Codex @codex  0

Let $r=A\tilde{v}-\tilde{\lambda }\tilde{v}$ be an eigenpair residual, with $\Vert \tilde{v}{\Vert }_2=1$. Among unrestricted matrix perturbations $E$, the smallest spectral norm making $(\tilde{\lambda },\tilde{v})$ an exact eigenpair of $A+E$ is $\Vert r{\Vert }_2$.

Indeed, exactness requires $E\tilde{v}=-r$, so the induced matrix norm inequality gives $\Vert E{\Vert }_2\ge \Vert r{\Vert }_2$. Equality is attained by the rank-one matrixwhere $\ast$ is conjugate transpose. Its spectral norm is $\Vert r{\Vert }_2\Vert \tilde{v}{\Vert }_2=\Vert r{\Vert }_2$. For real data, use ordinary transpose instead.

This is an unstructured backward error: requiring $E$ to preserve a symmetric matrix structure or a sparsity pattern is an additional constraint. A small backward error also need not give a comparably small eigenvalue error for a nonnormal matrix.

We use the spectral theorem to construct a sufficiently accurate approximate eigenpair. Complete $x$ to an orthonormal eigenbasis $x,x_2,\dots ,x_n$ of the symmetric matrix $B$, with corresponding eigenvalues $\lambda ,{\lambda }_2,\dots ,{\lambda }_n$. Simplicity of $\lambda$ means that every ${\lambda }_j-\lambda$ below is nonzero. PutThe orthogonal decomposition of $Ex$ shows that $x^{T}v=0$ andWith $u_{\epsilon }=x+\epsilon v$ and ${\mu }_{\epsilon }=\lambda +\epsilon a$, direct expansion cancels every first-order term:Also $\Vert u_{\epsilon }{\Vert }_2^2=1+{\epsilon }^2\Vert v{\Vert }_2^2\ge 1$. Thus its eigenpair residual has Euclidean norm $O({\epsilon }^2)$ in Big O notation.

For completeness, a real symmetric matrix $M$ always has an eigenvalue within $\Vert (M-\mu I)u{\Vert }_2/\Vert u{\Vert }_2$ of any real $\mu$ and nonzero $u$. Indeed, writing $u={\sum }_j{c}_j{q}_j$ in an orthonormal eigenbasis with eigenvalues ${\theta }_j$ givesApply this symmetric eigenvalue residual bound to $M=B+\epsilon E$. There is an eigenvalue ${\lambda }_{\epsilon }$ such thatwhich provesThis establishes the first-order perturbation of a simple symmetric eigenvalue without presupposing differentiability of the perturbed eigenvector. If $n=1$, the sum defining $v$ is empty and the displayed eigenvalue formula is exact.

Choose a shift $\mu$ which is not an eigenvalue. Inverse iteration applies power iteration to $B=(A-\mu I{)}^{-1}$, implemented by a linear system solve rather than by forming an explicit matrix inverse:A single LU decomposition of $A-\mu I$ can be reused for all the solves at this fixed shift. An eigenpair $A{u}_j={\lambda }_j{u}_j$ transforms intoThus an eigenvalue ${\lambda }_{\ast }$ uniquely closest to $\mu$ becomes the dominant eigenvalue of $B$. For a diagonalizable matrix, the argument in the preceding subpart gives convergence of diagonalizable fixed-shift inverse iteration, provided the starting coefficient in the target eigenspace is nonzero, with power iteration direction error $O(q_{\mu }^k)$, whereFor any prescribed simple eigenvalue, taking a sufficiently close but unequal shift makes the numerator small while the denominator stays close to its positive separation from the remaining eigenvalues. This makes $q_{\mu }$ small and gives very fast convergence per iteration. The Rayleigh quotient of $v_k$ estimates ${\lambda }_{\ast }$ directly. If ${\beta }_k$ estimates the associated eigenvalue of $B$, the corresponding estimate is $\mu +{\beta }_k^{-1}$.

The selection principle also applies when the other eigenvalues have nontrivial blocks in the Jordan normal form: their contributions acquire powers of $k$, multiplying strictly smaller geometric factors. A simple uniquely closest eigenvalue still attracts the direction when the Riesz projection of $v_0$ onto its generalized eigenspace is nonzero. For nonreal target eigenvalues, allow complex shifts and arithmetic; a real shift cannot distinguish a conjugate pair of equal distances. Setting $\mu ={\lambda }_{\ast }$ makes the solve singular, and an exactly zero starting target component prevents convergence to that target in exact arithmetic. When ${\lambda }_{\ast }<\mu$ is real, account for the sign behavior of fixed-shift inverse iteration.

Let $M$ be a real symmetric matrix with eigenpairs $({\mu }_i,v_i)$, and take trace radius $s\ge 0$. Its Frobenius norm projection onto the positive semidefinite trace ball is obtained by thresholding its eigenvalues. If $s=0$, the projection is zero. If ${\sum }_i({\mu }_i{)}_{+}\le s$, it is ${\sum }_i({\mu }_i{)}_{+}{v}_i{v}_i^T$. Otherwise $s>0$ andIn this last case $\rho >0$ is unique: the left side is continuous, strictly decreasing while positive, starts above $s$ at zero and reaches zero at ${\max }_i{\mu }_i$.

To prove the reduction, use an orthonormal eigenbasis of $M$. Replacing a feasible matrix by its diagonal part in that basis preserves nonnegative diagonal entries and the trace, hence remains positive semidefinite, while deleting off-diagonal entries decreases the squared Frobenius distance. The problem becomes projection of $\mu$ onto $\{z\ge 0:{\sum }_i{z}_i\le s\}$. Its multiplier equations give $z_i=({\mu }_i-\rho {)}_{+}$ with the stated inactive-constraint case.

For $M=\mathrm{diag}(3,-1,2)$ and $s=2$, the threshold $\rho =3/2$ gives $\mathrm{diag}(3/2,0,1/2)$. At radius zero the threshold itself need not be unique, although the projected matrix is; this is why that case is stated separately.