# Reordering numerically calculated eigenvalues assuming smooth dependence on a parameter

As was discussed in a different question here on SE (link) if you compute eigenvalues numerically of a matrix which depends on a parameter y, the resulting plot of the eigenvalues will yield non-smooth plots. Assuming that I know that my eigenvalues will transform smoothly under the change of y I would like to reorder them, so that when I plot them the color coding is correct.

Secondly I would also like to apply the reordering to the eigenvectors as I would like to have a look at how they transform under changes of y, too.

In this link george2079 does already partially answer my question by applying a minimization procedure which punishes large curvature via a cost function. However, I have trouble adapting it to my situation and I am not sure if this is the best approach (so if you have a better idea feel free to post it as an answer).

My Matrix:

m = {{-0.576 Cos[y], 0, 0, 0. + 0.06 I, 0, 0.369858, 0, 0, 0,    0, -0.0906385, 0, 0, -0.265868, 0.0366083, 0.0157771, -0.185737,    0.0349767, 0.0435434, -0.276945, (0.288 - 0.498831 I) Sin[y],    0. - 0.03 I, 0.03, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,    0}, {0, -0.576 Cos[y], 0. - 0.03 I, 0, 0, 0, -0.170921, 0, 0, 0, 0,    0.234664, 0.234164, 0, 0, -0.185737, 0.205344, -0.239136, -0.249447,    0.143353, 0. + 0.03 I, (0.288 - 0.498831 I) Sin[y], 0, -0.03,    0.0519615, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {0,    0. + 0.03 I, -0.576 Cos[y], 0, 0, 0, 0, 0.369858, 0, 0, 0,    0.234164, -0.0357261, 0, 0, 0.0349767, -0.239136, -0.0707859,    0.185737, 0.248295, -0.03, 0, (0.288 - 0.498831 I) Sin[y],    0. - 0.03 I, 0. - 0.0519615 I, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,    0, 0, 0}, {0. - 0.06 I, 0, 0, -0.576 Cos[y], 0, 0, 0,    0, -0.244138, -0.0422717, -0.265868, 0, 0, 0.216359, 0.0211358,    0.0435434, -0.249447, 0.185737, 0.0660567, 0.159894, 0, 0.03,    0. + 0.03 I, (0.288 - 0.498831 I) Sin[y], 0, 0, 0, 0, 0, 0, 0, 0, 0,    0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, -0.576 Cos[y], 0, 0,    0, -0.0422717, -0.195326, 0.0366083, 0, 0,    0.0211358, -0.195326, -0.276945, 0.143353, 0.248295, 0.159894,    0.293945, 0, -0.0519615, 0. + 0.0519615 I,    0, (0.288 - 0.498831 I) Sin[y], 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,    0, 0, 0}, {0.369858, 0, 0, 0, 0, -0.576 Cos[y], 0, 0, 0. + 0.06 I,    0, -0.0906385, 0, 0, 0.265868, -0.0366083, 0.0157771, 0.185737,    0.0349767, -0.0435434, 0.276945, 0, 0, 0, 0,    0, (0.288 + 0.498831 I) Sin[y], 0. - 0.03 I, 0.03, 0, 0, 0, 0, 0, 0,    0, 0, 0, 0, 0, 0}, {0, -0.170921, 0, 0, 0, 0, -0.576 Cos[y],    0. - 0.03 I, 0, 0, 0, 0.234664, -0.234164, 0, 0, 0.185737, 0.205344,    0.239136, -0.249447, 0.143353, 0, 0, 0, 0, 0,    0. + 0.03 I, (0.288 + 0.498831 I) Sin[y], 0, -0.03, 0.0519615, 0, 0,    0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0.369858, 0, 0, 0,    0. + 0.03 I, -0.576 Cos[y], 