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Put $t=\log (1+\sqrt{2})=\mathrm{arsinh}(1)$, so $c_K=2t/\pi$. The power series coefficients of the hyperbolic sine preprocessing areThus $f_j=|g_j|$ for every $j\ge 0$, and the series converge on the full interval. Applying coefficient-dominated entrywise positivity gives $Y⪰0$.

Every diagonal entry belongs to one of the diagonal blocks and equals $f(X_{ii})=f(1)=\sinh t$. Because $e^t=1+\sqrt{2}$ and $e^{-t}=\sqrt{2}-1$, this is $(e^t-e^{-t})/2=1$. ThereforeThe matrix lies in the elliptope and admits a Gram matrix representation by unit vectors. This preprocessing is the Krivine rounding scheme; the equality of absolute coefficients is what preserves positive semidefiniteness even though the cross-block sine coefficients alternate in sign.

The Gram matrix representation and $Y_{ii}=1$ give $\Vert v_i{\Vert }_2=1$. For a standard Gaussian random vector $Z$, each $⟨v_i,Z⟩$ is a standard normal scalar, so the zero event has probability zero. Choose either sign convention at zero; the resulting vector is almost surely a feasible sign vector. Consequently $y^{T}Ay\le v^{\ast }$ almost surely.

The Gaussian sign-correlation identity givesThe last equality uses symmetry of the real matrices. HenceThe inverse sine is applied entrywise, not through spectral matrix functional calculus.

For completeness, the correlation coefficient identity has a geometric proof. If the angle between two unit vectors is $\theta$, the isotropic Gaussian random vector direction in their two-dimensional span gives opposite signs on angular sectors with probability $\theta /\pi$. Thus the sign product has expectation $1-2\theta /\pi =(2/\pi )\arcsin (\cos \theta )$. Parallel and antiparallel pairs give the endpoint values $1$ and $-1$ directly. This is Gaussian hyperplane rounding, and needs no independence between the rounded coordinates.

For $z$ on the unit circle, use the conjugated monomial vectorThen the prescribed diagonal sums givePositive semidefiniteness provesThis is the Gram matrix representation of a trigonometric polynomial. The Hermitian condition also implies $p_{-k}=\overline{p_k}$, so its values on the unit circle are real. The conjugated monomial vector is required by the source's $i-j=k$ convention; the unconjugated vector would represent $p(z^{-1})$ instead.

In particular $p_0=\mathrm{tr}M\ge 0$. If this matrix trace is zero, all nonnegative eigenvalues vanish and $M=0$, so $p=0$. A general feasible Gram matrix need not have matrix rank one; the Fejér–Riesz theorem ensures a rank-one representative exists whenever the nonnegative trigonometric polynomial is nonzero.

The two predictors have correlation coefficient $\rho =0.9813846$, so there is severe multicollinearity. Their centered slope Gram matrix isThus either partial slope has variance ${\sigma }^2/[n(1-{\rho }^2)]$. Relative to a univariate regression with the same error variance, the two-predictor variance inflation factor and standard-error factor areThe small eigenvalue $n(1-\rho )$ makes the contrast between the two slope effects particularly poorly estimated. The univariate regressions test marginal association with the response; the bivariate tests ask whether one predictor adds an effect while holding its nearly identical companion fixed. Their much larger standard errors therefore make the partial tests insignificant without contradicting the marginal significance. The two fits also estimate slightly different residual variances, explaining why the reported standard-error ratio is not exactly 5.21.

Either essay option suffices; both are developed here to make the two mathematical constructions available.

Diffusion for images. Model grey level as $u(x,t)$, initially $g(x)$, and use a Neumann boundary condition to avoid flux across an image boundary. The linear heat equation $u_t=\Delta u$ smooths the data. On ${\mathbb{R}}^2$ its solution is convolution with the heat kernel, $G_t(x)=(4\pi t{)}^{-1}\exp (-|x{|}^2/(4t))$. Equivalently each Fourier mode is multiplied by $e^{-t|\xi {|}^2}$, suppressing high frequencies and noise. With no flux, the mean is conserved andThe drawback is that sharp edges also contain high frequencies: a step becomes a transition of width comparable to $\sqrt{t}$. Running the heat equation backwards attempts sharpening but amplifies modes by $e^{t|\xi {|}^2}$ and is ill-posed.

