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Nov 13, 2019 at 21:41 history edited Ludwig CC BY-SA 4.0
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Nov 13, 2019 at 18:11 history edited Ludwig CC BY-SA 4.0
I rewrote the entire question in order to improve clarity
Nov 13, 2019 at 16:38 history edited Ludwig CC BY-SA 4.0
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Nov 13, 2019 at 6:55 history edited Ludwig CC BY-SA 4.0
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Nov 13, 2019 at 6:32 comment added Ludwig @ChristianRemling yes it is not symmetric (there was a typo). However $g(A,t)$ is similar to a symmetric matrix.
Nov 13, 2019 at 6:26 history edited Ludwig CC BY-SA 4.0
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Nov 12, 2019 at 23:52 comment added Christian Remling I don't think $g(A,t)$ is still symmetric.
Nov 12, 2019 at 21:42 comment added Ludwig @IlyaBogdanov: yes thanks! I fixed the typo. I enumerate the eigenvalues in an increasing order: $\lambda_1$ is the smallest eigenvalue of $g(A,t)$. Basically, I wonder whether there are no crossings between the eigenvalues.
Nov 12, 2019 at 21:39 history edited Ludwig CC BY-SA 4.0
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Nov 12, 2019 at 20:58 comment added Ilya Bogdanov 1) Did you forget to take the inverse when defining $\Sigma(t)$? 2) How do you enumerate the eigenvalues? What is $\lambda_1$? Do you merely ask whether teo eigenvalues may become equal?
Nov 12, 2019 at 20:29 history asked Ludwig CC BY-SA 4.0