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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.

Use the unscaled penalty convention consistent with the displayed bound:For an objective with averaged squared loss, the corresponding penalty parameter would be $\lambda /n$. We retain the unscaled $\lambda$ throughout.

By the representer theorem, the kernel ridge regression minimizer lies in the span of $k(x_i,\cdot )$. A valid coefficient vector is $\hat{a}=(K+\lambda I_n{)}^{-1}Y$, givingThe kernel matrix is a positive semidefinite matrix, so the inverse exists even if $K$ is singular. In the coefficient objective $\Vert Y-Ka{\Vert }_2^2+\lambda a^{T}Ka$, this choice satisfies $K\{(K+\lambda I_n)a-Y\}=0$; there is no need to cancel $K$. The function itself is unique because of the strictly convex squared Hilbert norm. The matrix $H$ is the kernel-ridge hat matrix in the present normalization.

Let $m^0=(f^0(x_i){)}_{i=1}^n$. Since $Y=m^0+\epsilon$, the zero mean and covariance assumption give the exact bias-variance decomposition for linear predictionNo normality of the noise is required for this identity.

Choose an orthonormal eigenbasis $v_i$ of $K$, with eigenvalues $d_i\ge 0$. Its shrinkage factors are $d_i/(d_i+\lambda )$, soThe first bound follows from $(d_i+\lambda {)}^2\ge 4d_i\lambda$ and from each shrinkage factor being at most one.

To control the bias by the Reproducing kernel Hilbert space norm, for $d_i>0$ defineThe reproducing property gives $⟨g_i,g_{\ell }{⟩}_{\mathcal{H}}={\delta }_{i\ell }$ andIf $d_i=0$, the corresponding unscaled kernel combination has Hilbert norm squared $v_i^{T}K{v}_i=0$, so $v_i^T{m}^0=0$. There is therefore no bias component in a zero-eigenvalue direction. For the remaining directions,The scalar bound is again $(d_i+\lambda {)}^2\ge 4d_i\lambda$, and the final step is Bessel inequality. Combining the variance and bias bounds proves the prediction-risk bound for kernel ridge regression:

The kernel ridge regression estimator isBy the representer theorem, ${\hat{f}}_{\lambda }={\sum }_{j=1}^n{\alpha }_{j}k(x_j,\cdot )$. If $K_{ij}=k(x_i,x_j)$, substitution and differentiation giveThusThe matrix $H_{\lambda }$ is the kernel-ridge hat matrix.

For the unscaled objective ${\sum }_i(Y_i-f(x_i){)}^2+\lambda \Vert f{\Vert }_{\mathcal{H}}^2$, the kernel-ridge hat matrix is $H=K(K+\lambda I{)}^{-1}$. Its prediction variance is ${\sigma }^2\mathrm{tr}(H^2)/n$. For eigenvalues $d_i\ge 0$, use $d_i^2/(d_i+\lambda {)}^2\le \min \{d_i/(4\lambda ),1\}$ and ${\lambda }^2{d}_i/(d_i+\lambda {)}^2\le \lambda /4$ to bound the variance and the squared bias. The latter also uses the reproducing property and Bessel inequality, and does not require invertibility of $K$.

For any real design matrix $X$ and $\lambda >0$, $(X^{T}X+\lambda I{)}^{-1}{X}^T=X^T(X{X}^T+\lambda I{)}^{-1}$. Multiplication by $X$ identifies the primal ridge regression fitted values with $K(K+\lambda I{)}^{-1}Y$, $K=X{X}^T$, the kernel-ridge hat matrix for the linear kernel. No rank assumption is needed.