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On the finite formal span of symbols $e_x$, define $B(e_x,e_y)=k(x,y)$ and extend bilinearly. For a positive-semidefinite kernel, the null vectors of this positive semidefinite bilinear form form its radical of a bilinear form. Quotienting by that linear subspace gives an inner product space and feature map $x\mapsto [e_x]$ with $⟨[e_x],[e_y]⟩=k(x,y)$. This construction works for degenerate and zero kernels.

A real positive-semidefinite kernel satisfies $|k(x,y){|}^2\le k(x,x)k(y,y)$. This follows from positivity of its two-point kernel matrix. If a diagonal value is zero, the corresponding off-diagonal values must also vanish.

For $d$ input coordinates, the homogeneous quadratic polynomial kernel has a weighted quadratic feature map with $D=d(d+1)/2$ distinct monomials. Computing an inner product of already stored mapped vectors costs $O(D)$ arithmetic operations, whereas $(x^{T}z{)}^2$ costs $O(d)$ using the input vectors. Here $O$ denotes big O notation. Higher fixed degrees can make the dimension gap much larger.

Sample count still matters. With $n$ training examples, materializing a dense kernel matrix stores $n^2$ numbers and requires $n(n+1)/2$ distinct evaluations using symmetry. A dense direct solve for kernel ridge regression typically takes $O(n^3)$ operations after constructing the kernel matrix, while a dual prediction with $m$ nonzero coefficients uses $m$ kernel evaluations. Explicit finite feature maps can be preferable when $D$ is small relative to $n$, and a stored primal weight vector can make prediction depend on $D$ rather than on $n$. The kernel trick by Ryan Giordano (sections Ridge regression and The polynomial kernel) explains the change from feature-coordinate equations to training-example equations; neither representation is universally faster.

The kernel trick saves explicit feature construction when the positive-semidefinite kernel is economical to evaluate; it does not remove the cost of optimization, choose a useful positive-semidefinite kernel, or establish generalization. Random Fourier features illustrate a different tradeoff: replacing the exact Gaussian kernel with a finite approximate feature map can reduce dependence on the sample count while introducing approximation error.

For a real-valued stochastic process $(X_t)$ with finite second moments, write $m(t)=\mathbb{E}{X}_t$. Its covariance function isThe diagonal gives the variance at each time; off-diagonal values record linear dependence between times. For the independent standard normal random variables above, $C_X(m,n)=1_{\{m=n\}}$, whereas the shared Gaussian process $Y_n=Z$ has $C_Y(m,n)=1$ for all $m,n$. For a fair simple random walk on the integer line, $C(m,n)=\min (m,n)$; Brownian motion has the continuous-time counterpart $C(s,t)=\min (s,t)$. These formulas quantify how much random history the observations share.

Every covariance function is a positive-semidefinite kernel. Indeed, for any real coefficients $a_1,\dots ,a_k$,Thus a proposed covariance function must make every finite covariance matrix positive semidefinite. For a Gaussian process, the mean function and the covariance function determine every finite-dimensional distribution, including degenerate ones such as the shared process. Conversely, any real positive-semidefinite kernel satisfying $C(s,t)=C(t,s)$, together with a specified mean function, defines consistent multivariate normal distributions; the Kolmogorov extension theorem then constructs a Gaussian process. For a general stochastic process, covariance alone does not determine the full law. The exceptional completeness of this description is what makes Gaussian process regression possible with a mean and a covariance model.

If the weight vector lies in the linear span of the training feature map outputs, write $w={\sum }_i{\beta }_i\varphi (p_i)$. ThenThis is a dual representation: coefficients weight training examples rather than feature coordinates. They need not be positive, and need not be unique when the kernel matrix is singular. Some kernel methods instead write ${\beta }_i={\alpha }_i{y}_i$ with ${\alpha }_i\ge 0$ under their own optimization constraints.^[ref] The formula is an algebraic conversion for a weight vector in this linear span, not a claim about every possible weight vector. The representer theorem gives a broader reason this form appears in regularized kernel methods.