0, 0, 0, -0.234164, -0.0357261, 0, 0,    0.0349767, 0.239136, -0.0707859, -0.185737, -0.248295, 0, 0, 0, 0,    0, -0.03, 0, (0.288 + 0.498831 I) Sin[y], 0. - 0.03 I,    0. - 0.0519615 I, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0,    0, -0.244138, -0.0422717, 0. - 0.06 I, 0, 0, -0.576 Cos[y], 0,    0.265868, 0, 0, 0.216359,    0.0211358, -0.0435434, -0.249447, -0.185737, 0.0660567, 0.159894, 0,    0, 0, 0, 0, 0, 0.03, 0. + 0.03 I, (0.288 + 0.498831 I) Sin[y], 0,    0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, -0.0422717, -0.195326, 0,    0, 0, 0, -0.576 Cos[y], -0.0366083, 0, 0, 0.0211358, -0.195326,    0.276945, 0.143353, -0.248295, 0.159894, 0.293945, 0, 0, 0, 0, 0,    0, -0.0519615, 0. + 0.0519615 I, 0, (0.288 + 0.498831 I) Sin[y], 0,    0, 0, 0, 0, 0, 0, 0, 0, 0}, {-0.0906385, 0, 0, -0.265868,    0.0366083, -0.0906385, 0, 0, 0.265868, -0.0366083, -0.576 Cos[y], 0,    0, 0. + 0.06 I, 0, 0.0911964, 0, 0.356683, 0, 0, 0, 0, 0, 0, 0, 0,    0, 0, 0, 0, -0.576 Sin[y], 0. - 0.03 I, 0.03, 0, 0, 0, 0, 0, 0,    0}, {0, 0.234664, 0.234164, 0, 0, 0, 0.234664, -0.234164, 0, 0,    0, -0.576 Cos[y], 0. - 0.03 I, 0, 0, 0, -0.208851, 0,    0.0722586, -0.286706, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,    0. + 0.03 I, -0.576 Sin[y], 0, -0.03, 0.0519615, 0, 0, 0, 0, 0}, {0,    0.234164, -0.0357261, 0, 0, 0, -0.234164, -0.0357261, 0, 0, 0,    0. + 0.03 I, -0.576 Cos[y], 0, 0, 0.356683, 0, 0.343409, 0, 0, 0, 0,    0, 0, 0, 0, 0, 0, 0, 0, -0.03, 0, -0.576 Sin[y], 0. - 0.03 I,    0. - 0.0519615 I, 0, 0, 0, 0, 0}, {-0.265868, 0, 0, 0.216359,    0.0211358, 0.265868, 0, 0, 0.216359, 0.0211358, 0. - 0.06 I, 0,    0, -0.576 Cos[y], 0, 0, 0.0722586, 0, -0.00936269, -0.319789, 0, 0,    0, 0, 0, 0, 0, 0, 0, 0, 0, 0.03, 0. + 0.03 I, -0.576 Sin[y], 0, 0,    0, 0, 0, 0}, {0.0366083, 0, 0, 0.0211358, -0.195326, -0.0366083, 0,    0, 0.0211358, -0.195326, 0, 0, 0, 0, -0.576 Cos[y], 0, -0.286706,    0, -0.319789, 0.293945, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -0.0519615,    0. + 0.0519615 I, 0, -0.576 Sin[y], 0, 0, 0, 0,    0}, {0.0157771, -0.185737, 0.0349767, 0.0435434, -0.276945,    0.0157771, 0.185737, 0.0349767, -0.0435434, 0.276945, 0.0911964, 0,    0.356683, 0, 0, 1.02645, 0, 0, 0. + 0.53 I, 0, 0, 0, 0, 0, 0, 0, 0,    0, 0, 0, 0, 0, 0, 0, 0, 0, 0. - 0.265 I, 0.265, 0, 0}, {-0.185737,    0.205344, -0.239136, -0.249447, 0.143353, 0.185737, 0.205344,    0.239136, -0.249447, 0.143353, 0, -0.208851, 0,    0.0722586, -0.286706, 0, 1.02645, 0. - 0.265 I, 0, 0, 0, 0, 0, 0, 0,    0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0. + 0.265 I, 0, 0, -0.265,    0.458993}, {0.0349767, -0.239136, -0.0707859, 0.185737, 0.248295,    0.0349767, 0.239136, -0.0707859, -0.185737, -0.248295, 0.356683, 0,    0.343409, 0, 0, 0, 0. + 0.265 I, 1.02645, 0, 0, 0, 0, 0, 0, 0, 0, 0,    0, 0, 0, 0, 0, 0, 0, 0, -0.265, 0, 0, 0. - 0.265 I,    0. - 0.458993 I}, {0.0435434, -0.249447, 0.185737, 0.0660567,    0.159894, -0.0435434, -0.249447, -0.185737, 