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A coarse-grid correction improves a fine-grid approximation by solving a smaller equation for the part of its approximation error representable on a coarse mesh. Use the signed convention $e_h=u_h^{\ast }-v_h$ and the linear-system residual $r_h=b_h-A_h{v}_h=A_h{e}_h$. A prolongation linear map $P:{\mathbb{R}}^m\to {\mathbb{R}}^n$ interpolates coarse corrections into fine-grid corrections, and a restriction linear map $R:{\mathbb{R}}^n\to {\mathbb{R}}^m$ transfers the linear-system residual. Solve $A_H{z}_H=R{r}_h$ exactly or approximately and update $v_h\leftarrow v_h+P{z}_H$. The plus sign follows from defining the approximation error as exact minus approximate. It is the correction equation's right-hand side $R{r}_h$, rather than merely restricted solution values, that determines the update.

For the multigrid method model $A_h=h^{-2}\mathrm{tridiag}(-1,2,-1)$, take $n=2m+1$, so $H=2h$ and the coarse nodes coincide with every second fine node. Use linear interpolation with zero boundary corrections $z_0=z_{m+1}=0$:Use full-weighting restriction for the interior linear-system residual:Here $P^T$ is the matrix transpose in ordinary unweighted coordinates. With these specific transfers, the Galerkin coarse-grid operator isThus it equals the rediscretized Poisson equation operator on this constant-coefficient one-dimensional mesh. The equality is special to the stated model and transfers. In particular, dropping $h^{-2}$ from the fine matrix while using $H^{-2}$ on the coarse mesh changes the correction unless the right-hand sides and transfers are rescaled consistently.

Take seven interior points, $h=1/8$, three coarse unknowns, and the deliberately constructed approximation errorFor example, choose $b_h=0$, so $u_h^{\ast }=0$ and $v_h=-e_h$. Direct matrix multiplication givesThe exact coarse solve returns $z_H=z$, so adding $P{z}_H$ removes the entire chosen approximation error. A broad triangular error can have a sharply localized linear-system residual: the latter measures discrete curvature, not error height. This is why the two should not be confused. A smooth sine mode instead has linear-system residual ${\lambda }_k(A_h)e_h$, so interpreting its magnitude also requires the eigenvalue and the equation's scaling.

The algebra explains the exact cancellation beyond this example. Assume $P$ has linearly independent columns and choose an invertible Galerkin coarse-grid operator $A_H=R{A}_{h}P$. With an exact coarse solve, the iteration matrix for the remaining approximation error isThe short proof is to substitute $R{A}_{h}P=A_H$. Thus every vector in the column space of $P$ is eliminated, and the corrected linear-system residual restricts to zero. These are exact projection identities, not a statement that every visually smooth vector belongs to that column space. If $A_h$ is a symmetric matrix and a positive-definite matrix, and $R=c{P}^T$ with $c>0$, then $A_H$ is also a positive-definite matrix. Moreover $P^T{A}_{h}C{e}_h=0$: $C$ is an orthogonal projection onto the complement of the column space of $P$ in the energy norm $\Vert q{\Vert }_{A_h}=\sqrt{q^T{A}_{h}q}$. Equivalently, the added correction is the best approximation to $e_h$ from that column space in this energy norm. Without these symmetry and transfer assumptions, the projection need not be orthogonal.^[ref]

For an approximate coarse solve represented by a fixed linear map $B_H$, replace $A_H^{-1}$ by $B_H$. ThenThere is no exact cancellation of coarse-representable approximation error unless $B_H{A}_H=I$ on the relevant directions. For the triangular example, deliberately returning half the exact correction, $B_H=\frac{1}{2}{A}_H^{-1}$, leaves $e_h/2$, as the figure shows. This controlled mathematical comparison illustrates an incomplete coarse solve; it is not a measured performance result or a prescription for a geometric multigrid V-cycle. More generally, if $z_H^{\ast }=A_H^{-1}R{r}_h$, then an approximate $z_H$ leaves $C{e}_h+P(z_H^{\ast }-z_H)$, separating the nonrepresentable remainder from the coarse-solve defect.

