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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?

Appliqué
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  • 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 $t_{k+1}$. – Hui Zhang Mar 20 '12 at 12:00
  • 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. – Appliqué Mar 20 '12 at 13:07
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    Related question on MO: http://mathoverflow.net/questions/80059/numerically-track-spectrum-curves-of-a-parameter-dependent-linear-operator/85558#85558 – David Ketcheson Mar 20 '12 at 14:17
  • 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. – Geoff Oxberry Mar 20 '12 at 14:45
  • 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). – Appliqué Mar 20 '12 at 15:04
  • If $\Lambda(t)$ is degenerate, $S(t)$ is not unique. If this is a quantum problem, you may want to characterize $A$ by operators that conjugate with $A$, i.e., $CG_A={B|(AB-BA)=0}$. Then you can put constraints on $S$ to map to simultaneous eigenvectors of $CG_A$. – Deathbreath Mar 22 '12 at 16:44

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