0.0660567, 0.159894, 0,    0.0722586, 0, -0.00936269, -0.319789, 0. - 0.53 I, 0, 0, 1.02645, 0,    0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0.265,    0. + 0.265 I, 0, 0}, {-0.276945, 0.143353, 0.248295, 0.159894,    0.293945, 0.276945, 0.143353, -0.248295, 0.159894, 0.293945,    0, -0.286706, 0, -0.319789, 0.293945, 0, 0, 0, 0, 1.02645, 0, 0, 0,    0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -0.458993, 0. + 0.458993 I,    0, 0}, {(0.288 + 0.498831 I) Sin[y], 0. - 0.03 I, -0.03, 0, 0, 0, 0,    0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0.576 Cos[y], 0, 0,    0. - 0.06 I, 0, 0.369858, 0, 0, 0, 0, -0.0906385, 0, 0, -0.265868,    0.0366083, 0.0157771, -0.185737, 0.0349767,    0.0435434, -0.276945}, {0. + 0.03 I, (0.288 + 0.498831 I) Sin[y], 0,    0.03, -0.0519615, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,    0.576 Cos[y], 0. + 0.03 I, 0, 0, 0, -0.170921, 0, 0, 0, 0, 0.234664,    0.234164, 0, 0, -0.185737, 0.205344, -0.239136, -0.249447,    0.143353}, {0.03, 0, (0.288 + 0.498831 I) Sin[y], 0. - 0.03 I,    0. - 0.0519615 I, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,    0. - 0.03 I, 0.576 Cos[y], 0, 0, 0, 0, 0.369858, 0, 0, 0,    0.234164, -0.0357261, 0, 0, 0.0349767, -0.239136, -0.0707859,    0.185737, 0.248295}, {0, -0.03,    0. + 0.03 I, (0.288 + 0.498831 I) Sin[y], 0, 0, 0, 0, 0, 0, 0, 0, 0,    0, 0, 0, 0, 0, 0, 0, 0. + 0.06 I, 0, 0, 0.576 Cos[y], 0, 0, 0,    0, -0.244138, -0.0422717, -0.265868, 0, 0, 0.216359, 0.0211358,    0.0435434, -0.249447, 0.185737, 0.0660567, 0.159894}, {0, 0.0519615,    0. + 0.0519615 I, 0, (0.288 + 0.498831 I) Sin[y], 0, 0, 0, 0, 0, 0,    0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0.576 Cos[y], 0, 0,    0, -0.0422717, -0.195326, 0.0366083, 0, 0,    0.0211358, -0.195326, -0.276945, 0.143353, 0.248295, 0.159894,    0.293945}, {0, 0, 0, 0, 0, (0.288 - 0.498831 I) Sin[y],    0. - 0.03 I, -0.03, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0.369858, 0,    0, 0, 0, 0.576 Cos[y], 0, 0, 0. - 0.06 I, 0, -0.0906385, 0, 0,    0.265868, -0.0366083, 0.0157771, 0.185737, 0.0349767, -0.0435434,    0.276945}, {0, 0, 0, 0, 0, 0. + 0.03 I, (0.288 - 0.498831 I) Sin[y],    0, 0.03, -0.0519615, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -0.170921, 0,    0, 0, 0, 0.576 Cos[y], 0. + 0.03 I, 0, 0, 0, 0.234664, -0.234164,    0, 0, 0.185737, 0.205344, 0.239136, -0.249447, 0.143353}, {0, 0, 0,    0, 0, 0.03, 0, (0.288 - 0.498831 I) Sin[y], 0. - 0.03 I,    0. - 0.0519615 I, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0.369858, 0,    0, 0, 0. - 0.03 I, 0.576 Cos[y], 0, 0, 0, -0.234164, -0.0357261, 0,    0, 0.0349767, 0.239136, -0.0707859, -0.185737, -0.248295}, {0, 0, 0,    0, 0, 0, -0.03, 0. + 0.03 I, (0.288 - 0.498831 I) Sin[y], 0, 0, 0,    0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -0.244138, -0.0422717, 0. + 0.06 I,    0, 0, 0.576 Cos[y], 0, 0.265868, 0, 0, 0.216359,    0.0211358, -0.0435434, -0.249447, -0.185737, 