$Assumptions = a > 0 && b > 0 && c > 0 && t > 0 && {x,y} \[Element] Reals;
R[angle_] = {{Cos[angle], -Sin[angle]},{Sin[angle], Cos[angle]}};
W = {
{-a, x - I y, Exp[-I 2 Pi/3] b, Exp[I 2 Pi/3] b, b},
{x + I y, -a, b, b, b},
{Exp[I 2 Pi/3] b, b, Dot[{x, y}, R[t].{0, 1}] + c, 0, 0},
{Exp[-I 2 Pi/3] b, b, 0, Dot[{x, y}, R[2 Pi/3].R[t].{0, 1}] + c, 0},
{b, b, 0, 0, Dot[{x, y}, R[4 Pi/3].R[t].{0, 1}] + c}

We seek 5 closed-form expressions for the eigenvalues of W, which should appear as functions of real variables x,y and positive parameters a,b,c,t. W is Hermitian so these eigenvalues must be real.

Equivalently, we might attempt to find the roots of

P = (-1)*CharacteristicPolynomial[W, \[Lambda]];  

Which is a degree-5 polynomial in \[Lambda] with leading coefficient (+1) whose roots are the eigenvalues of W.

Additional Constraints and previous work

For (physical) reasons outside of the definition of this matrix, we know that the spectrum is constrained by a rotational $C_{3}$ symmetry, i.e., the spectrum must be invariant under rotation by $2\pi/3$.

I have attempted to utilize this information to simplify the problem, see my previous post on math.SE. For this, I imagined that I might factor the characteristic polynomial $P$ into the product of ([lambda]-root) and write roots 4 and 5 in terms of root 3. Nothing has come of this, yet.

I have also posted previously on mma.SE about solving such a factored nonlinear equation.


If closed-form eigenvalues can be found, what are they?

If not, can we prove if $P$ is of the form of a "Solvable quintic" or an "unsolvable" one?

  • $\begingroup$ All of this sounds rather like a mathematical problem than a Mathematica one. The fact that it is formulated in Mathematica code may not be sufficient here. Perhaps you could highlight the points with which you seek code help? $\endgroup$ – MarcoB May 23 '16 at 19:43
  • $\begingroup$ In some sense, that might be fair, and I can repost this problem to math.SE instead, but in reality, no one is likely to solve this problem by hand. The expanded characteristic polynomial is hundreds of characters long and basically intractable without some sort of mathematica manipulation, which might exclude plenty of programatic solutions, such as CoefficientList, Expand, Simplify, Series, Solve, Reduce, SolveAlways, etc., some of which methods have already had indications of success for me. $\endgroup$ – Steve May 23 '16 at 20:32
  • $\begingroup$ Moreover, please see my linked post on mma.SE which is clearly a mathematica question rather than a mathematical one. However, the solutions generated there were insufficient for this problem, which led to people asking me to post this, genuine question. Thus, I seek a programmatic solution to find the roots, possibly by using mathematica to solve a nonlinear equation for the roots by equating the standard and factored forms of the characteristic polynomial. mathematica.stackexchange.com/questions/115757/… $\endgroup$ – Steve May 23 '16 at 20:39
  • $\begingroup$ What exactly gets rotated by 2*Pi/3? $\endgroup$ – Daniel Lichtblau May 23 '16 at 20:44
  • $\begingroup$ The spectrum is symmetric under $C_{3}$, i.e. the eigenvalues are invariant under rotation by $2\pi/3$. A conjecture from this, that I suspect may be true, is that, given function $f(x,y)$ which specifies eigenvalue "3", eigenvalues "4" and "5" might be given by $f(x',y')$ and $f(x'',y'')$ where (x',y') is the point (x,y) rotated by $2\pi/3$ and (x'',y'') is the point (x,y) rotated by $4\pi/3$. $\endgroup$ – Steve May 23 '16 at 20:48

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