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Starting from zero, the Landweber iteration isFor $K\ne 0$, take the step size $0<\tau <2/\Vert K{\Vert }^2$. Iterating the linear update givesand its Landweber spectral filter in a singular system of a compact operator isThe strict upper step size bound makes $|1-\tau {\sigma }_j^2|<1$ for every positive singular value. A common more restrictive choice is $0<\tau \le \Vert K{\Vert }^{-2}$, giving nonnegative damping factors. Each finite iterate is bounded and linear; early stopping of Landweber iteration controls the amplification of noise as smaller singular values are progressively inverted. If $K=0$, every iterate is zero for any positive step size.

The singular value decomposition of $B$ is $B=U\Sigma V^{†}$, where $U,V$ are unitary and $\Sigma =\mathrm{diag}({\sigma }_1,\dots ,{\sigma }_n)$, ${\sigma }_1\ge \cdots \ge {\sigma }_n>0$ since $B$ is invertible. Its right singular vector $v_j$ and left singular vector $u_j$ are the corresponding unit columns of $V$ and $U$, withEquivalently $B^{†}B{v}_j={\sigma }_j^2{v}_j$ and $B{B}^{†}{u}_j={\sigma }_j^2{u}_j$. The positive numbers ${\sigma }_j$ are the singular values. Degenerate singular subspaces allow any orthonormal choice of paired vectors.

Write $y={\psi }_s$ and retain the Landweber relaxation parameter $\tau$. The printed $\Vert A\Vert <2$ allows the fixed choice $\tau =1/2$, since then $\tau \Vert A{\Vert }^2<2$. It does not justify a unit step: the scalar linear operator $A=3/2$ has norm below $2$, but unit-step error is multiplied by $1-|A{|}^2=-5/4$ and diverges. We use $0<\tau <2/\Vert A{\Vert }^2$ throughout.

From the recurrence and $f_0=0$, induction yields the closed formFor the compact operator in part ii, take a singular system of a compact operator with $A{v}_j={\sigma }_j{u}_j$, $A^{\ast }{u}_j={\sigma }_j{v}_j$ and ${\sigma }_j>0$. Thus ${\sigma }_j^2$ are the eigenvalues of $A^{\ast }A$; this fixes the notation implicit in the printed hint. In the singular component $j$, the recurrence isSumming the geometric series gives the Landweber spectral filterThere is no contribution from data in $\ker A^{\ast }$ or from $\ker A$ in the reconstruction. Starting at zero is what selects the minimum-norm least-squares solution.

For fixed ${\sigma }_j>0$, $|1-\tau {\sigma }_j^2|<1$, so the numerator tends to $1$. The assumption $y\in D(A^{†})$ is the Picard criterionwith an arbitrary additional component in $\ker A^{\ast }$. The squared reconstruction error isEach term tends to zero and is bounded by a summable term from the Picard criterion. The dominated convergence theorem therefore proves $R_{n}y\to A^{†}y$, where $A^{†}$ is the Moore–Penrose inverse of an operator.

For each finite $n$, the regularization of an inverse problem is stable. Set $t=\tau {\sigma }^2\in (0,2)$. The geometric series and $|1-t|\le 1$ implyConsequently the Landweber noise amplification bound in this general step-size convention is $\Vert R_n\Vert \le \sqrt{2\tau n}$. If additionally $\tau \Vert A{\Vert }^2\le 1$, the sharper numerator bound $\min (nt,1)$ gives $\Vert R_n\Vert \le \sqrt{\tau n}$.

To see the regularization parameter directly, put $\alpha =1/(\tau n)$. Modes with ${\sigma }^2\ll \alpha$ have numerator approximately $n\tau {\sigma }^2$, so their inverse coefficient is approximately $\sigma /\alpha$ rather than $1/\sigma$. Each fixed nonzero mode is eventually restored as $\alpha \downarrow 0$. ThusThe finite iterates suppress unstable small singular values; infinitely many iterations remove this suppression. For noisy data $\Vert y^{\delta }-y\Vert \le \delta$, the noise-bias decomposition for linear regularization givesChoosing $n(\delta )\to \infty$ and ${\delta }^{2}n(\delta )\to 0$ proves noisy-data convergence. The reciprocal iteration index is a regularization parameter, and early stopping controls noise amplification.