0.0660567,    0.159894}, {0, 0, 0, 0, 0, 0, 0.0519615, 0. + 0.0519615 I,    0, (0.288 - 0.498831 I) Sin[y], 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,    0, -0.0422717, -0.195326, 0, 0, 0, 0, 0.576 Cos[y], -0.0366083, 0,    0, 0.0211358, -0.195326, 0.276945, 0.143353, -0.248295, 0.159894,    0.293945}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -0.576 Sin[y],    0. - 0.03 I, -0.03, 0, 0, 0, 0, 0, 0, 0, -0.0906385, 0,    0, -0.265868, 0.0366083, -0.0906385, 0, 0, 0.265868, -0.0366083,    0.576 Cos[y], 0, 0, 0. - 0.06 I, 0, 0.0911964, 0, 0.356683, 0,    0}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0. + 0.03 I, -0.576 Sin[y], 0,    0.03, -0.0519615, 0, 0, 0, 0, 0, 0, 0.234664, 0.234164, 0, 0, 0,    0.234664, -0.234164, 0, 0, 0, 0.576 Cos[y], 0. + 0.03 I, 0, 0,    0, -0.208851, 0, 0.0722586, -0.286706}, {0, 0, 0, 0, 0, 0, 0, 0, 0,    0, 0.03, 0, -0.576 Sin[y], 0. - 0.03 I, 0. - 0.0519615 I, 0, 0, 0,    0, 0, 0, 0.234164, -0.0357261, 0, 0, 0, -0.234164, -0.0357261, 0, 0,    0, 0. - 0.03 I, 0.576 Cos[y], 0, 0, 0.356683, 0, 0.343409, 0,    0}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -0.03,    0. + 0.03 I, -0.576 Sin[y], 0, 0, 0, 0, 0, 0, -0.265868, 0, 0,    0.216359, 0.0211358, 0.265868, 0, 0, 0.216359, 0.0211358,    0. + 0.06 I, 0, 0, 0.576 Cos[y], 0, 0, 0.0722586,    0, -0.00936269, -0.319789}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,    0.0519615, 0. + 0.0519615 I, 0, -0.576 Sin[y], 0, 0, 0, 0, 0,    0.0366083, 0, 0, 0.0211358, -0.195326, -0.0366083, 0, 0,    0.0211358, -0.195326, 0, 0, 0, 0, 0.576 Cos[y], 0, -0.286706,    0, -0.319789, 0.293945}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,    0, 0, 0. - 0.265 I, -0.265, 0, 0, 0.0157771, -0.185737, 0.0349767,    0.0435434, -0.276945, 0.0157771, 0.185737, 0.0349767, -0.0435434,    0.276945, 0.0911964, 0, 0.356683, 0, 0, 1.02645, 0, 0, 0. - 0.53 I,    0}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0. + 0.265 I, 0,    0, 0.265, -0.458993, -0.185737, 0.205344, -0.239136, -0.249447,    0.143353, 0.185737, 0.205344, 0.239136, -0.249447, 0.143353,    0, -0.208851, 0, 0.0722586, -0.286706, 0, 1.02645, 0. + 0.265 I, 0,    0}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0.265, 0, 0,    0. - 0.265 I, 0. - 0.458993 I, 0.0349767, -0.239136, -0.0707859,    0.185737, 0.248295, 0.0349767,    0.239136, -0.0707859, -0.185737, -0.248295, 0.356683, 0, 0.343409,    0, 0, 0, 0. - 0.265 I, 1.02645, 0, 0}, {0, 0, 0, 0, 0, 0, 0, 0, 0,    0, 0, 0, 0, 0, 0, 0, -0.265, 0. + 0.265 I, 0, 0,    0.0435434, -0.249447, 0.185737, 0.0660567,    0.159894, -0.0435434, -0.249447, -0.185737, 0.0660567, 0.159894, 0,    0.0722586, 0, -0.00936269, -0.319789, 0. + 0.53 I, 0, 0, 1.02645,    0}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0.458993,    0. + 0.458993 I, 0, 0, -0.276945, 0.143353, 0.248295, 0.159894,    0.293945, 0.276945, 0.143353, -0.248295, 0.159894, 0.293945,    0, -0.286706, 0, -0.319789, 0.293945, 0, 0, 0, 0, 1.02645}}
(*empty line*)