The Perona-Malik equation instead uses a decreasing diffusivity,Small gradients are smoothed strongly and large ones less strongly. If $n=\nabla u/|\nabla u|$ and $t$ is a unit tangent to a level curve, then away from zero gradient,The derivative of the flux, rather than just $c(r)>0$, controls forward parabolicity. Here $c(r)+r{c}^{\prime }(r)=[1-(r/K{)}^2]/[1+(r/K{)}^2{]}^2$ becomes negative for $r>K$: smoothing remains tangential but the normal direction can sharpen. This formal edge enhancement comes with forward-backward ill-posedness, so existence and stability of the unregularized continuum equation must not be presumed. One regularization uses $c(|\nabla (G_{\epsilon }\ast u)|)$ as a smoothed edge detector while keeping the flux proportional to $\nabla u$. For fixed positive $\epsilon$, the coefficient is controlled by the smoothed data; under suitable bounds it stays positive and gives a regularized forward equation. It preserves edges through reduced cross-edge transport, without the same local backward-diffusion calculation.

Diffusion from a gradient energy connects these equations with variational regularization. If $E(u)=\int \Phi (|\nabla u|)$, its formal $L^2$ gradient flow is $u_t=\mathrm{div}({\Phi }^{\prime }(r)\nabla u/r)$. Linear diffusion corresponds to $\Phi (r)=r^2/2$. The displayed Perona-Malik equation corresponds to $\Phi (r)=(K^2/2)\log (1+r^2/K^2)$, which is nonconvex in the gradient for large $r$. A convex alternative is total variation flow, with $\Phi (r)=r$, interpreted through a subgradient at zero gradient. Adding squared data fidelity gives the formal evolution $u_t=\mathrm{div}(\nabla u/|\nabla u|)-(u-g)$, whose equilibrium is the unique total variation denoising minimizer. Convex TV preserves sharp interfaces more effectively than the heat energy, but it may produce piecewise constant plateaux, known as staircasing in total variation denoising. A stopping time or fidelity weight controls the smoothing scale. Linear diffusion is stable but blurs edges; nonlinear diffusion must balance edge selectivity with parabolicity and regularization.

Wavelets on an interval. Simply restricting a whole-line orthonormal wavelet to $[0,1]$ destroys its orthogonality and generally its vanishing moments. Periodizing the basis restores orthogonality on the circle, but treats the two endpoints as neighbors. This is suitable for periodic data and can create an artificial endpoint jump for nonperiodic data.

A localized interval-adapted wavelet basis instead uses unchanged interior functions and finitely many special boundary wavelets at each endpoint. Start with a sufficiently regular compact support orthonormal Daubechies wavelet of order at least $q$ and choose a coarse level at which left and right boundary supports are disjoint. At each boundary, take appropriate finite combinations of the scaling functions that meet the endpoint, restricted to the interval. Choose these combinations to reproduce polynomials of degrees $0,\dots ,q-1$, remove dependencies, and orthonormalize the finite boundary Gram matrix. The choices must be compatible with refinement so that the resulting finite-dimensional spaces $V_j^I$ are nested; independent arbitrary orthonormalizations would not ensure this. Interior functions retain their whole-line filters, while the boundary functions use finite boundary refinement matrices.

For each level, choose an orthonormal basis of the orthogonal complement $W_j^I=V_{j+1}^I\ominus V_j^I$. It consists of interior wavelets and a bounded number of left and right boundary wavelets. Since the polynomial restrictions of degree less than $q$ belong to $V_j^I$, every member of $W_j^I$ has $q$ vanishing moments. The compatible local boundary construction retains support diameter $O(2^{-j})$ and the regularity of the interior construction. Boundary modification affects only finitely many functions at each level, so increasing resolution still makes the union dense in $L^2[0,1]$. Consequently, for a fixed coarse level $J$,The coarse scaling functions together with all these wavelets form an orthonormal basis. This construction is the Cohen-Daubechies-Vial interval wavelet construction. It preserves localization, polynomial cancellation and stable coefficient extraction without imposing periodic or zero boundary data. The finite boundary refinement matrices also permit a fast transform; a mere restriction followed by one unrelated Gram-Schmidt process at each scale does not establish all these properties.

For a proper three-graph colouring $c$, associate the colors with the three roots of unity $d(i)=\exp (2\pi i(c(i)-1)/3)$, viewed as unit vectors in ${\mathbb{R}}^2$. Define $U_{ij}=2⟨d(i),d(j)⟩$ using the real inner product.