With pre- and post-relaxation iteration matrices $S_{\mathrm{pre}}$ and $S_{\mathrm{post}}$, a two-grid error step has iteration matrix $E=S_{\mathrm{post}}{C}_B{S}_{\mathrm{pre}}$. A geometric multigrid V-cycle supplies $B_H$ by one recursive coarse cycle from zero, with a direct solve only at the bottom. Effective smoothing, a coarse column space that approximates the surviving approximation error, and a sufficiently accurate coarse correction must work together. Coarsening alone supplies no convergence guarantee, and correcting an arbitrary fine-grid vector exactly would require a coarse column space large enough to contain it.

Let self-adjoint orthogonal projections $P_i$ sum to the identity. For $v$ in the range of $P_j$,The $j$th term already equals $\Vert v{\Vert }^2$, so $P_{i}v=0$ for every $i\ne j$. Hence $P_i{P}_j=P_j{P}_i=0$ and the projective measurement operators commute.

// Target: quantum-theory.bigb

The conditional variance measures the remaining squared uncertainty after specified information has been used to form a conditional mean. For a real square-integrable random variable $X$ and a sub-sigma-algebra $\mathcal{G}$, let $m=\mathbb{E}[X\mid \mathcal{G}]$. ThenIt is a $\mathcal{G}$-measurable random variable, nonnegative almost surely, rather than necessarily a single number. With the information $\mathcal{G}=\sigma (Y)$, its value is the variance in the group identified by $Y$. Its expected value is the mean squared error of predicting $X$ by its conditional mean.^[ref]

In the exact mixture distribution in law of total expectation, group $A$ contains $0,4$ around mean two, and group $B$ contains $6,10$ around mean eight. Both groups have conditional variance four because their deviations from their own means are $-2,2$ with equal probabilities. Therefore $\mathbb{E}[\mathrm{Var}(X\mid Y)]=4$. Meanwhile, the two group means have variance $\frac{1}{4}(2-13/2{)}^2+\frac{3}{4}(8-13/2{)}^2=27/4$. The law of total variance givesDirect calculation gives $\mathbb{E}[X^2]=53$ and $\mathrm{Var}(X)=53-(13/2{)}^2=43/4$, checking the decomposition.

For any square-integrable $\mathcal{G}$-measurable predictor $Z$, the error identity isThe cross term vanishes because $\mathbb{E}[X-m\mid \mathcal{G}]=0$. Thus using the conditional mean minimizes mean squared error, while conditional variance quantifies what this information still cannot predict. This is the orthogonal projection interpretation in L2 space.^[ref]

In a unitary representation of a finite group, if a cyclic vector $v$ satisfies $\rho (g)v=av$ for some $a\ne 1$, then there are no invariant vectors. Indeed the averaging orthogonal projection $P=|G{|}^{-1}{\sum }_h\rho (h)$ satisfies $P\rho (g)=P$, so $Pv=aPv=0$. All translates of $v$ also project to zero, and they span the representation.

A finite-dimensional density operator $\rho$ represents a quantum state on a Hilbert space. It is Hermitian, positive semidefinite, and has trace one:The density matrix is its matrix in a chosen orthonormal basis. In dimension $N$ it has $N^2-1$ real parameters: a Hermitian matrix has $N^2$, and fixing its trace removes one; positivity restricts the allowed values.

A quantum state ensemble of normalized pure states $|{\psi }_i⟩$ with positive probabilities $p_i$ givesZero-weight terms can be omitted. Conversely, the spectral theorem gives such an ensemble for every density operator, using its orthonormal eigenvectors and nonnegative eigenvalues, with zero eigenvalues omitted. A pure state has $\rho =|\psi ⟩⟨\psi |$; a mixed state needs at least two nonzero eigenvalues. ^[ref]

For a projective measurement with orthogonal projections ${\Pi }_a$, the Born rule isThis is the average of the component pure state probabilities. Positivity makes each outcome probability nonnegative, and ${\sum }_a{\Pi }_a=I$ makes them sum to one. The same density operator therefore predicts every measurement on the system without requiring a unique preparation history. ^[ref]

For a closed system, Schrödinger picture evolution under a unitary operator iswhen ${\rho }_H={\rho }_S(0)$ denotes the time-independent Heisenberg picture state and $U(0)=I$. In the Schrödinger picture the density operator evolves; in the Heisenberg picture the observables evolve instead, with identical measurement predictions. This unitary operator preserves the eigenvalues and hence the purity of a density operator. ^[ref]

A historical source for the probabilistic formulation of quantum mechanics is von Neumann's 1927 paper Wahrscheinlichkeitstheoretischer Aufbau der Quantenmechanik. ^[ref]

In a nearest-neighbour frustration-free Hamiltonian chain, multiply the bond orthogonal projections in two disjoint alternating layers. The resulting product $K$ fixes the common ground state space but contracts its orthogonal complement by a factor bounded away from one in terms of the spectral gap. A bound $\Vert K{|}_{\mathrm{supp}H}\Vert \le (1+\Delta /2{)}^{-1/3}$ implies exponentially convergent powers. The product is generally neither Hermitian nor normal.

