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 $\mathbb{R}^3$.
[Math] Relationship between the derivative of a matrix and its eigenvalues
ca.classical-analysis-and-odeslinear algebramatricesreal-analysis
Best Answer
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.