To first order in $n_{ϵ}$, replace $q$ by $2n_{ϵ}$ and use the known incident field in the outgoing integral. The Born approximation for scalar wave scattering givesBoth the $n_{ϵ}^2$ contrast term and the internal-field correction are second order for fixed geometry in the perturbative regime established in part i.

Specify the data space before taking an adjoint operator. For example, let $X=L^2(B)$ and $Y=L^2(\Gamma )$, where $\Gamma$ is a bounded measurement surface separated from $B$. Define the linear operatorFor bounded incident field this is a Hilbert-Schmidt operator, hence a compact operator. The equation is $A{n}_{ϵ}=y$, with $y={\psi }_s{|}_{\Gamma }$. Other sampling geometries give corresponding data spaces and weights; the paper does not specify one. Using the usual complex $L^2$ inner products, its adjoint operator isThe complex conjugations are required by the adjoint operator identity, not by wave reciprocity alone.

The Landweber iteration starts from $f_0=0$ and applies gradient descent to $\frac{1}{2}\Vert Af-y{\Vert }^2$:Each step back-propagates the data residual. The Landweber relaxation parameter controls stability, and early stopping of Landweber iteration prevents small singular values from amplifying measurement errors. For an explicitly real-valued index contrast, use the real Hilbert space structure and replace $A^{\ast }$ by $\mathrm{Re}{A}^{\ast }$ in this update. Additional sign or support constraints require corresponding projections; none are assumed here.

At each fixed time, the maximum over nonzero initial conditions is the optimal energy amplification of a linear system:Write $a=e^{{\lambda }_{1}t}$, $d=e^{{\lambda }_{2}t}$ and $b=(a-d)/({\lambda }_1-{\lambda }_2)$. The symmetric matrix $A^{T}A$ has trace $\tau =a^2+b^2+d^2$ and determinant $a^2{d}^2$. Its largest eigenvalue isAn optimal initial condition is a right singular vector of $A(t)$ associated with its largest singular value.

If ${\lambda }_2>{\lambda }_1$, let $\alpha =1/({\lambda }_2-{\lambda }_1)$. Then $A(t)/e^{{\lambda }_{2}t}\to \left(\begin{array}{cc}0& 0\\ \alpha & 1\end{array}\right)$, so the precise long-time asymptotic statement isThe Cauchy-Schwarz inequality shows that the unit initial conditions achieving this leading factor areThis is an adjoint eigenvector, since $L^T(\alpha ,1{)}^T={\lambda }_2(\alpha ,1{)}^T$. It differs from the right eigenvector $v_2=(0,1{)}^T$: the initial condition maximizes the projection onto the slow mode, whereas the eventual state aligns with that right eigenvector. For ${\lambda }_2<0$, even the optimal energy ultimately decays to zero; the prefactor describes enhanced excitation of the slow mode.

On each left singular vector $u_j$, use $A^{\ast }{u}_j={\sigma }_j{v}_j$ and $(I-A^{\ast }A)v_j=(1-{\sigma }_j^2)v_j$. The finite geometric series givesThus with $\alpha =1/n$, the dimensionless Landweber spectral filter isThe full inverse-coefficient convention instead uses $[1-(1-{\sigma }^2{)}^{1/\alpha }]/\sigma$. With general step $\tau$, replace ${\sigma }^2$ inside the power by $\tau {\sigma }^2$.

In the normalized problem, $0\le f_n(\sigma )\le \min (n{\sigma }^2,1)$, so the inverse coefficient is at most $\min (n\sigma ,1/\sigma )\le \sqrt{n}$. This verifies directly why finite iteration is stable and why admitting progressively smaller singular values eventually amplifies noise.

The maximum of $E(t)/E(0)$ is the square of the largest singular value of $A(t)$, hence the largest eigenvalue $\lambda$ of $A(t{)}^{t}A(t)$. Its determinant is one and its trace givesFor $0<\beta <1$, this is largest when $|\sin \Gamma t|=1$, namely $\Gamma t=\pi /2$ modulo $\pi$. ThenAt such a time $A(t)$ is off diagonal, and the maximizing initial condition is $\boxed{x(0)=0}$ with $y(0)\ne 0$. For negative $\beta$, the axes interchange and the formula uses $|\beta |$.

