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For an $n\times p$ design matrix, distinguish the coefficient vector from the fitted-response vector. The ridge regression objective and its solution areThe gradient is $2(X^{T}X+\lambda I_p)\beta -2X^{T}Y$. For $\lambda >0$, $X^{T}X+\lambda I_p$ is a positive-definite matrix, so this is the unique minimizer even if $X$ is rank deficient. The expression with an additional leading $X$ is instead the $n$-vector of fitted values,This distinction corrects the dimensional mismatch in the printed coefficient-estimator notation.

The quadratic penalty discourages large coefficients, exchanging some bias for lower variance. If $X=UD{V}^T$ is a reduced singular value decomposition, with positive singular values $d_k$, thenThe principal-component shrinkage by ridge regression therefore shrinks directions with small $d_k^2$ most strongly. These are precisely the poorly determined directions created by multicollinearity, where ordinary least-squares coefficients can have large variance or fail to be uniquely determined. A positive penalty stabilizes their inversion and sets coefficients in the null space to zero. It does not make the predictors independent or provide a universal risk improvement for every signal. Because the penalty depends on coefficient units, predictors are ordinarily standardized before choosing $\lambda$ by cross-validation. An unpenalized intercept in ridge regression is handled separately by centering; the displayed objective either has no intercept or applies the penalty to every component included in $\beta$.

With the same squared-error normalization as in Question 2, the ridge estimator solvesThe unpenalized intercept in ridge regression is $\hat{a}=\overline{Y}$ because the columns of $X$ are centered. Differentiating the remaining quadratic objective gives $(X^{T}X+n\lambda I_p){\hat{\beta }}_{\lambda }^R=X^T{Y}_c$. Since $\lambda >0$, the left matrix is positive definite even when $X$ is rank-deficient. Thus the closed-form ridge regression estimator isThe second equality again uses centering. If the penalty convention is $\lambda \Vert b{\Vert }_2^2/2$ with an unnormalized squared-error term $\Vert Y_c-Xb{\Vert }_2^2/2$, replace $n\lambda$ by $\lambda$ throughout; the limiting assertion is unchanged.

For the full-column-rank case, take a singular value decomposition $X=UD{V}^T$ with positive singular values $d_1,\dots ,d_p$. The ordinary least squares estimator, which is the slope maximum-likelihood estimator in this normal linear model, and the ridge regression estimator are respectivelyConsequently principal-component shrinkage by ridge regression multiplies the least-squares coordinate in direction $v_j$ by $d_j^2/(d_j^2+n\lambda )$. Ridge shrinks most strongly in directions with the smallest singular values. These are combinations of predictors that the data distinguish least well.

Near multicollinearity creates small singular values, and the factor $1/d_j$ in ordinary least squares greatly amplifies noise in those directions. Its coefficient variance there is ${\sigma }^2/d_j^2$, while the ridge variance iswhich is smaller. The price is bias: the expectation of the ridge coordinate is $d_j^2/(d_j^2+n\lambda )$ times the corresponding true coefficient. This bias-variance tradeoff can substantially reduce estimation and prediction error when the poorly identified directions do not contain an excessively large signal. It is not a guarantee of improvement for every coefficient vector. Adding $n\lambda$ to every eigenvalue also improves the condition number of the normal matrix and makes numerical inversion more stable.

To handle every rank and every relation between $n$ and $p$, put $G=X^{T}X$ and use its spectral theorem for real symmetric matrices. Choose an orthonormal eigenbasis $v_1,\dots ,v_p$ with eigenvalues ${\gamma }_j\ge 0$, and put $z=X^{T}Y$. ThenIf ${\gamma }_j=0$, then $\Vert X{v}_j{\Vert }_2^2=v_j^{T}G{v}_j=0$, so $X{v}_j=0$ and $v_j^{T}z=(X{v}_j{)}^{T}Y=0$. Thus these terms are exactly zero for every positive $\lambda$; there is no divergent component in the null space. Each remaining term converges to $(v_j^{T}z/{\gamma }_j)v_j$. By the definition of the Moore-Penrose pseudoinverse,and thereforeThis proves the vanishing-penalty ridge limit without an invertibility assumption. The limit is the minimum-norm least-squares solution: every other least-squares solution differs by a vector in $\ker X$, orthogonal to the displayed solution, and therefore has at least as large a Euclidean norm.

Using the estimator from part (b) and the spectral decomposition of $X^{T}X$,ThusUnder the stated separation of scales, the shrinkage factor is approximately one for $i\le q$ and approximately zero for $i>q$. The principal-component shrinkage by ridge regression formula therefore reduces approximately tothe orthogonal projection of $Y$ onto the span of the $q$ normalized sample principal components with greatest variance.

