Is there any relationship between the derivative of a matrix and its eigenvalues? If, for example, the derivative is strictly positive definite, can I say that the eigenvalues are strictly increasing?
In particular my matrix is
$$-\frac{ik}{4\pi}I+diag[a_1,\ldots,a_n]+A(k)$$ where $I$ is the identity matrix, $\alpha_i$ are real constants and
$$[A(k)]_{jj}=0 $$
$$[A(k)]_{j\,l}= -\frac{e^{ik|y_j-y_l|}}{4\pi|y_j-y_l|}\, j\neq l $$
with $y_i$ point in $\mathbb{R}^3$.
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Look at: Andreas Kriegl, Peter W. Michor, Armin Rainer: Denjoy-Carleman differentiable perturbation of polynomials and unbounded operators. Integral Equations and Operator Theory 71,3 (2011), 407-416, (pdf). There is an overview on available results. |
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Sorry for posting this as an answer, I cannot leave comment. Are you sure your matrix is $$[A(k)]_{j\,l}= -\frac{e^{ik|y_j-y_l|}}{4\pi|y_j-y_l|}\, j\neq l$$ and not $$[A(k)]_{j\,l}= -\frac{e^{ik|y_j-y_l|}-1}{4\pi|y_j-y_l|}\, j\neq l$$ If you want to extend some properties of the derivative of a matrix to its eigenvalue, the eigenvectors have to be independent of the derivation variable. |
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