For an orthogonal $R$, unitary invariance of the Hilbert-Schmidt norm givesThe product $L_2^{\ast }{L}_1$ is trace-class. Its polar decomposition of a bounded operator and trace duality givewhere $({\sigma }_k)$ are its singular values. Taking the infimum proves

For factorizations $C_i=L_i{L}_i^{\ast }$ and $C_j=L_j{L}_j^{\ast }$, the Procrustes distance between covariance operators is the following infimum over unitary operators $R$:Unitary invariance of the Hilbert-Schmidt norm givesThe polar decomposition of a bounded operator and trace duality implywhere the last equality expresses the trace norm as the sum of the singular values. Taking the infimum proves

The HHL algorithm requires coherent, efficient and repeatable preparation of the normalized state $|b⟩$, normally through a known preparation circuit and its inverse; possession of a single unknown physical specimen does not supply that access. The component of $b$ on any discarded or unresolved small-eigenvalue subspace must also be negligible. Here $U$ is a unitary operator, so it is invertible and all its singular values equal one, giving condition number $\kappa =1$.

Standard HHL is stated for a Hermitian matrix with an efficient sparse-access or block encoding oracle. A non-Hermitian $U$ can be embedded in the Hermitian block matrixpart (a) supplies efficient access to $U^{†}$. With inverse-polynomial target precision, $m=O(\log n)$ phase bits, and an efficient preparation oracle for $|b⟩$, the runtime is $\mathrm{poly}(n)$. The output is the normalized quantum state proportional to the solution $x$, rather than a classical list of all its amplitudes.

WriteThe extremal squared amplification factors are the largest and smallest singular values of $A$ squared, equivalently the eigenvalues $G$ ofSince its trace is $a^2+b^2+c^2$ and its determinant is $a^2{c}^2$, the requested quadratic is

For a biregular graph of edge density of a bipartite graph $\gamma$, small rectangular discrepancy, a box norm fourth power close to ${\gamma }^4$, and a small second singular value of the normalized adjacency operator of a bipartite graph are equivalent notions of quasirandomness, with quantitative bounds independent of the part sizes. Centering the adjacency function removes the constant singular component.

A unit right singular vector of a matrix or compact operator $A$ is an eigenvector of $A^{\ast }A$ with eigenvalue ${\sigma }_j^2$. For ${\sigma }_j>0$, $u_j=A{v}_j/{\sigma }_j$ is the corresponding left singular vector. The largest singular value is the operator norm of $A$, attained on its corresponding right singular vectors.

This range condition puts the exact solution in the operator range of the adjoint operator. It controls the small-singular value components of the solution and supplies convergence rates for spectral regularization methods. It is the quadratic-penalty case of a source condition in variational regularization.

If $\Vert f(x){\Vert }_{\mathrm{op}}\le 1$ and $\Vert f{\Vert }_{U^2}^4\ge cn>0$, select the singular values ${\lambda }_p\ge \sqrt{c/2}$ of the matrix Fourier blocks. Their weighted count $m={\sum }_p{d}_{{\rho }_p}$ lies between $cn/2$ and $2n/c$. Scaling unit right singular vectors and their corresponding unit left singular vectors by $\sqrt{d_{{\rho }_p}}$ gives matrices $U(p),V(p)$ satisfying ${\mathbb{E}}_{x}f(x)U(p){\rho }_p(x{)}^{\ast }={\lambda }_{p}V(p)$ and Hilbert-Schmidt inner product orthogonality within each chosen irreducible representation. The Schur averaging of rectangular matrices then gives ${\mathbb{E}}_x\Vert UP(x{)}^{\ast }b{\Vert }_2^2=\Vert b{\Vert }_2^2$ for the concatenated $U$ and the corresponding block diagonal matrix $P(x)$.

The trace norm of a finite-dimensional operator $A$ is the sum of its singular values:For a Hermitian operator $A=A_{+}-A_{-}$, it equals $\mathrm{Tr}{A}_{+}+\mathrm{Tr}{A}_{-}$.

A complex matrix $Z=-Z^T$ can be put by unitary congruence into two-by-two skew blocks and zero blocks; the magnitudes $|z_r|$ are its singular values, each repeated twice. One proof uses the antilinear map $A(v)=Z\bar{v}$. Because $Z^{†}=-\bar{Z}$, $A^2=-Z{Z}^{†}$. On a nonzero eigenspace of $Z{Z}^{†}$ with eigenvalue $s^2$, any unit vector $v$ has orthogonal partner $A(v)/s$, and $A(A(v)/s)=-sv$. These orthonormal pairs give the blocks; the orthogonal complement is invariant, so the procedure repeats. Phases of the paired vectors set the block phases.