Consider an $n \times n$ real, time-dependent matrix $A(t)$ such that $A(t) = A(t)^{T} > 0$ and $A(t)$ is continuous on $[a,b]$. Then it is posible to specify a matrix $S(t) \in SO(n)$ such that $S(t)A(t)S^{T}(t) = \Lambda(t)$, where $\Lambda$ is a diagonal matrix. For instance, we can obtain such $S(t)$ using Jacobi rotation method.

I'm looking for time-continuous on $[a,b]$ matrices $S(t)$ and $\Lambda(t)$. Lets consider a net $$ a = t_0 < t_1 < \ldots < t_n = b $$ Assume that $S(t_k)$, $\Lambda(t_k)$ are computed. How to receive than $S(t_{k+1})$ and $\Lambda(t_{k+1})$ nearest possible to $S(t_{k})$ and $\Lambda(t_k)$? Which algorithm should I use?

  • $\begingroup$ You can order the eigenvalues from small to big. If your time net is fine enough, the order of your eigenvalues should not change from $t_k$ to $t_{k+1}$. $\endgroup$
    – Hui Zhang
    Mar 20, 2012 at 12:00
  • $\begingroup$ I can't let my net be small enough, $A(t)$ may represent an oscillating ellipsoid such that it's axes may frequently change their relational ratio. But I tried to sort it's eigenvectors: it works, but it is very time consuming. Now I'm trying to deal with Rayleigh Quotient Iteration (RQI) algorithm to find eigenvectors on the next step using their preimages. $\endgroup$
    – Appliqué
    Mar 20, 2012 at 13:07
  • 2
    $\begingroup$ Related question on MO: mathoverflow.net/questions/80059/… $\endgroup$ Mar 20, 2012 at 14:17
  • $\begingroup$ Do you know that your eigenvectors are continuous? Eigenvalues are continuous functions of matrix entries, but eigenvectors are not necessarily continuous, which would present an issue in your problem formulation. $\endgroup$ Mar 20, 2012 at 14:45
  • $\begingroup$ Of course, eigenvectors aren't continuous in general case. But I think if matrix $A(t) = A^{T} > 0$ is continuous they can be choosen continuous because they represent the main axes directions of an ellipsoid $\langle x, A(t) x \rangle \leqslant 1$ (correct me if it isn't true). $\endgroup$
    – Appliqué
    Mar 20, 2012 at 15:04


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