For each $m\in\{1, 2, ..., n\}$, is there a transformation $\phi_m$ that I can apply to a matrix $M\in \mathbb{R}^{n\times n}$ such that the $m^{th}$ largest eigenvalue $\lambda_m$ of $M$ is the smallest eigenvalue $\lambda’_n$ of the matrix $M’ = \phi_m (M)$?

Edited for generality.

  • 1
    $\begingroup$ In absolute value or not ? Because for $M'=-M$ you get to invert the ordering on eigenvalues... If kernel is empty, you can take $M'=M^{-1}$ it transforms $\lambda$ in $\frac 1\lambda$. But in both these examples we do not have $\lambda_1=\lambda'_n$, just some relation between them. Do you want equality for just this one eigenvalue ? $\endgroup$ – zwim May 14 at 23:28
  • $\begingroup$ Note that if you know $\lambda_1$ then $M'=2\lambda_1I-M$ works. Maybe if you are able to have an upper bound $m$ for the eigenvalues, then $M'=2mI-M$ would also suits your needs (for an algorithm for instance). $\endgroup$ – zwim May 14 at 23:40
  • $\begingroup$ Edited post for generality. $\endgroup$ – Josh Payne May 17 at 7:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.