A unital subalgebra $A\subseteq \mathrm{End}(V)$ closed under the adjoint operator, on a finite-dimensional real or complex inner product space, equals its double commutant of an operator algebra. Here is a proof that does not require a classification of algebras. For $N=\dim V$, choose a basis $(v_1,\dots ,v_N)$ and let $W=\{(a{v}_1,\dots ,a{v}_N):a\in A\}\subseteq V^N$. The diagonal action of $A$ preserves $W$ and $W^{\perp }$, so the orthogonal projection onto $W$ commutes with that action. Each matrix entry of this projection belongs to $A^{\prime }$. Hence the diagonal action of $b\in A^{\prime \prime }$ commutes with the projection and preserves $W$. Since $1\in A$, the tuple $(v_1,\dots ,v_N)$ belongs to $W$, and therefore $(b{v}_1,\dots ,b{v}_N)=(a{v}_1,\dots ,a{v}_N)$ for one $a\in A$. This implies $b=a$. Without the unital hypothesis the conclusion is $A^{\prime \prime }=A+\mathbb{F}1$.

The efficient score is the residual after orthogonal projection of the parametric score function onto the nuisance tangent space. It is centered and orthogonal to every nuisance direction. Its squared L2 norm is the efficient information.

With unknown independent covariate and centered error distributions, take the regular mean-preserving error tangent space, let $h={\partial }_{\theta }{g}_{\theta }$, $\rho =-f^{\prime }/f$ and ${\tau }^2=E_f{\epsilon }^2>0$. Under $E_f\rho =0$, $E_f(\epsilon \rho )=1$ and finite second moments, orthogonal projection removes $(E_{v}h)(\rho -\epsilon /{\tau }^2)$ from $h\rho$. Thus the efficient score and efficient information are $\tilde{\ell }=(h-E_{v}h)\rho +(E_{v}h)\epsilon /{\tau }^2$ and $\tilde{I}={\mathrm{Var}}_v(h)E_f{\rho }^2+(E_{v}h{)}^2/{\tau }^2$. The reduction to $h\epsilon /{\tau }^2$ is valid for a normal distribution of errors or for constant $h$, but need not hold otherwise. Centered covariates and logistic distribution errors give the counterexample $\tilde{\ell }=X\tanh (\epsilon /2)$.

For an elliptically polarized wave, the tip of its transverse vector traces an ellipse at a fixed point during each wave period. For polarization of an electromagnetic wave, this is the electric field vector. With the real components in polarization of waves, assume $A_x,A_y>0$ and $\sin \delta \ne 0$. Eliminating time givesA short derivation uses $\varphi =\omega t$ and the cosine addition formula to obtain $E_y/A_y=(E_x/A_x)\cos \delta -\sin \varphi \sin \delta$; squaring and using the sine and cosine identity ${\sin }^2\varphi =1-{\cos }^2\varphi$ gives the displayed equation. Outside phase quadrature, the mixed term gives a tilted ellipse; $A_x,A_y$ are component wave amplitudes and need not be its semi-major and semi-minor axes.

In phase quadrature, $\delta =\pm \pi /2$, the mixed term vanishes. For $(A_x,A_y)=(2,1)V/m$, the ellipse is $E_x^2/4+E_y^2=1$, its semi-major and semi-minor axes have lengths $2$ and $1V/m$, and its electric field magnitude varies between $1$ and $2V/m$. At $\varphi =\pi /4$ with $\delta =\pi /2$, the electric field is $(\sqrt{2},-1/\sqrt{2})V/m$ and has magnitude $\sqrt{5/2}V/m$. Equal wave amplitudes instead give circular polarization, a special case of elliptical polarization with constant magnitude. As $\delta$ approaches $0$ or $\pi$, the ellipse collapses to a line segment and gives linear polarization; this limit is degenerate rather than a nondegenerate ellipse.