For the XOR separation by a product feature, let $\Phi$ have rows $\varphi (p_i{)}^T$. The Gram matrix of those rows, equivalently the kernel matrix, isIts eigenvalues are $8,8,0,0$, so it is a positive semidefinite matrix of rank two, not an invertible matrix. More generally, for any real coefficient vector $a$ and any finite set of inputs,This proves the positive-semidefinite kernel condition for the displayed polynomial kernel on all of ${\mathbb{R}}^2$, rather than checking only this square.

Here $Ky=8y$, so the explicit choice $\beta =y/8$ gives $K\beta =y$ and ${\Phi }^T\beta =w$. ConsequentlyFor the new input $z=(1,1/2)$, the four polynomial kernel evaluations are $(9/4,1/4,1/4,9/4{)}^T$. Their weighted sum is $-1/2$, and the sign function predicts $-1$. No explicit feature map is needed for this calculation. These coefficients were constructed to reproduce the chosen separator; no fitting algorithm or generalization guarantee is being asserted.

The Gaussian density kernel is the standard normal probability density used as a unit-integral kernel for density estimation. Its second moment is one and $R(K_N)=1/(2\sqrt{\pi })$. Its canonical kernel has scale $(2\sqrt{\pi }{)}^{-1/5}$. It is a function of one smoothing displacement, distinct from the two-argument Gaussian kernel used as a positive-semidefinite kernel.

The Gaussian kernel is a translation-invariant positive-semidefinite kernel whose canonical feature map takes values in an infinite-dimensional vector space.

Suppose $k_{\tau }$ is a positive-semidefinite kernel for every $\tau$ and its diagonal entries are integrable. Positivity of each $2\times 2$ kernel matrix givesThe Cauchy-Schwarz inequality makes every entry integrable, and integrating each nonnegative finite quadratic form shows that $\int k_{\tau }d\tau$ is again a positive-semidefinite kernel.

A kernel method represents data through a positive-semidefinite kernel, allowing algorithms to use inner products in an implicit feature map.

The kernel trick evaluates inner products of feature map outputs through $k(x,z)=⟨\varphi (x),\varphi (z)⟩$ without explicitly constructing those vectors. An algorithm expressed entirely through these inner products can therefore operate on nonlinear derived coordinates using only evaluations of a positive-semidefinite kernel on the original inputs. A classifier linear in the derived coordinates can have a nonlinear boundary in the original coordinates.

The polynomial kernel gives a finite, fully visible example: $k(x,z)=(x^{T}z{)}^2$ corresponds to $\varphi (x)=(x_1^2,\sqrt{2}{x}_1{x}_2,x_2^2{)}^T$. The XOR separation by a product feature below makes the interaction coordinate $x_1{x}_2$ useful, while dual prediction with a kernel shows how to predict without calculating it explicitly.

An arbitrary score called a similarity need not be a positive-semidefinite kernel. Symmetry alone, even with nonnegative entries, is insufficient: the symmetric matrixhas $a^{T}Sa=-2$ for $a=(1,-1{)}^T$, so it cannot be a Gram matrix of real inner products. A positive-semidefinite kernel requires a positive semidefinite matrix for every finite choice of inputs, not just the observed training data.^[ref] The existing Moore-Aronszajn theorem connects this requirement to a Reproducing kernel Hilbert space; this example needs only finite linear algebra.

The Laplace kernel on ${\mathbb{R}}^d$ is a translation-invariant positive-semidefinite kernel. Independent standard Cauchy random variables give its Random Fourier feature representation through their characteristic function $e^{-|t|}$.

