals its transpose $x^T{K}^{T}x=-x^{T}Kx$, hence vanishes. Thus a real quadratic form $x^{T}Ax$ depends only on $S$.

For a divergence-free drift $\mathbf{b}$ and a function $v$ with zero boundary Sobolev trace, integration by parts gives ${\int }_{\Omega }v\mathbf{b}\cdot \nabla v=\frac{1}{2}{\int }_{\partial \Omega }{v}^2\mathbf{b}\cdot \mathbf{n}-\frac{1}{2}{\int }_{\Omega }(\nabla \cdot \mathbf{b})v^2=0$. Consequently the advection matrix in a conforming Galerkin method has zero quadratic form; its symmetric part vanishes. Adding a diffusion stiffness matrix gives a nonpositive spatial generator and a nonincreasing L2 norm in the semidiscrete advection-diffusion equation. The same identity holds with a suitable zero normal-flux boundary term instead of zero trace.

Take a star graph $G$ with center $c$ and four leaves $1,2,3,4$. Let $L_G$ be its Laplacian matrix and $I$ the identity matrix. In the order $(c,1,2,3,4)$, the original example isFor a nonzero vector $u$, its quadratic form isThus $A$ is a symmetric positive-definite matrix and either ordering below admits a Cholesky decomposition. Changing the ordering changes the work, rather than the mathematical solvability.

Center first. Eliminating $c$ with pivot element $5$ leaves the Schur complementEvery pair of leaves now has coupling $-1/5$: six new edges, hence six undirected fill-in positions. The four remaining vertices form a clique. In the Cholesky decomposition, the first column has $C_{cc}=\sqrt{5}$ and $C_{ic}=-1/\sqrt{5}$. The rest of the lower triangular matrix is dense. To check that later cancellation does not remove its entries, after $r$ leaves have also been eliminated, the active leaf block isIndeed a leaf step sends $\alpha$ to $\alpha +{\alpha }^2/(2-\alpha )$. All active off-diagonal entries remain negative. The successive pivot elements in ordinary symmetric Gaussian elimination, equivalently the squared diagonals in the Cholesky decomposition, areThe Cholesky decomposition therefore stores $5+4+3+2+1=15$ nonzero entries, including its diagonal.

Leaves first. Permute to $(1,2,3,4,c)$. Each leaf has only one remaining neighbor, so its elimination changes only the center's diagonal, reducing it by $(-1{)}^2/2=1/2$. The successive pivot elements are $2,2,2,2,3$, and the exact Cholesky decomposition isThere is no fill-in, and only $5+4=9$ nonzero entries in $C$. This is a concrete instance of leaf elimination of a tree-pattern positive-definite matrix; its general proof explains why the same idea works on every tree pattern. The minimum degree algorithm also chooses a leaf at every nontrivial stage of this particular example, but that success gives no general optimality guarantee.

The input has 13 nonzero matrix elements when both symmetric halves are counted, or nine in its lower triangle. The six added undirected edges correspond to six added positions in one triangular factor, or 12 positions if both symmetric halves of the filled pattern are displayed. Comparing all 13 input entries directly with the 15 entries of $C$ would mix counting conventions.

A real symmetric matrix whose quadratic form is positive at every nonzero vector in the nonnegative orthant. By compactness of the nonnegative unit sphere, this is equivalent to a positive uniform lower bound $x^{T}Ax\ge \delta \Vert x{\Vert }_2^2$ on that orthant. It need not be a positive-definite matrix.