Relationship between the derivative of a matrix and its eigenvalues 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
$$-\frac{ik}{4\pi} I + \mathrm{diag}[a_1,\ldots,a_n]+A(k)$$ where $I$ is the identity matrix, $\alpha_i$ are real constants and
$$ [A(k)]_{jj}=0 $$ 
$$ [A(k)]_{jl}= -\frac{e^{ik|y_j-y_l|}}{4\pi|y_j-y_l|}\quad \text{for } j\neq l $$
with each $y_i$ a point in $\mathbb{R}^3$.
 A: The relation you are looking for is in the article "On Eigenvalues of Matrices Dependent on Parameter" by P.Lancaster (1964), theorem 5.
It states, that for any matrix $A$:
$$
\frac{\mathrm{d} \lambda^{(j)}_t}{\mathrm{d} t} = \frac{ y_t^{(j)T} A'_t x^{(j)}_t }{ y_t^{(j)T} x^{(j)}_t }
$$
For real parameter $t$ and $y^{(j)}_t$, $x^{(j)}_t$ being left and right eigenvectors of $A_t$ corresponding to j-th eigenvalue $\lambda^{(j)}_t$.
If $A_t$ is additionally symmetric, then $y^{(j)}_t = x^{(j)}_t$ and it can be chosen real-valued. If also $A'_t$ is positive definitive, then the right side of equation is strictly positive and so is the derivative of eigenvalue.
In your particular case in the book you mentioned $k$ is set: $k=i\chi$ and $\chi$ is real-valued parameter.
A: 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.  
