0
$\begingroup$

I have a $12 \times 12$ matrix $K$ depending on 2 parameters, $k$ and $\beta$ ($k=0.1,0.2,0.3,0.4$ and $\beta=0.1,0.2,0.3$). The analytical expressions of its eigenvalues are too cumbersome to evaluate. Hence, I would like to evaluate numerically its minimum eigenvalue as $k$ and $\beta$ vary in their ranges, and plot the results.

How can I do this?

K={{-2 + 2/k + (
   50000000000 Tan[β]^2)/(250000 + 250000 Tan[β]^2)^(
   3/2), -(2/k), 
  1 - (25000000000 Tan[β]^2)/(250000 + 250000 Tan[β]^2)^(
   3/2), 1 - (
   25000000000 Tan[β]^2)/(250000 + 250000 Tan[β]^2)^(3/2),
   0, 0, 0, 
  0, -((25000000000 Tan[β])/(250000 + 250000 Tan[β]^2)^(
   3/2)), (25000000000 Tan[β])/(250000 + 
    250000 Tan[β]^2)^(3/2), 0, 
  0}, {-(2/k), -2 + 2/k + (
   50000000000 Tan[β]^2)/(250000 + 250000 Tan[β]^2)^(3/2),
   0, 0, 1 - (
   25000000000 Tan[β]^2)/(250000 + 250000 Tan[β]^2)^(3/2),
   1 - (25000000000 Tan[β]^2)/(250000 + 250000 Tan[β]^2)^(
   3/2), 0, 0, 0, 
  0, -((25000000000 Tan[β])/(250000 + 250000 Tan[β]^2)^(
   3/2)), (25000000000 Tan[β])/(250000 + 
    250000 Tan[β]^2)^(
  3/2)}, {1 - (
   25000000000 Tan[β]^2)/(250000 + 250000 Tan[β]^2)^(3/2),
   0, -(1/2) - 1/(-2 + k) + 10/Sqrt[Tan[β]^2] + (
   25000000000 Tan[β]^2)/(250000 + 250000 Tan[β]^2)^(
   3/2), -(1/2) - 10/Sqrt[Tan[β]^2], 0, 
  1/(-2 + k), -((
   25000000000 Tan[β])/(250000 + 250000 Tan[β]^2)^(3/2)), 
  0, (25000000000 Tan[β])/(250000 + 250000 Tan[β]^2)^(
  3/2), 0, 0, 
  0}, {1 - (
   25000000000 Tan[β]^2)/(250000 + 250000 Tan[β]^2)^(3/2),
   0, -(1/2) - 10/Sqrt[Tan[β]^2], -(1/2) - 1/(-2 + k) + 10/Sqrt[
   Tan[β]^2] + (
   25000000000 Tan[β]^2)/(250000 + 250000 Tan[β]^2)^(3/2),
   1/(-2 + k), 0, (
  25000000000 Tan[β])/(250000 + 250000 Tan[β]^2)^(3/2), 0,
   0, -((25000000000 Tan[β])/(250000 + 250000 Tan[β]^2)^(
   3/2)), 0, 0}, {0, 
  1 - (25000000000 Tan[β]^2)/(250000 + 250000 Tan[β]^2)^(
   3/2), 0, 
  1/(-2 + k), -(1/2) - 1/(-2 + k) + 10/Sqrt[Tan[β]^2] + (
   25000000000 Tan[β]^2)/(250000 + 250000 Tan[β]^2)^(
   3/2), -(1/2) - 10/Sqrt[Tan[β]^2], 
  0, -((25000000000 Tan[β])/(250000 + 250000 Tan[β]^2)^(
   3/2)), 0, 0, (
  25000000000 Tan[β])/(250000 + 250000 Tan[β]^2)^(3/2), 
  0}, {0, 1 - (
   25000000000 Tan[β]^2)/(250000 + 250000 Tan[β]^2)^(3/2),
   1/(-2 + k), 
  0, -(1/2) - 10/Sqrt[Tan[β]^2], -(1/2) - 1/(-2 + k) + 10/Sqrt[
   Tan[β]^2] + (
   25000000000 Tan[β]^2)/(250000 + 250000 Tan[β]^2)^(3/2),
   0, (25000000000 Tan[β])/(250000 + 250000 Tan[β]^2)^(
  3/2), 0, 0, 
  0, -((25000000000 Tan[β])/(250000 + 250000 Tan[β]^2)^(
   3/2))}, {0, 
  0, -((25000000000 Tan[β])/(250000 + 250000 Tan[β]^2)^(
   3/2)), (25000000000 Tan[β])/(250000 + 
    250000 Tan[β]^2)^(3/2), 0, 
  0, -2 + 2/k + 20/Sqrt[k^2] + 
   50000000000/(250000 + 250000 Tan[β]^2)^(3/2), -(2/k) - 20/
   Sqrt[k^2], 1 - 25000000000/(250000 + 250000 Tan[β]^2)^(3/2), 
  1 - 25000000000/(250000 + 250000 Tan[β]^2)^(3/2), 0, 0}, {0, 
  0, 0, 0, -((
   25000000000 Tan[β])/(250000 + 250000 Tan[β]^2)^(
   3/2)), (25000000000 Tan[β])/(250000 + 
    250000 Tan[β]^2)^(
  3/2), -(2/k) - 20/Sqrt[k^2], -2 + 2/k + 20/Sqrt[k^2] + 
   50000000000/(250000 + 250000 Tan[β]^2)^(3/2), 0, 0, 