Changing $\delta$ from $\pi /2$ to $-\pi /2$ reverses traversal of the same ellipse. In the displayed $x,y$ plot, $\delta =\pi /2$ moves from $(A_x,0)$ toward negative $y$ as time increases. A label such as right- or left-handed circular polarization additionally requires a stated viewing direction and naming convention: viewing along propagation and viewing against propagation reverse the apparent rotation. ^[ref].

The ellipse records both the relative wave amplitudes and the phase difference of two transverse oscillations. This separates a change of polarization of waves from a change of overall wave amplitude: multiplying both components by the same positive factor enlarges the ellipse while preserving its shape and orientation.

A rotating ideal linear polarizer can probe this geometry. For the $2:1$ ellipse in phase quadrature, orthogonal projection onto a pass axis at angle $\theta$ gives $E_{\mathrm{out}}(t)=2\cos \theta \cos \omega t-\sin \theta \sin \omega t$, in $V/m$. Averaging its square over a wave period gives $(4{\cos }^2\theta +{\sin }^2\theta )/2$, while the incoming squared electric field magnitude averages to $5/2$. Thus the transmitted fraction of light intensity is $(4{\cos }^2\theta +{\sin }^2\theta )/5$: it is $4/5$ at $0^{\circ }$, $1/2$ at $45^{\circ }$ and $1/5$ at $90^{\circ }$. There is no single incident line of linear polarization to insert into Malus's law; projecting the electric field and averaging its square is the appropriate generalization. This measurement cannot distinguish $\delta =\pi /2$ from $\delta =-\pi /2$, because reversing the traversal leaves all these squared averages unchanged.

The expectation value of a quantum observable is the expected value of its measurement outcome random variable in a specified quantum state. It predicts the mean of outcomes from many identical quantum state preparations. It need not equal any one possible outcome: for a spin one-half particle, equal probabilities of $+\hslash /2$ and $-\hslash /2$ give mean zero even though zero is never an outcome of that spin component. Here $\hslash$ is the reduced Planck constant.

Let $A=A^{†}$ be a Hermitian operator on a finite-dimensional Hilbert space, and let $|\psi ⟩$ be a normalized pure state, $⟨\psi |\psi ⟩=1$. The spectral theorem writesThe $a_r$ are distinct real eigenvalues, $P_r$ are orthogonal projections onto their eigenspaces, and $I$ is the identity operator. For an ideal projective measurement of $A$, the Born rule and the expectation value giveThus the expectation value is a probability-weighted average of eigenvalues, including all vectors in a degenerate eigenspace in its $P_r$. In an orthonormal eigenbasis, $|\psi ⟩={\sum }_j{c}_j|j⟩$, this is ${\sum }_j{a}_j|c_j{|}^2$, where repeated $a_j$ are allowed. The weighted average lies between the smallest and largest eigenvalues; it can lie in a gap between them. The statistical meaning and operator expression are connected in this way in the authored treatment of average energies.^[ref]

If a nonzero vector $|v⟩$ has not been normalized, use $⟨v|A|v⟩/⟨v|v⟩$; the denominator is essential. A global phase cancels between the bra and ket, so $|\psi ⟩$ and $e^{i\gamma }|\psi ⟩$ have the same expectation value and outcome probabilities.

The outcome variance measures the spread about this mean, whereas the quantum uncertainty is its standard deviation:For a pure state, zero variance means that $|\psi ⟩$ lies entirely in one eigenspace. That eigenspace need not be one-dimensional. Knowing only a mean generally does not determine the entire outcome probability distribution.

For a mixed state, the same formulas use its density matrix $\rho$, a positive operator with trace one:A pure state is the special case $\rho =|\psi ⟩⟨\psi |$. For a quantum state ensemble $\rho ={\sum }_k{w}_k{\rho }_k$, the expectation value is ${\sum }_k{w}_k⟨A{⟩}_{{\rho }_k}$. This linearity is useful when preparations vary between trials; different quantum state ensembles can have the same density matrix.

These formulas require no operator-domain qualifications in finite dimension. For an unbounded self-adjoint operator, the spectral theorem instead gives a spectral probability measure; a finite mean requires an integrable absolute first moment, and finite variance requires a finite second moment. Writing $⟨\psi |A|\psi ⟩$ as an operator inner product additionally requires $|\psi ⟩$ in the operator domain of $A$. The quadratic form of a positive quantum Hamiltonian explains the useful wider form-domain version.