This is twice a Gram matrix, so it is a positive semidefinite matrix. Its diagonal is $2$. On every edge, the endpoint colors differ, so their vectors make angle $2\pi /3$ or $4\pi /3$ and $U_{ij}=2\cos (2\pi /3)=-1$. Thus $t=3$ and this $U$ are feasible in the second semidefinite program, provingThis is the regular simplex construction of a strict vector coloring.

For an edge $ij$, put $\rho =⟨v_i,v_j⟩$ and $\theta =\arccos \rho$. Since $\rho \le -1/2$, we have $\theta \ge 2\pi /3$.

The random normal $a$ has the isotropic Gaussian distribution $N(0,I_p)$. Its distribution is invariant under orthogonal transformations; if $v_i,v_j$ are linearly independent, its projection onto their plane has a uniformly distributed direction. The signs of its inner products with $v_i,v_j$ differ in two sectors of total angle $2\theta$, out of $2\pi$. Thus random hyperplane rounding givesIf $v_j=-v_i$, the signs differ with probability one and the same formula holds with $\theta =\pi$. Zero inner products have probability zero, since each $v_i$ is a unit vector.

If the graph has an edge, its corresponding principal $2\times 2$ block of $U$ forces $t\ge 2$, so the division by $t-1$ used to obtain the Gram matrix is valid. An edgeless graph can instead be colored with one color directly.

Because the sphere passes through the origin and has centre $P$, its radius is $|P|$. A point $X$ lies on it exactly whenor equivalentlyWith $P=\alpha A+\beta B+\gamma C$, the three required equations are the Gram matrix systemFor the specified vectors this becomesThusand the centre is

Again let $v_i\in {\mathbb{F}}_2^n$ be the characteristic vectors. They all lie in the even-weight subspacewhich has dimension $n-1$. Their Gram matrix isbecause its diagonal entries are zero and its off-diagonal entries are one. If $Gx=0$ and $s={\sum }_i{x}_i$, then $x=s1$. ConsequentlySince $G=V{V}^T$ for a matrix whose rows lie in $E$, its rank is at most $n-1$. If $n$ is even, an even $m$ cannot reach $n$, while an odd $m$ is at most $n-1$; hence $m\le n-1$. If $n$ is odd, the same calculation gives $m\le n$.

Both bounds are sharp. For odd $n$, take $A_i=[n]\setminus \{i\}$ for $1\le i\le n$: each set has even size and two distinct sets meet in the odd number $n-2$. For even $n$, apply the same construction to $[n-1]$ and regard the resulting $n-1$ sets as subsets of $[n]$.

Suppose first that the stated unitary and environment states exist. Preservation of the inner product givesThe environment states are normalized, so the Cauchy-Schwarz inequality gives $|⟨e_0|e_1⟩|\le 1$. Therefore

Conversely, writeand assume $|a|\le |b|$. If $b\ne 0$, put $c=a/b$, so $|c|\le 1$, and chooseThen $⟨e_0|e_1⟩=c$ and the desired output vectors have inner product $bc=a$, exactly matching the input vectors. If $b=0$, the inequality forces $a=0$, and one may take $|e_0⟩=|e_1⟩=|0⟩$.

In either case the input pair and output pair have the same Gram matrix. The isometry taking one pair to the other extends to a unitary matrix, so the required $U$ exists. Hencewhich is the environment-assisted two-state pure-state transformation criterion.

Substitute the B-spline expansion $s={\sum }_{j=1}^n{a}_j{M}_j$ into $(M_i,s)=k!{\gamma }_i$. The coefficients solvewith the Gram matrix and right-hand sideThe linearly independent B-splines have a positive-definite Gram matrix, so this system has a unique coefficient vector.

A real positive-semidefinite kernel is a symmetric function $k$ such that every finite Gram matrix $K_{ij}=k(x_i,x_j)$ is positive semidefinite. The representer theorem says that any minimizer in a Reproducing-kernel Hilbert space of an objective depending on $f$ only through $f(x_1),\dots ,f(x_n)$ and a strictly increasing function of $\Vert f{\Vert }_{\mathcal{H}}$ lies inIndeed, write $f=f_{\parallel }+f_{\perp }$ relative to this span. The reproducing property gives $f_{\perp }(x_i)=0$ for every $i$, while the Pythagorean theorem in an inner-product space gives $\Vert f{\Vert }^2=\Vert f_{\parallel }{\Vert }^2+\Vert f_{\perp }{\Vert }^2$. Removing a nonzero perpendicular component preserves all data values and strictly decreases the penalty, proving the theorem.

