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 $\mathbb{R}^3$.
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1$\begingroup$ You need to explain what you mean by a derivative of a matrix, since as defined they are not functions. $\endgroup$– Chris GodsilCommented Mar 6, 2013 at 11:53
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$\begingroup$ I have a matrix wich depends by a parameter $k$; so I have to derive the entries to respect to $k$. $\endgroup$– MarioCommented Mar 6, 2013 at 11:56
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$\begingroup$ 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 RichardCommented Mar 6, 2013 at 12:48
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$\begingroup$ 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. $\endgroup$– MarioCommented Mar 6, 2013 at 13:09
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1$\begingroup$ Page 116 theorem 1.1.4 $\endgroup$– MarioCommented Mar 6, 2013 at 16:25
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2 Answers
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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.
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1$\begingroup$ Theorem 5 assumes that the eigenvalues $\lambda_{t}^{(j)}$ are simple. It is possible to generalize this result to multiple eigenvalues? $\endgroup$– TadashiCommented May 4, 2015 at 20:48
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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.
answered Mar 6, 2013 at 16:31