For equal finite-dimensional closed subspaces of a Hilbert space $T,S$, positivity of $\cos {\theta }_{T,S}$ makes $P_S{|}_T$ invertible. The displayed reconstruction is the unique element of $T$ whose orthogonal projection onto $S$ matches that of $f$. It is the oblique projection onto $T$ along $S^{\perp }$. Its operator norm is the value of the secant function at the directed subspace angle, and its error is at most that secant function value times the best orthogonal projection error. The lower error bound follows from the Pythagorean identity. Zero-dimensional cases are handled directly.

The finite element method replaces an infinite-dimensional weak formulation by finitely many unknowns attached to a finite element mesh. Simple local functions, often piecewise polynomial functions, join to form a global approximation. The attraction is that each finite element can be integrated separately, while the assembled equations enforce a global balance. A piecewise-linear hat function only meets its immediate neighbors, so a large problem can have a sparse matrix rather than a dense one. Gilbert Strang's MIT lecture develops this connection from the one-dimensional weak formulation through local hats to the stiffness matrix and load integrals.^[ref]

For the conforming Galerkin method, let $V$ be a real Hilbert space, $a$ a bounded coercive bilinear form, and $\ell$ a bounded linear functional. Given a conforming finite element space $V_h\subset V$, the continuous and discrete problems areWriting $u_h={\sum }_j{c}_j{\varphi }_j$ in a basis of $V_h$ gives the linear systemHere $K$ is the stiffness matrix and $F$ is the vector of loads tested against the basis functions. The Galerkin method uses the same trial and test space in this construction. Finite element methods also use the Petrov-Galerkin method, with different trial and test spaces; testing against the trial space is therefore a property of this construction, rather than a definition of every finite element method.

The finite element solution of the one-dimensional Poisson equation below gives every entry explicitly. Refining the finite element mesh then improves the approximation by adding local degrees of freedom. For a symmetric coercive bilinear form, Galerkin orthogonality makes the discrete solution the best approximation in the energy norm; the geometry of orthogonal projection explains why this local construction yields a globally meaningful answer.

An orthonormal set ${\varphi }_0,\dots ,{\varphi }_{m-1}$ in an $L^2$ space defines the integral kernel of the orthogonal projection onto its span by this sum. It satisfies $\int K(x,y)K(y,z)d\mu (y)=K(x,z)$ and $\int K(x,x)d\mu (x)=m$. Its evaluation matrices are positive semidefinite Gram matrices.

Fitted values are the predictions at the predictor values used to fit a regression model. In linear regression with an explicit regression intercept,Here $X$ is the slope design matrix, and $r$ is the regression residual vector. The fitted values estimate the regression function at the observed predictors; they need not equal the observed responses, which may also contain random errors. Prediction at a new predictor vector $x_{\ast }$ uses $\hat{\alpha }+x_{\ast }^T\hat{\beta }$ with the same centering and scaling used in fitting. A prediction interval addresses uncertainty in a new response rather than just displaying this point prediction.

For the five constructed observations in linear regression, ordinary least squares gives fitted values $(4/5,9/5,14/5,19/5,24/5)$. The ridge regression fit at $\lambda =10$ instead gives $(9/5,23/10,14/5,33/10,19/5)$. Both fitted vectors have sample mean $14/5$ because the regression intercept is unpenalized; their regression residuals differ and their residual sums of squares are $4/5$ and $33/10$, respectively.

For ordinary least squares, the fitted values are the orthogonal projection of $Y$ onto the column space of the complete design matrix, including its regression intercept column. They remain unique even in rank-deficient ordinary least squares, although the regression coefficients can be nonunique. The regression residuals are then orthogonal to every column.^[ref] With centered slopes, ridge regression instead satisfies $X_c^{T}r=\lambda {\hat{\beta }}_{\lambda }$, so the regression residuals need not be orthogonal to the slope columns. Their sum is still zero because the regression intercept remains free.

For fixed predictors and scales and a fixed positive penalty, ridge regression gives a linear smoother,For $\lambda >0$ with at least one nonzero slope direction, this matrix is not an orthogonal projection matrix: applying it twice shrinks that direction again. Its effective degrees of freedom are $1+{\sum }_{j=1}^r{d}_j^2/(d_j^2+\lambda )$, where $d_j$ are the positive singular values of $X_c$; the extra $1$ counts the free regression intercept. If the penalty is selected using $Y$, the resulting fitting rule need not be a linear smoother, since its chosen $H_{\lambda }$ also depends on $Y$.