Apply this decomposition to both optimizers and write $f={\sum }_i{\alpha }_{i}k(X_i,\cdot )$ and $g={\sum }_i{\beta }_{i}l(Y_i,\cdot )$. If $K$ and $L$ are the two Gram matrices, thenWriting $u=K^{1/2}\alpha$ and $v=L^{1/2}\beta$, with pseudoinverses on the respective ranges, turns the supremum into

The Gram matrix of the canonical height pairing of rational points gives a robust independence certificate. For two points, set $a=\hat{h}(P)$, $b=\hat{h}(Q)$ and $c=(\hat{h}(P+Q)-a-b)/2$. If bounds from rounded measurements imply $a>0$ and $ab-c^2>0$, the matrix is positive-definite. Any integer relation $rP+sQ=O$ would give $a{r}^2+2crs+b{s}^2=0$, forcing $r=s=0$. Thus the points generate a free abelian group of rank two. Checking an interval lower bound for the determinant, rather than just its approximate central value, makes the certificate insensitive to rounding. It certifies independence, not that the points are a basis of the entire Mordell-Weil group.

For coefficient column $q=(q_0,\dots ,q_d{)}^T$, $M_{ij}=q_i\overline{q_j}$ is a Hermitian matrix and a positive semidefinite matrix because $x^{\ast }Mx=|q^{\ast }x{|}^2$. Expanding $|q(z){|}^2$ on the unit circle gives coefficient $p_k={\sum }_{i-j=k}{M}_{ij}$. Thus $M$ has matrix rank one when $q\ne 0$ and zero otherwise. General feasible Gram matrices need not have matrix rank one.

For real symmetric $A$, replace sign-vector lifts $x{x}^T$ by all matrices in the elliptope:Every sign vector gives a feasible rank-one matrix with the same objective, so the relaxed maximum is an upper bound. The omitted rank-one condition is substantive: an arbitrary feasible Gram matrix need not come from signs.

With the nonnegative Laplace-Beltrami operator, a frequency $w$ in the dual lattice gives the eigenfunction $e^{2\pi i⟨w,x⟩}$ and eigenvalue $4{\pi }^2|w{|}^2$. Its multiplicity is the number of dual vectors with that norm. Fourier series prove completeness. In dimension two the shortest vector, shortest independent vector and covolume determine a reduced Gram matrix, proving spectral rigidity.

Let $b_i=N{e}_i$ be the interpolating cardinal basis and set ${\Gamma }_{ij}={\int }_a^b{b}_i^{\prime \prime }{b}_j^{\prime \prime }dx$. Linearity gives $J(N\mathbf{v})={\mathbf{v}}^T\Gamma \mathbf{v}$, and this Gram matrix is a positive semidefinite matrix. Its null space consists precisely of vectors $(\alpha +\beta x_i{)}_i$, by the null-space statement for the second derivative roughness penalty. Thus its rank is $n-2$, including the zero matrix when $n=2$.

An order-$k$ B-spline is supported on $[t_i,t_{i+k}]$. Below this interval its truncated-power knot data agree with a degree-$k-1$ polynomial, annihilated by the order-$k$ divided difference; above it all data vanish. Therefore its ordinary or mixed Gram matrix has entries zero when $|i-j|\ge k$. The half-bandwidth is $k-1$, and for strictly increasing knots this bound is sharp.

If two simple roots have angle $5\pi /6$, normalize them to unit vectors $u,v$ with $(u,v)=-\sqrt{3}/2$. For any other unit simple root $w$, set $r=(u,w)\le 0$, $s=(v,w)\le 0$. Its three-root Gram matrix has determinant $1/4-r^2-s^2-\sqrt{3}rs$. Any nonzero $r$ or $s$ has absolute value at least $1/2$, contradicting positive definiteness. Thus all other simple roots are orthogonal to $u,v$. An irreducible root system containing this triple bond must be the rank-two G2 root system.

The vector-length multiset of a rank-two Euclidean lattice determines it up to an orthogonal map. Let $a$ be the shortest nonzero length. Removing two copies of each positive integer multiple of $a$ removes one primitive lattice line; the shortest remaining length $b$ is the shortest independent-vector length. Those two vectors form a lattice basis: otherwise a nonzero point in the centred basis parallelogram would be an independent lattice vector of length less than $b$. Lattice-point density determines the covolume $A$. The basis Gram matrix then has diagonal entries $a^2,b^2$ and absolute off-diagonal entry $\sqrt{a^2{b}^2-A^2}$. This proves two-dimensional spectral rigidity for a flat torus through its dual lattice.