Ridge regression fits a linear regression model while penalizing the size of its slope regression coefficients. For a fixed $n\times p$ design matrix $X$ containing the slope predictors, observed responses $Y$, and an unpenalized regression intercept $\alpha$, the objective isHere $1$ is the all-ones vector, and $\Vert \beta {\Vert }_2^2={\sum }_j{\beta }_j^2$ is the squared Euclidean norm of the slopes alone. The first term is the residual sum of squares; the second implements regularization. No random-error assumption is needed to define this fit. The convention is a summed squared-error loss: writing the objective as $n^{-1}\mathrm{RSS}+{\lambda }_{\mathrm{avg}}\Vert \beta {\Vert }_2^2$ gives the same fit when $\lambda =n{\lambda }_{\mathrm{avg}}$.^[ref]

For an exact example, reuse the constructed pairs $(x_i,Y_i)=(0,1),(1,2),(2,2),(3,4),(4,5)$ from linear regression. Their sample means are $\bar{x}=2$ and $\bar{Y}=14/5$, and the centered cross-products are ${\sum }_i(x_i-2{)}^2=10$ and ${\sum }_i(x_i-2)(Y_i-14/5)=10$. ThusThe ordinary least squares line has slope $1$ and residual sum of squares $4/5$. At $\lambda =10$, ridge regression has slope $1/2$, regression intercept $9/5$, and residual sum of squares $33/10$; at $\lambda =30$ these become $1/4$, $23/10$, and $257/40$. Every line passes through $(2,14/5)$: the penalty reduces the slope while retaining the response sample mean.

The computational benefit is visible after predictor centering. Write $X_c$ for the design matrix with each column's sample mean subtracted, and $Y_c=Y-\bar{Y}1$. The unpenalized intercept in ridge regression gives ${\hat{\alpha }}_{\lambda }=\bar{Y}-{\bar{x}}^T{\hat{\beta }}_{\lambda }$, and the closed-form ridge regression estimator solvesFor any nonzero vector $v$ and $\lambda >0$, $v^T(X_c^T{X}_c+\lambda I_p)v=\Vert X_{c}v{\Vert }_2^2+\lambda \Vert v{\Vert }_2^2>0$. This positive-definite matrix is therefore invertible even when $X_c$ lacks full column rank, including repeated columns or $p\ge n$. The unique penalized slopes resolve the ambiguity of rank-deficient ordinary least squares; they do not supply evidence about slope directions absent from the data. This makes ridge regression useful for multicollinearity.^[ref]

Predictor centering and scaling serve different purposes. Centering handles the free regression intercept. Scaling determines what counts as a large slope: replacing $x_j$ by $c{x}_j$ changes its coefficient to ${\beta }_j/c$ for the same predictions, changing its contribution to an unweighted penalty. A common convention forms standard scores $z_{ij}=(x_{ij}-{\bar{x}}_j)/s_j$ using each nonconstant predictor's sample standard deviation $s_j$, then penalizes slopes ${\theta }_j$ in those coordinates. Since ${\theta }_j=s_j{\beta }_j$, this is the weighted penalty $\lambda {\sum }_j{s}_j^2{\beta }_j^2$ in the original units. Constant columns have $s_j=0$ and must be removed before this scaling. Scaling is a modelling choice; equal penalties in original units and in standardized units define different fits.^[ref]

The singular value decomposition explains the shrinkage. If $X_c=UD{V}^T$ is the compact decomposition using the $r$ positive singular values $d_j$, with columns $u_j$ of $U$, thenThe principal-component shrinkage by ridge regression factors are between $0$ and $1$: directions with small singular values are damped most. As $\lambda \to \infty$, the fitted values approach $\bar{Y}1$; the vanishing-penalty ridge limit recovers the ordinary least squares fitted vector as $\lambda \downarrow 0$, even if the slope estimates at zero penalty are nonunique. The covariance and bias of a ridge regression estimator describes the resulting bias-variance tradeoff. A penalty can reduce variance of an estimator while introducing bias of an estimator; the unbounded directional risk of fixed ridge shrinkage rules out a uniform improvement for a fixed positive penalty over all possible signals.^[ref]

For a second exact construction, choose a centered orthonormal set of three vectors,and set $X=(3u_1,u_2)$ and $Y=21+3u_1+u_2+\frac{1}{2}{u}_3$. The squared singular values are $9$ and $1$. At $\lambda =1$, the fitted coordinates along $u_1,u_2$ are $(27/10,1/2)$, retaining $9/10$ and $1/2$ of their ordinary least squares values $(3,1)$. At $\lambda =9$ they become $(3/2,1/10)$, retaining $1/2$ and $1/10$. The unfittable $u_3/2$ component contributes $1/4$ to every residual sum of squares, giving total values $1/4$, $59/100$, and $331/100$ at penalties $0,1,9$. These are constructed fits, not measured prediction results.

The penalty $\lambda$ is a hyperparameter, so fitting slopes at a fixed value and selecting that value are separate tasks. Use a validation set or K-fold cross-validation to compare candidate penalties by held-out squared-error loss. In each split, compute predictor sample means and scales from that split's training data, apply the same transformation to its validation set, and refit the slopes; using held-out information in preprocessing introduces data leakage. Minimum training residual sum of squares cannot select useful shrinkage: ordinary least squares already attains the smallest possible value.^[ref] After model selection, refit on all training data with the selected penalty. A separate test set or nested cross-validation assesses the entire selection procedure; a low selected validation score alone is no performance guarantee.^[ref]