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The implicit iteration $(I+\tau K^{\ast }K)u_{k+1}=u_k+\tau K^{\ast }f$, starting from zero, multiplies each data singular vector by $[1-(1+\tau {\sigma }^2{)}^{-k}]/\sigma$. Its operator norm is at most $\sqrt{k\tau }$.

If $\lambda$ is a dominant integral weight, the maximal proper submodule of $M(\lambda )$ is generated by the simple-root singular vectorsThe generator associated with ${\alpha }_i$ has weight $s_i(\lambda +\rho )-\rho$.

The adjoint operators, orthonormal singular vectors and squared-norm normal equation for a linear inverse problem conditions require Hilbert spaces. For arbitrary normed spaces the Banach-space adjoint takes values in the dual spaces and need not produce this formula. Interpret the given singular value system in Hilbert spaces, with $u_j$ in the unknown space and $v_j$ in the data space, as in the printed identities. Positive singular values tending to zero provide the compact ill-posed case. A general bounded linear operator need not have this property; the identity operator is a well-conditioned counterexample. Components in the kernel must be fixed by a minimum-norm convention, and the unregularized inverse is defined only for data satisfying the corresponding range condition.

Let $\eta =f^{\delta }-f$, so $\Vert \eta \Vert \le \delta$. By linearity, $u_{\delta }^{(k)}-u^{(k)}=R_{1/k}\eta$. The preceding operator norm estimate givesMoreover, $K{R}_{1/k}$ has factors $1-(1+\tau {\sigma }_j^2{)}^{-k}$ on the data singular vectors, and is zero on $\mathcal{R}(K{)}^{\perp }$. Every factor is in $[0,1]$, henceThus, for any fixed $k$, one can take $C=\sqrt{k\tau }$ and $\gamma =1$.

The first constant cannot generally be chosen independently of the iteration count. On the l2 sequence space, take $(Ku{)}_n=u_n/n$, $\tau =1$, and noise $\eta =\delta e_n$. At $k=n^2$ the solution perturbation has normThis disproves a uniform-in-$k$ interpretation of the stated first estimate. For a closed operator range, boundedness of the Moore–Penrose inverse of an operator instead supplies the uniform constant $\Vert K^{†}\Vert$.

Let $H$ be the measured wave scattering response> linear operator, with direct background propagation removed if present. Its time reversal operator is the positive operator $T=H^{\ast }H$. For a normalized emitted signal $h$,The maximizing signal is a right singular vector, equivalently an eigenvector of $T$ with its largest eigenvalue. This is a precise intensity statement independent of a scatterer model.

Under the Born approximation for scalar wave scattering, neglect interactions between point scatterers and writewhere $b_j$ is the receiver response to scatterer $j$, $c_j^{\ast }h$ is its illumination by the source array, and ${\alpha }_j$ is its scattering wave amplitude. Well-resolved point scatterers have approximately orthogonal vectors $b_j$ and $c_j$. In the ideal orthogonal limit,Thus the most reflective, geometrically weighted scatterer gives the largest eigenvalue, and its normalized steering vector $c_j/\Vert c_j\Vert$ gives the corresponding wave focusing signal. If illumination and reception factors are equal for all scatterers, this is precisely the scatterer with largest $|{\alpha }_j{|}^2$. Geometric size by itself is not the quantity being ranked.

Repeated adjoint time reversal acoustics is power iteration on $T$. If its top eigenvalue is simple and the initial signal has a nonzero component along its eigenvector, thenA tied largest eigenvalue leaves a combination in the leading eigenspace, and an initial signal orthogonal to that eigenspace cannot excite it. This is the basis of the DORT method. Mere physical separation is insufficient if the array cannot resolve the scatterers; coherent steering-vector overlap or multiple scattering can mix the modes. Also, if $H$ includes unrestricted homogeneous transmission, its largest eigenvalue need not identify any individual scatterer. The correspondence concerns the resolved wave scattering response>. The ideal correspondence and selective wave focusing are analyzed by Prada and Fink.

The Poincaré-Birkhoff-Witt theorem shows that the weights of the Verma module $M({\omega }_2)$ arewith multiplicities given by the corresponding Kostant partition function.

The Dynkin labels of ${\omega }_2$ are $(0,1)$. Hence the two simple-root singular vectors are $f_1{v}_{{\omega }_2}$, of weight ${\omega }_2-{\alpha }_1$, and $f_2^2{v}_{{\omega }_2}$, of weight ${\omega }_2-2{\alpha }_2$. They generate the Maximal proper submodule of a dominant integral Verma module. Its set of weights is consequently

For ${\mathfrak{sl}}_2$, the Verma module has basisby the Poincaré-Birkhoff-Witt theorem. If $hv=\lambda v$, thenFor $\lambda =-d$ with $d>0$, the coefficient isfor every $r\ge 1$. Thus no $f^{r}v$ with $r>0$ is a singular vector.

Any nonzero submodule contains a weight vector $f^{r}v$ because the $h$-weights are distinct. Applying $e^r$ gives a nonzero multiple of $v$, after which applying powers of $f$ generates all of $M(-d)$. HenceThis is also the negative-highest-weight case of Reducibility of an sl2 Verma module.

For the sl2 Lie algebra, let $v_{\lambda }$ be the highest-weight vector. The Poincaré-Birkhoff-Witt theorem gives the basis $(f^r{v}_{\lambda }{)}_{r\ge 0}$, and the defining Lie brackets implyA positive-degree basis vector is singular exactly when $r=\lambda +1$ is a positive integer. Therefore $M_{\lambda }$ is irreducible when $\lambda \notin {\mathbb{Z}}_{\ge 0}$.

If $\lambda =m\in {\mathbb{Z}}_{\ge 0}$, the vector $f^{m+1}{v}_{\lambda }$ has weight $-m-2$ and generates a submodule isomorphic to $M_{-m-2}$. The latter is irreducible because $-m-2\notin {\mathbb{Z}}_{\ge 0}$. Every nonzero submodule contains a singular vector by the preceding part, and the displayed coefficient shows that this is the only possible proper singular vector. Henceis the unique proper nonzero submodule, as summarized by the Reducibility of an sl2 Verma module.

The weights of $M_{\lambda }$ lie below $\lambda$ in the positive-root order, and every weight space is finite-dimensional. A nonzero submodule $V$ is stable under the Cartan subalgebra, so it is a direct sum of its weight spaces. Choose a maximal weight $\mu$ occurring in $V$. Every positive-root operator would raise its weight; maximality therefore makes it kill any nonzero $v\in V_{\mu }$. Thus $v$ is a singular vector.

The Casimir element is central and acts throughout $M_{\lambda }$ byThe same element acts on the highest-weight vector $v$ of weight $\mu$ byBoth are the action of one operator on the same module, so the scalars agree and