Trying to simplify Root expressions from the output of Eigenvalues

Question

I am trying to calculate eigenvalues of a sparse matrix with only two distinct non-zero elements, here Alpha and Beta, which are both negative reals. Mathematica returns some complex expressions with Root[] values when using the Eigenvalues[] command on the following matrixA:

In all cases the matrices are symmetric and real and hence have real eigenvalues.

matrixA={
        {α, β, 0, 0, 0, 0, β, 0, 0, β},
        {β, α, β, 0, 0, 0, 0, 0, 0, 0},
        {0, β, α, β, 0, 0, 0, 0, 0, 0},
        {0, 0, β, α, β, 0, 0, 0, 0, 0},
        {0, 0, 0, β, α, β, 0, 0, 0, 0},
        {0, 0, 0, 0, β, α, β, 0, 0, 0},
        {β, 0, 0, 0, 0, β, α, β, 0, 0},
        {0, 0, 0, 0, 0, 0, β, α, β, 0},
        {0, 0, 0, 0, 0, 0, 0, β, α, β},
        {β, 0, 0, 0, 0, 0, 0, 0, β, α}
        }

For comparison, with all the other similar matrices I've tried (see below e.g. matrixB) Mathematica will put out simple decimal approximations (using Eigenvalues[matrixB] // N // Simplify)

Can anyone point out a way to get expressions for the matrixA as simple as for matrixB?

And yes, the desired simple answers for matrixA do exist, I can get them with other programs, but I want to use Mathematica!

---

I should add that I already have already used $Assumptions = α<0 && β <0 at the top of my worksheet.

matrixB={
        {α, β, 0, 0, 0, 0, 0, 0, 0, β},
        {β, α, β, 0, 0, 0, 0, 0, 0, 0},
        {0, β, α, β, 0, 0, 0, β, 0, 0},
        {0, 0, β, α, β, 0, 0, 0, 0, 0},
        {0, 0, 0, β, α, β, 0, 0, 0, 0},
        {0, 0, 0, 0, β, α, β, 0, 0, 0},
        {0, 0, 0, 0, 0, β, α, β, 0, 0},
        {0, 0, β, 0, 0, 0, β, α, β, 0},
        {0, 0, 0, 0, 0, 0, 0, β, α, β},
        {β, 0, 0, 0, 0, 0, 0, 0, β, α}
        }

Answer 1

The reason why you can't get a simple expression for eigenvalues is that the characterisitc polynomial of matrixA is not factorizable (in general) to lower order polynomials, unlike for matrixB.

CharacteristicPolynomial[matrixA, x] // Factor

CharacteristicPolynomial[matrixB, x] // Factor

There is no general method of solving sixth order polynomial equations, unlike for forth order ones. In general, you can still simplify a bit the expression for eigenvalues of matrixA adding an option Quartics -> True to Eigenvalues :

Eigenvalues[matrixA, Quartics -> True]

Answer 2

Well, I figured out how the other programs do get numeric answers. Of course the trick is to eliminate the symbols. Since matrixA is so simply structured it can be massaged into a non-symbolic form, calculate numerically the eigenvalues of that, and then unmassage them to recover the symbolic eigenvalues. Divide the whole matrix by β then "re-zero" the main diagonal to α/β.

For reference,

reducedmatrixA=({
      {0, 1, 0, 0, 0, 0, 1, 0, 0, 1},
      {1, 0, 1, 0, 0, 0, 0, 0, 0, 0},
      {0, 1, 0, 1, 0, 0, 0, 0, 0, 0},
      {0, 0, 1, 0, 1, 0, 0, 0, 0, 0},
      {0, 0, 0, 1, 0, 1, 0, 0, 0, 0},
      {0, 0, 0, 0, 1, 0, 1, 0, 0, 0},
      {1, 0, 0, 0, 0, 1, 0, 1, 0, 0},
      {0, 0, 0, 0, 0, 0, 1, 0, 1, 0},
      {0, 0, 0, 0, 0, 0, 0, 1, 0, 1},
      {1, 0, 0, 0, 0, 0, 0, 0, 1, 0}
     } )

numericeigenvalues = Sort[Eigenvalues[reducedmatrixA] // Simplify // N]
symboliceigenvalues = α + β numericeigenvalues

does the trick. Thanks everyone for your pointers on the algebra.

Answer 3

Presumably those other systems are making some assumptions about the values of α and β, which Mathematica does not. Mathematica's symbolic engine does not assume that symbols represent real-valued quantities.

You can probably get some simplification using the Assuming construct or the Assumptions option to Simplify and FullSimplify, like this:

FullSimplify[Eigenvalues[matrixA], 
 Assumptions -> {Element[α, Reals], Element[β, Reals]}]

Or even:

FullSimplify[Eigenvalues[matrixA], 
 Assumptions -> {α > 0, β > 0}]

The latter simplifies the first eigenvalue to:

α - 1/4 (1 + Sqrt[5] + Sqrt[2 (11 + Sqrt[5])]) β

However, some of the others are still higher-order expressions represented by Root expressions.