For a Fredholm operator between Hilbert spaces, the index is the difference of the finite dimensions of its kernel and cokernel. Since the range is closed, its cokernel identifies with $\ker T^{\ast }$ by orthogonal projection. An invertible operator has index zero. Conversely, an index-zero Fredholm operator becomes invertible after adding a finite-rank operator: choose an isomorphism $J:\ker T\to \ker T^{\ast }$, extend it by zero on $(\ker T{)}^{\perp }$, and observe that $T+J$ maps the two summands bijectively onto $\mathrm{ran}T\oplus \ker T^{\ast }$.

Let $V_h\subset V$ be a conforming finite element space. Suppose the exact solution satisfies $a(u,v)=\ell (v)$ for every $v\in V$ and its Galerkin method approximation satisfies $a(u_h,v_h)=\ell (v_h)$ for every $v_h\in V_h$, using the same bilinear form and linear functional. Subtracting the two identities gives Galerkin orthogonality:This identity itself does not require symmetry. Its interpretation as an orthogonal projection in an energy norm does require a bounded symmetric coercive bilinear form: then $a$ is an inner product, $e=u-u_h$ is perpendicular to every available discrete correction, andConsequently $u_h$ is the unique closest point in $V_h$ in the energy norm. This Pythagorean theorem in a function space explains the beauty of the Galerkin method: integrated local equations compute a global best approximation. The Céa lemma supplies the corresponding estimate in a chosen reference norm.

For the finite element solution of the one-dimensional Poisson equation, $a(w,v)={\int }_0^1{w}^{\prime }{v}^{\prime }dx$ and $\Vert w{\Vert }_a=\Vert w^{\prime }{\Vert }_{L^2(0,1)}$. Testing the error against the $i$th piecewise-linear hat function on a uniform mesh givesIn this model $e$ vanishes at every node, so each of these integrals is individually zero, although $e$ is positive inside every element. Galerkin orthogonality controls the derivative-weighted balance; it does not say that the pointwise error vanishes.

The exact identity assumes conformity and exact evaluation of the continuous forms, including any numerical integration that is used. If assembly instead uses $a_h$ and ${\ell }_h$, subtraction gives the consistency defectThus inexact numerical integration need not preserve the projection. In the Petrov-Galerkin method the error is tested against a different test space, and for a nonsymmetric bilinear form the displayed identity need not describe perpendicularity in any energy norm.

Draw a $k\times D$ random matrix $G$ whose entries are independent random variables with the standard normal distribution, and setThe factor $1/\sqrt{k}$ makes the expected value of the squared L2 norm equal to the original squared L2 norm. This Gaussian Johnson–Lindenstrauss transform is a linear map, generally not an orthogonal projection: its rows are not required to be orthogonal vectors. ^[ref]

For the fixed finite set $S$, let $0<\delta <1$ be an allowed failure probability, and defineWith $\log$ denoting the natural logarithm, a sufficient integer target dimension isThen one draw of this specified random matrix satisfies all the squared Euclidean distance bounds simultaneously with probability at least $1-\delta$. Since $c(\epsilon )\ge {\epsilon }^2/12$, the simpler sufficient bound $k\ge \lceil 12{\epsilon }^{-2}\log (n(n-1)/\delta )\rceil$ also works. Both are sufficient bounds for this argument; they do not claim the smallest possible target dimension.

This is a probability statement about a map drawn independently of the fixed finite set. It implies existence because the success probability is positive, but permits failed draws. The set cannot be chosen after inspecting the map with this same guarantee. In particular, no single draw preserves every possible finite set in the ambient Euclidean space when $k<D$.

For $m=Au+\eta$, with independent $u\sim \mathcal{N}(0,I_d)$ and $\eta \sim \mathcal{N}(0,{\delta }^2{I}_k)$, the multivariate normal distribution of the Bayesian posterior has posterior mean $b_{\delta }=(A^{T}A+{\delta }^2{I}_d{)}^{-1}{A}^{T}m$ and covariance matrix $C_{\delta }={\delta }^2(A^{T}A+{\delta }^2{I}_d{)}^{-1}$. The singular value decomposition shows that the variance on a right singular vector of nonzero singular value $s$ is ${\delta }^2/(s^2+{\delta }^2)$, while the variance along the null space is one. Thus, for fixed data, the mean tends to the minimum-norm least-squares solution and the covariance tends to the orthogonal projection onto the null space. Unobserved components retain their prior distribution even as the noise vanishes.