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I have a $3\times 3$ matrix, depending on a single parameter. My job is to find the eigenvectors for the matrix. I tried two ways, shown below

Uw = {{-0.198669 E^(I t), 0. + 0.980067 I, 0}, {0,   0, -I}, {(-0.96053 + 0.194709 I) E^(   I t - I (2.94159 + t)), (-0.0394695 - 
 0.194709 I) E^(-I (2.94159 + t)), 0}}
ti = -π/2 ;  (*Initial value of t *)
tf = 3 π/2 ;  (*Final value of t *)
tinc = π/2; (*Increment in t*)


eigUw = Table[Eigenvectors[Uw], {t, ti, tf, tinc}];

Out: eigUw[[1]]=
  {{-0.448931 - 0.240756 I, 0.608481, 0.506865 - 0.336655 I},
  {0.642748, 0.45931 + 0.287179 I, -0.287179 - 0.45931 I},
  {0.570899 - 0.0380308 I, -0.0769228 - 0.574801 I, 0.579926}}

Another way I tried was,

Eigenvectors[Uw /. {t -> -π/2}]

Out: {
   {-0.0380308 + 0.570899 I, 0.579926 + 0. I, -0.0769228 + 0.574801 I},
   {-0.448931 - 0.240756 I, 0.608481 + 0. I, 0.506865 - 0.336655 I},
   {0.642748 + 0. I,   0.45931 + 0.287179 I, -0.287179 - 0.45931 I}}

I know the eigenvectors are same upto a phase, but it doesn't look the here. I am referring to 3$^{rd}$ of first Out and 1$^{st}$ of second Out

Please let me know if i am missing something basic(I think so)

In[3]:= $Version

Out[3]= "11.2.0 for Linux x86 (64-bit) (September 11, 2017)"
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  • $\begingroup$ The value for k is missing here? With k=1 on mac with mathematica 11.2, the results are identical. Another possibility is degenerate eigenvalues. $\endgroup$ – egwene sedai Feb 21 '18 at 19:59
  • $\begingroup$ @egwenesedai Sorry it was t. Ooh $\endgroup$ – L.K. Feb 21 '18 at 20:03
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    $\begingroup$ Still identical on mac with mathematica 11.2, result is the 1st one as eigUw[[1]]; wolframalpha.com gives the same result with Eigenvectors[{{-0.198669 E^(I t), 0. + 0.980067 I, 0}, {0, 0, -I}, {(-0.96053 + 0.194709 I) E^(I t - I (2.94159 + t)), (-0.0394695 - 0.194709 I) E^(-I (2.94159 + t)), 0}}/.t->-Pi/2] $\endgroup$ – egwene sedai Feb 21 '18 at 20:10
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    $\begingroup$ Concerning the output you gave, if I let x be the first result and y be the second, then 3rd of first Out and 1st of second Out is same up to a phase: y[[1]].x[[3]]\[Conjugate]/(Norm[y[[1]]] Norm[x[[3]]]) // Abs gives 1. $\endgroup$ – egwene sedai Feb 21 '18 at 20:14
  • $\begingroup$ @egwenesedai My ignorance. Thanks a lot for the clarification $\endgroup$ – L.K. Feb 22 '18 at 10:39

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