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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)}}
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  • $\begingroup$ Manipulate[Eigenvalues[mat /. {k -> k1, Beta -> b1}, -1], {k1, 0, 1}, {b1, 0, 1}]? $\endgroup$
    – Michael E2
    Commented Sep 14, 2023 at 13:16
  • $\begingroup$ Your matrix seems to have a three-dimensional null-space. Maybe characterize it and project it out? $\endgroup$
    – Roman
    Commented Sep 14, 2023 at 19:03
  • $\begingroup$ @Roman yes, it has 3 null eigenvalues...my question is related to the minimum non-null one $\endgroup$
    – Gae P
    Commented Sep 15, 2023 at 7:57

1 Answer 1

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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"]
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  • $\begingroup$ Further request: how can I use DiscretePlot3D such that null points will not be shown? $\endgroup$
    – Gae P
    Commented Sep 14, 2023 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
    Commented Oct 14, 2023 at 20:09

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