ListPlot[Transpose@Table[Eigenvalues[m], {y, -Pi, Pi, .01}]]


Original code (m is a 3x3 matrix):

alle = Table[Eigenvalues[m], {y, -Pi, Pi, .01}];
original = Show[
MapIndexed[ListPlot[Flatten[Take[alle, All, {#}] , 2],
Joined -> True, PlotRange -> All,
PlotStyle -> Hue[First@#2/3]] &, Range[3]]];
v = Table[V[i], {i, Length[alle]}];
SetOptions[NMinimize, MaxIterations -> 500];
Do[
alle = MapThread[ Prepend[  Drop[ #2 , {#1}]  , #2[[#1]] ]  &,
{ (v /. Last@Minimize[{Total[((#[[1]] + #[[3]] - 2 #[[2]])^2 & /@
Partition[ Indexed @@@ Transpose[{alle, v}], 3, 1, {1, 3}, {}])],
Table[k <= V[i] <= 3, {i, Length[alle]}]}, v, Integers]) ,
alle}] , {k, 2}];
Show[original,
ListPlot[Flatten@Take[alle, All, {#}], Joined -> True,
PlotStyle -> {Thick, Black, Dashed}] & /@ Range[3]]


My adoption (m is my matrix):

alle = Table[Eigenvalues[m], {y, -Pi, Pi, .01}];
original = Show[
MapIndexed[ListPlot[Flatten[Take[alle, All, {#}] , 2],
Joined -> True, PlotRange -> All,
PlotStyle -> Hue[First@#2/40]] &, Range[40]]];
v = Table[V[i], {i, Length[alle]}];
SetOptions[NMinimize, MaxIterations -> 500];
Do[
alle = MapThread[ Prepend[  Drop[ #2 , {#1}]  , #2[[#1]] ]  &,
{ (v /. Last@Minimize[{Total[((#[[1]] + #[[3]] - 2 #[[2]])^2 & /@
Partition[ Indexed @@@ Transpose[{alle, v}], 40, 1, {1, 40}, {}])],
Table[k <= V[i] <= 40, {i, Length[alle]}]}, v, Integers]) ,
alle}] , {k, 40-1}];
Show[original,
ListPlot[Flatten@Take[alle, All, {#}], Joined -> True,
PlotStyle -> {Thick, Black, Dashed}] & /@ Range[40]]


My Questions:

1. Can someone explain george2079's code, because I still have trouble to figure out exactly what is going on?

• What is the v table for?

• I understand the idea of the cost function but how is it used to reorder the eigenvalues?

2. Is my adoption of the 3x3 example correct for my 40x40 example?
3. How do I add my eigenvectors so that I still know at the end which eigenvectors belongs to which eigenvalue?
4. Does the above code work on your system and if it does how long does the job take to finish?
5. If it does not finish within reasonable time (for me that is < 40min): Is there a way to increase performance?

Note If you have a better idea how to solve my problem, this is of course also great. I am not bound in any way to use the above code.

• One thing you might want to do, if it is not computationally troublesome, is to find points where the characteristic polynomial of the matrix, and its derivative with respect to the parameter, simultaneously vanish. Such points are where you have multiple roots to the char poly, that is, parametrized eigenvalue branch function crossings. – Daniel Lichtblau May 20 '15 at 19:30
• @Daniel, Mathematica refuses to give me the characteristic polynomial if I still have my parameter y in place. I have opened another question for that: mathematica.stackexchange.com/questions/83994/… – NOhs May 21 '15 at 8:05
• @Daniel, ok Mathematica takes forever to compute the characteristic polynomial I don't know if it will finish, ever (the matrix is 40x40) – NOhs May 21 '15 at 14:29
• sorry to not be much help, but I'm not surprised the linked solution doesn't scale well to such a large system. – george2079 May 27 '15 at 20:22

Apologies for the procedural approach, but this seems to work. Trace out one line at a time, using Interpolation to extrapolate where the next point is likely to be:

c = {};
frames = {};
xvals = Pi Range[-1, 1, 1/100];
alle = Table[({y, #} & /@ Eigenvalues[m]), {y, xvals}] ;
colors = Table[Hue[RandomReal[{0, 2/3}]], {Length@m}];
Clear[g];
Monitor[
Do[
line =
First@Position[alle, First@SortBy[#, #[[2]] &]] & /@ alle[[;; 2]];
MapIndexed[(nextx = #;
proj = {nextx,
Quiet[ Interpolation[alle[[Sequence @@ #]] & /@ line]@nextx ]};
AppendTo[line, Position[alle, First@Nearest[alle[[First@#2 + 2]],
proj]][[1]]] ) &, xvals[[3 ;;]]];
AppendTo[c, alle[[Sequence @@ #]] & /@ line];
alle = Delete[alle, line];
AppendTo[frames,
g = Show[ {
If[Length@First@alle > 0, ListPlot[Flatten[alle, 1]],
Graphics[]] ,
Graphics[
MapIndexed[{Thick, colors[[First@#2]], Line[#]} &, c]]},
PlotRange -> {-2, 2}, AxesOrigin -> {0, 0}]],
{nrem, Length@m, 1, -1}], g]


