MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $\mathbb{R}^3$.

share|cite|improve this question
You need to explain what you mean by a derivative of a matrix, since as defined they are not functions. – Chris Godsil Mar 6 '13 at 11:53
I have a matrix wich depends by a parameter $k$; so I have to derive the entries to respect to $k$. – Mario Mar 6 '13 at 11:56
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 ?) – Thomas Richard Mar 6 '13 at 12:48
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. – Mario Mar 6 '13 at 13:09
What's $i$? Is this $\sqrt{-1}$? Then it what sense your derivative is positive-definite? – Alexandre Eremenko Mar 6 '13 at 13:35

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.

share|cite|improve this 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.

share|cite|improve this answer
Theorem 5 assumes that the eigenvalues $\lambda_{t}^{(j)}$ are simple. It is possible to generalize this result to multiple eigenvalues? – Shamisen May 4 '15 at 20:48

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.