In the compact-domain spectral setting, a Mercer kernel is a continuous symmetric positive-semidefinite kernel on a compact space equipped with a finite full-support measure. Mercer's theorem yields nonnegative eigenvalues and an expansion $k(x,z)={\sum }_j{\lambda }_j{e}_j(x)e_j(z)$, giving features $(\sqrt{{\lambda }_j}{e}_j(x){)}_j$ in ${\ell }^2$. The finite-Gram positivity condition on arbitrary domains is more general than these spectral hypotheses; all such kernels still have a Reproducing kernel Hilbert space realization.

Use the normalizationThis is ridge regression; the penalty is the squared Euclidean norm. The centering removes the need for an intercept. Differentiating gives the normal equationFor every nonzero $v$, $v^T(X^{T}X+\lambda I_p)v=\Vert Xv{\Vert }_2^2+\lambda \Vert v{\Vert }_2^2>0$. The coefficient matrix is therefore a positive-definite matrix, even if the design matrix has deficient matrix rank. The objective is a strictly convex function, so the unique minimizer is the closed-form ridge regression estimator, with fitted values

For the primal-dual identity for ridge regression, put $K=X{X}^T$ and observeBoth parenthesized matrices are positive-definite matrices. Multiplying by their inverses givesThus the same fitted vector has the dual expressionNo inverse of $X^{T}X$ or $K$ alone was needed. This is also the kernel-ridge hat matrix for the linear kernel.

Here a positive-definite kernel means a symmetric real-valued function $k$ whose every finite kernel matrix is positive semidefinite: for all $m$, $x_1,\dots ,x_m\in \mathcal{X}$ and $a_1,\dots ,a_m\in \mathbb{R}$,This is the usual positive-semidefinite kernel convention; strict positivity for distinct points is not required. In particular $k(x,x)\ge 0$. Write $a=k(x,x)$, $b=k(x^{\prime },x^{\prime })$ and $r=k(x,x^{\prime })$. Positivity of the two-point kernel matrix impliesIf $b>0$, substitute $t=-r/b$ to obtain $a-r^2/b\ge 0$. If $b=0$, the affine expression $a+2tr$ can be nonnegative for every $t$ only if $r=0$. Both cases give the Cauchy-Schwarz inequality for positive-semidefinite kernels:The zero-diagonal case is essential when the positive-definite kernel is not strictly positive.

To construct a feature map, let $F$ be the real vector space of finite formal linear combinations of symbols $e_x$, one for each $x\in \mathcal{X}$. Define a symmetric bilinear form byThe positive-definite kernel property makes this a positive semidefinite bilinear form. It need not yet be an inner product: distinct formal vectors can have zero squared length. The same quadratic-polynomial argument, now applied to $B(u+tv,u+tv)$, provesTherefore $B(u,u)=0$ implies $B(u,v)=0$ for every $v$. The setis the radical of a bilinear form, hence a linear subspace. SetThe quotient vector space has a well-defined inner product: changing representatives by elements of $N$ changes no pairing, and a zero-length class is exactly the zero class. It is therefore an inner product space.

Define $\varphi (x)=e_x+N$. By construction,Every positive-definite kernel thus has a feature representation, including degenerate kernels and the zero kernel. This is its canonical feature space of a positive-semidefinite kernel. The construction already gives the requested inner product space.

Construct a mean-square derivative of a Gaussian process on the original probability space before choosing its continuous modification. The real covariance function satisfies $K(h)=K(-h)$, because covariance is symmetric. Three-times differentiability implies that $K^{\prime \prime }$ is continuous everywhere and differentiable at zero. Since $K^{\prime \prime }$ is an even function,This uses differentiability of $K^{\prime \prime }$ at zero, not a bound on $K^{\prime \prime \prime }$ throughout a neighborhood.

