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^{iky_jy_l}}{4\piy_jy_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: DenjoyCarleman differentiable perturbation of polynomials and unbounded operators. Integral Equations and Operator Theory 71,3 (2011), 407416, (pdf). There is an overview on available results. 


Sorry for posting this as an answer, I cannot leave comment. Are you sure your matrix is $$[A(k)]_{j\,l}= \frac{e^{iky_jy_l}}{4\piy_jy_l}\, j\neq l$$ and not $$[A(k)]_{j\,l}= \frac{e^{iky_jy_l}1}{4\piy_jy_l}\, j\neq l$$ 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. 