  1 - 25000000000/(250000 + 250000 Tan[β]^2)^(3/2), 
  1 - 25000000000/(250000 + 250000 Tan[β]^2)^(3/2)}, {-((
   25000000000 Tan[β])/(250000 + 250000 Tan[β]^2)^(3/2)), 
  0, (25000000000 Tan[β])/(250000 + 250000 Tan[β]^2)^(
  3/2), 0, 0, 0, 
  1 - 25000000000/(250000 + 250000 Tan[β]^2)^(3/2), 
  0, -(1/2) + 10000/Sqrt[(1000 - 500 k)^2] - 1/(-2 + k) + 
   25000000000/(250000 + 250000 Tan[β]^2)^(3/2), -(1/2), 0, 
  1/(-2 + k) + (
   10000 (-1000 + 500 k))/((1000 - 500 k) Sqrt[(1000 - 500 k)^2])}, {(
  25000000000 Tan[β])/(250000 + 250000 Tan[β]^2)^(3/2), 0,
   0, -((25000000000 Tan[β])/(250000 + 250000 Tan[β]^2)^(
   3/2)), 0, 0, 
  1 - 25000000000/(250000 + 250000 Tan[β]^2)^(3/2), 
  0, -(1/2), -(1/2) + 10000/Sqrt[(1000 - 500 k)^2] - 1/(-2 + k) + 
   25000000000/(250000 + 250000 Tan[β]^2)^(3/2), 
  1/(-2 + k) + (
   10000 (-1000 + 500 k))/((1000 - 500 k) Sqrt[(1000 - 500 k)^2]), 
  0}, {0, -((
   25000000000 Tan[β])/(250000 + 250000 Tan[β]^2)^(3/2)), 
  0, 0, (25000000000 Tan[β])/(250000 + 250000 Tan[β]^2)^(
  3/2), 0, 0, 1 - 25000000000/(250000 + 250000 Tan[β]^2)^(3/2), 
  0, 1/(-2 + k) + (
   10000 (-1000 + 500 k))/((1000 - 500 k) Sqrt[(1000 - 500 k)^2]), -(
    1/2) - 1/(-2 + k) + (10000 (-1000 + 500 k)^2)/((1000 - 500 k)^2)^(
   3/2) + 25000000000/(250000 + 250000 Tan[β]^2)^(3/2), -(1/
   2)}, {0, (
  25000000000 Tan[β])/(250000 + 250000 Tan[β]^2)^(3/2), 0,
   0, 0, -((
   25000000000 Tan[β])/(250000 + 250000 Tan[β]^2)^(3/2)), 
  0, 1 - 25000000000/(250000 + 250000 Tan[β]^2)^(3/2), 
  1/(-2 + k) + (
   10000 (-1000 + 500 k))/((1000 - 500 k) Sqrt[(1000 - 500 k)^2]), 
  0, -(1/2), -(1/2) - 1/(-2 + k) + (
   10000 (-1000 + 500 k)^2)/((1000 - 500 k)^2)^(3/2) + 
   25000000000/(250000 + 250000 Tan[β]^2)^(3/2)}}
Improve this question
$\endgroup$
3
  • $\begingroup$ Manipulate[Eigenvalues[mat /. {k -> k1, Beta -> b1}, -1], {k1, 0, 1}, {b1, 0, 1}]? $\endgroup$
    – Michael E2
    Sep 14 at 13:16
  • $\begingroup$ Your matrix seems to have a three-dimensional null-space. Maybe characterize it and project it out? $\endgroup$
    – Roman
    Sep 14 at 19:03
  • $\begingroup$ @Roman yes, it has 3 null eigenvalues...my question is related to the minimum non-null one $\endgroup$
    – Gae P
    Sep 15 at 7:57

1 Answer 1

Reset to default
0
$\begingroup$

I found the solution:

DiscretePlot3D[Min@Chop@Eigenvalues[K /. h -> 500], {k, 0.1, 0.9, 0.1}, {\[Beta], Pi/18, Pi/2.5, Pi/36}, PlotTheme -> "Detailed"]
Improve this answer
$\endgroup$
2
  • $\begingroup$ Further request: how can I use DiscretePlot3D such that null points will not be shown? $\endgroup$
    – Gae P
    Sep 14 at 15:58
  • 1
    $\begingroup$ this “further request” will not readily reach anyone whilst buried in the comments of your own answer. It might get slightly more attention as a comment on your question, but even that can be easily missed. Instead, it sounds like an entirely new question. It would be better, then, to post this “further request” as a new question, linking to this QA for reference. Hope this helps! $\endgroup$
    – CA Trevillian
    Oct 14 at 20:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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