Let $D_{h}X(t)=(X(t+h)-X(t))/h$ for $h\ne 0$. Twice applying the fundamental theorem of calculus to the covariance function givesOriented integrals make this identity valid for negative as well as positive $h,k$. For a fixed $t$, the right side tends to $-K^{\prime \prime }(0)$ as $h,k\to 0$. In particular,By completeness of L2 space, the mean-square derivative of a Gaussian processexists. Finite vectors consisting of values of $X$ and $Y$ have a centered multivariate normal distribution: approximate the $Y$ coordinates by difference quotients and use mean-square convergence and characteristic functions. Thus $Y$ is a centered Gaussian process jointly constructed with $X$, withIn particular $-K^{\prime \prime }(0)\ge 0$, and $-K^{\prime \prime }$ is a positive-semidefinite kernel because it is an actual covariance function. Merely constructing an unrelated Gaussian process with that covariance function would not yet construct a modification of a stochastic process of $X$.

The increment variance of $Y$ isFor small $|t-s|$, differentiability of $K^{\prime \prime }$ at zero bounds this by $C|t-s|$. For the remaining distances in $[0,1]$, continuity of $K^{\prime \prime }$ enlarges $C$ to give the same bound. The fourth moment of a centered normal distribution is three times its variance squared, soThe one-dimensional Kolmogorov continuity theorem states that $\mathbb{E}|U(t)-U(s){|}^p\le C|t-s{|}^{1+\eta }$ with $p,\eta >0$ gives a continuous modification, with sample-path Hölder continuity of every order less than $\eta /p$. Apply it here with $p=4,\eta =1$. We obtain a continuous modification $\tilde{Y}$ on $[0,1]$. On its null exceptional event, set the entire path equal to zero.

Define, on this same probability space,The path integral exists because its integrand is a continuous function. It agrees in L2 space with the integral of $Y$: Riemann sums converge in L2 space because $Y$ is mean-square continuous; sums of $\tilde{Y}$ have the same values almost surely at each finite collection of mesh points and converge pathwise to the integral. These two limits therefore agree almost surely for each fixed $t$. Equivalently, the integral is the Bochner integral of the L2 space-valued map $s\mapsto Y(s)$.

To verify the modification of a stochastic process property directly, put $A_t=X(t)-X(0)$ and $I_t={\int }_0^{t}Y(s)ds$. The preceding covariance identities and the fundamental theorem of calculus giveTherefore $\mathbb{E}(A_t-I_t{)}^2=0$ and $\tilde{X}(t)=X(t)$ almost surely for every fixed $t$. This is also the mean-square fundamental theorem of calculus in this stationary setting. If a version on all of $\mathbb{R}$ is desired, retain the original $X(t)$ outside $[0,1]$; equality at every fixed parameter still holds.

The ordinary fundamental theorem of calculus now gives the simultaneous pathwise conclusionThus the requested differentiable modification of a stationary Gaussian process exists. Integrability of $K$ is compatible with the hypotheses but is unnecessary for this direct argument. It also avoids any assumption that a third absolute moment of a spectral density follows from three-times differentiability.

In the terminology used for positive-definite kernels, positive definiteness means positive semidefiniteness of every finite kernel matrix: $k$ is symmetric and, for every $m$, points $x_1,\dots ,x_m$ and real $c_1,\dots ,c_m$,The linear kernel is symmetric, andThus it is a positive-semidefinite kernel, with the identity as its feature map. It need not be strictly positive definite: a zero vector or linearly dependent points can make this quadratic form zero for nonzero $c$. This distinction is essential when interpreting the terminology.

A positive-semidefinite kernel on $\mathcal{X}$ is a symmetric function $k:{\mathcal{X}}^2\to \mathbb{R}$ such that, for every $x_1,\dots ,x_n$ and $c_1,\dots ,c_n$,If $k(x,x^{\prime })=⟨\varphi (x),\varphi (x^{\prime })⟩$ for a feature map into an inner product space, thenThus every feature-map inner product is a kernel. If ${\alpha }_1,{\alpha }_2\ge 0$, then each finite quadratic form for ${\alpha }_1{k}_1+{\alpha }_2{k}_2$ is the corresponding nonnegative linear combination, so nonnegative linear combinations of kernels are kernels.