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 $$ [A(k)]_{jj}=0 $$ $$ [A(k)]_{jl}= \frac{e^{iky_jy_l}}{4\piy_jy_l}\quad \text{for } j\neq l $$ with each $y_i$ a point in $\mathbb{R}^3$.

1$\begingroup$ You need to explain what you mean by a derivative of a matrix, since as defined they are not functions. $\endgroup$– Chris GodsilMar 6, 2013 at 11:53

$\begingroup$ I have a matrix wich depends by a parameter $k$; so I have to derive the entries to respect to $k$. $\endgroup$– MarioMar 6, 2013 at 11:56

$\begingroup$ If I understand well, your matrix is really explicit and of size 3. Have you tried computations, either symbolic with a CAS, or numerical ? That may give you a good idea of what is happening. (Maybe plot the eigenvalues against k, for some particular points y ?) $\endgroup$– Thomas RichardMar 6, 2013 at 12:48

$\begingroup$ My matrix is square of size $n$; I don't have tried any calculation because in the book where I'm studying the author invokes the calculation of the derivative of the matrix, but I don't really understand his reasoning. $\endgroup$– MarioMar 6, 2013 at 13:09

$\begingroup$ What's $i$? Is this $\sqrt{1}$? Then it what sense your derivative is positivedefinite? $\endgroup$– Alexandre EremenkoMar 6, 2013 at 13:35
2 Answers
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 jth eigenvalue $\lambda^{(j)}_t$.
If $A_t$ is additionally symmetric, then $y^{(j)}_t = x^{(j)}_t$ and it can be chosen realvalued. 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 realvalued parameter.

1$\begingroup$ Theorem 5 assumes that the eigenvalues $\lambda_{t}^{(j)}$ are simple. It is possible to generalize this result to multiple eigenvalues? $\endgroup$– TadashiMay 4, 2015 at 20:48
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.