• I am currently running a different approach but the estimated time it takes is about 23 hours (and only 12 hours have passed so far). How fast was your solution? (BTW, looks really nice and short (compared to my idea :D)) – NOhs May 29 '15 at 7:28
• Ok, I see, it is fast! I think the only trouble you will get is if there are two possible values which have both (apart from numerical imprecision) the same distance to the extrapolated function. This is why I again employed a Minimization procedure that would "grow" all 40 lines at the same time so that in the above situation one of the two cases would lead to a higher penalty of a neighboring line. However, the minimization in 40 variables is awfully slow :(. – NOhs May 29 '15 at 7:43
• if it doesn't need to be fully automatic, you might be best off to do some manual selection if that situation arises. – george2079 May 29 '15 at 15:58
• I guess so. I just encountered a different issue. For a different matrix I have non-smooth dependence (sqrt(abs))-like behaviour :D – NOhs May 29 '15 at 16:03
• How long is fast for you all? I couldn't get alle to compute in a reasonable amount of time. Is it trying to make the Eigenvalues computation before variable assignment? – CA Trevillian Aug 14 '19 at 0:06

I wanted to give an answer that is based on George's one but uses the Eigenvectors instead of the Eigenvalues, as I noticed that in the crossing regions the Eigenvalue method has issues (jumps in the Eigenvector plot) whereas the Eigenvectors even at intersections of the Eigenvalues are orthogonal and thus can be easily distinguished.

What I did is simply find the next Eigenvector that is closest (Norm of difference). The only trouble with the Eigenvectors is the free phase one has to consider. To overcome this I simply multiplied all possible next Eigenvectors by {1,I,-1,-I}. After that I looked which one of them is closest. For at least one of the 4 phases the correct Eigenvector is closer than the rest.

generateSortedEigensystem[matrix_,theta_,thetaMin_,thetaMax_,numXVals_]:=Module[
{ thetaStep, initialData, outputData, xvals,
line, proj, nrem, x, lastEv, ind, dd },

thetaStep   = ( thetaMax - thetaMin ) / If[ numXVals == 1, 1, numXVals - 1 ];
outputData  = {};
xvals       = Range[ thetaMin, thetaMax, thetaStep ];

initialData = Table[ ( Flatten[ { y, # }, 1 ] &/@ ( Thread @ Eigensystem[ matrix/.theta->y ] ) ), { y, xvals } ];
Do[
line = Position[ initialData, First @ SortBy[ #, #[[2]] & ] ] &@ initialData[[1]];
( lastEv = initialData[[ Sequence @@ (Last @ line), 3 ]];
proj   = { #,
Last @ First @ Position[ #, Min@# ] &@
( Map[ Norm@( lastEv - # ) &, #, {2} ] &@
Array[ Function[ dd, {1,I,-1,-I}[[ dd ]]*initialData[[ #, All, 3 ]] ], 4 ] )
};
AppendTo[ line, proj ] )
&/@ Range[2,Length@xvals];

AppendTo[ outputData, initialData[[ Sequence @@ # ]] &/@ line ];
initialData = Delete[ initialData, line ];

, { nrem, Length @ matrix, 1, -1 }
];
Return[outputData]


]

• I think a more easily criteria would be take the projection $v(t)^* \cdot v(t+dt)$ on to the previous eigenvectors. The one has the largest magnitude would be the correct one. – xslittlegrass Oct 1 '15 at 14:32
• That is true. This way the phase is irrelevant. But you would need to add Abs [] of course to you formula as the result of the projection might be complex. – NOhs Oct 1 '15 at 15:03
• Yes, exactly. The plot in the question looks like a band structure for me, and I think that's the standard way we deal with the fake avoided crossings in the band structure. – xslittlegrass Oct 1 '15 at 15:07
• Great answer. Is there a way to parallelize this? – sintetico Feb 26 '16 at 17:27
• @xslittlegrass How one can modify the definition of the function "generateSortedEigensystem" above in order to use the criteria of the projection that you suggest? – sintetico Feb 26 '16 at 17:40