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This seems a counterexample: $c_+=1$ (which you can assume without loss of generality), $n=2$, $A = \begin{bmatrix}2 & -\varepsilon \\\ -\varepsilon & 2\end{bmatrix}$, $x = \begin{bmatrix}1 \\ 1\end{b …

That quantity $s = \frac{|u^Tv|}{\|u\|\|v\|}$ is the inverse of the eigenvalue condition number. The smaller it is, the more sensitive to perturbation the Perron value is. More precisely, any perturbe …

EDIT: just a partial answer that does not settle the question completely. This is a variant of symplectic matrices. Your matrices $M_\mu$ are orthogonal wrt the indefinite scalar product induced by $$ …

[This is an answer to the first version of the question, which asked if $\min \sigma(S(x)) \leq 0 \leq \max \sigma(S(x)) \text{ a.e on } \mathbb{R}^d$ implies $\min \sigma \left( \int u S \right)\leq …

If you have a matrix with a rapidly growing diagonal, you can check alternatively if $DAD$ is diagonally dominant with a positive diagonal, where $D= \operatorname{diag}(1,\rho, \rho^2, \dots, \rho^{n …

This is true for (EDIT: only the Euclidean norm) by the Bauer-Fike theorem, since for a symmetric / normal matrix the eigenvalue matrix is unitary and hence has condition number 1.

In numerical analysis lingo, you are more or less asking if matrix multiplication is backward stable. The answer seems to be no: see Section 3.5 of Higham, Accuracy and stability of numerical algorith …

An interlacing result is in Theorem 4.3.4 of Horn, Johnson, Matrix Analysis (first edition - sorry, I don't have a copy of the second): for each Hermitian $B$, $$ \lambda_k(B) \leq \lambda_{k+1}(B\pm …

Not true in general, as noted by @SergeiIvanov, but true for (element-wise) nonnegative matrices. Note that if $\rho(S) < b$, then $b(bI-S)^{-1}=(I-\frac{S}{b})^{-1}=\sum_{i=0}^\infty \frac{S^i}{b^i} …

I am not sure if you are allowed to change your objective function, but a natural alternative for measuring "rank-1-ness" is $$ \frac{\lambda_1^2}{\lambda_1^2+\lambda_2^2+\dots+\lambda_n^2}. $$ This …

Computing a Cholesky factor is a much easier task than computing the eigendecomposition. For instance, you can do Cholesky while staying in the original field (modulo some square roots on the diagonal …

If you take a Schur form $A=QTQ^T$, then $H=QT(I-T)^{-1}Q^T$, and you can ignore the orthogonal factors $Q$. You might also want to set $N=I-T$, so that $Q^THQ=N^{-1}-I$. Now the problem looks much si …

Just a random idea: The standard method for getting a small part of the spectrum in large and sparse symmetric problems is the restarted Lanczos method. Essentially, you run some iterations of the La …

This subject has been studied in literature --- it turns out that for some class of matrices (such as M-matrices) you can obtain componentwise accurate solutions. A good starting point is the section …

In general, there is no simple way to relate the eigenvalues of a sum to the eigenvalues of the original matrices. What you can do computationally is switching to iterative algorithms (Arnoldi method …