Question

If one of the roots of `x^3-3x+1=0` is given by the rational (companion) matrix `((0,0,-1),(1,0,3),(0,1,0))`, then what rational(?) matrices represent the other two roots?

I think the other two roots are probably not in the field generated by `((1, 0, 0),(0,1,0),(0,0,1))` and `((0,0,-1),(1,0,3),(0,1,0))`. They satisfy a quadratic equation with coefficients in that field, but are probably not rational. For example, if a cubic has one Real and two Complex zeros then the field generated from `QQ` by adding the Real root will clearly not include the Complex roots.

If it helps, the Real roots of `x^3-3x+1=0` are:

`x_1 = omega^(1/3)+omega^(-1/3)`
`x_2 = omega^(4/3)+omega^(-4/3)`
`x_3 = omega^(7/3)+omega^(-7/3)`

where `omega = -1/2+sqrt(3)/2` is the primitive Complex cube root of `1`

Answer 1

We seem to need `6xx6` matrices in order to have rational coefficients.

Explanation:

I think I'm barking up the wrong tree with this question.

It seems like we need `6xx6` matrices in order to be able to use rational coefficients.

Given:

`x^3-3x+1 = 0`

Use Cardano's method to solve the cubic:

Let `x = u+v`

Then:

`u^3+v^3+3(uv-1)(u+v) + 1 = 0`

Add the constraint `v = 1/u` to eliminate the term in `(u+v)` and get:

`u^3+1/u^3+1 = 0`

Multiply through by `u^3` to get the quadratic in `u^3`:

`(u^3)^2+(u^3)+1 = 0`

This is recognisable as `t^2+t+1 = 0`, which has roots:

`u^3 = omega" "` and `u^3 = omega^2`

where `omega = -1/2+sqrt(3)/2` is the primitve Complex cube root of `1`

Hence the roots of our cubic are:

`x_1 = root(3)(omega) + 1/(root(3)(omega)) = omega^(1/3)+omega^(8/3)`

`x_2 = omega root(3)(omega) + 1/(omega root(3)(omega)) = omega^(4/3)+omega^(5/3)`

`x_3 = omega^2 root(3)(omega) + 1/(omega^2 root(3)(omega)) = omega^(7/3)+omega^(2/3)`

Note that `root(3)(omega)` has minimum polynomial `Phi_9 = x^6+x^3+1`

So to represent all three roots concurrently with rational matrices will almost certainly require `6xx6` matrices.

First represent `root(3)(omega)` by the `3xx3` matrix:

`((0, 1, 0),(0, 0, 1),(omega, 0, 0))`

Starting with the `0`th power, this has powers:

`((1, 0, 0),(0, 1, 0),(0, 0, 1))`, `((0, 1, 0),(0, 0, 1),(omega, 0, 0))`, `((0, 0, 1),(omega, 0, 0),(0, omega, 0))`

`((omega, 0, 0),(0, omega, 0),(0, 0, omega))`, `((0, omega, 0), (0, 0, omega), (omega^2, 0, 0))`, `((0, 0, omega),(omega^2,0,0),(0, omega^2,0))`

`((omega^2, 0, 0),(0, omega^2, 0),(0, 0, omega^2))`, `((0, omega^2, 0),(0, 0, omega^2),(1, 0, 0))`, `((0, 0, omega^2),(1,0,0),(0,1,0))`

which makes the roots:

`x_1 = ((0, 1, omega^2),(1,0,1),(omega,1,0))`

`x_2 = ((0, omega, omega),(omega^2,0,omega),(omega^2,omega^2,0))`

`x_3 = ((0, omega^2,1),(omega,0,omega^2),(1,omega,0))`

Then note that `omega` can be represented by the companion matrix of `t^2+t+1 = 0`:

`omega = ((0, -1),(1, -1))`

`omega^2 = ((-1, 1),(-1,0))`

Along with `0 = ((0, 0),(0,0))` and `1 = ((1,0),(0,1))` we can substitute these values into our `3xx3` matrices to get:

`x_1 = ((((0, 0),(0,0))"", ((1,0),(0,1))"", ((-1, 1),(-1,0))""),(((1,0),(0,1))"",((0, 0),(0,0))"",((1,0),(0,1))""),(((0, -1),(1, -1))"",((1,0),(0,1))"",((0, 0),(0,0))"")) = ((0,0,1,0,-1,1),(0,0,0,1,-1,0),(1,0,0,0,1,0),(0,1,0,0,0,1),(0,-1,1,0,0,0),(1,-1,0,1,0,0))`

`x_2 = ((((0, 0),(0,0))"", ((0, -1),(1, -1))"", ((0, -1),(1, -1))""),(((-1, 1),(-1,0))"",((0, 0),(0,0))"",((0, -1),(1, -1))""),(((-1, 1),(-1,0))"",((-1, 1),(-1,0))"",((0, 0),(0,0))"")) = ((0,0,0,-1,0,-1),(0,0,1,-1,1,-1),(-1,1,0,0,0,-1),(-1,0,0,0,1,-1),(-1,1,-1,1,0,0),(-1,0,-1,0,0,0))`

`x_3 = ((((0, 0),(0,0))"", ((-1, 1),(-1,0))"",((1,0),(0,1))""),(((0, -1),(1, -1))"",((0, 0),(0,0))"",((-1, 1),(-1,0))""),(((1,0),(0,1))"",((0, -1),(1, -1))"",((0,0),(0,0))"")) = ((0,0,-1,1,1,0),(0,0,-1,0,0,1),(0,-1,0,0,-1,1),(1,-1,0,0,-1,0),(1,0,0,-1,0,0),(0,1,1,-1,0,0))`

Answer 2

Here's another way to construct rational `6xx6` matrices for the `3` roots...

Explanation:

Another way to construct `6xx6` rational solutions is to start with the companion matrix of `Phi_9 = x^6+x^3+1` which is the minimal polynomial of `root(3)(omega)`

Noting that the Real roots of `x^3-3x+1 = 0` are:

`x_1 = omega^(1/3) + omega^(8/3)`

`x_2 = omega^(4/3) + omega^(5/3)`

`x_3 = omega^(7/3) + omega^(2/3)`

we will be able to construct rational matrix solutions from a matrix which behaves like `root(3)(omega)`

The companion matrix of `Phi_9` is:

`A = ((0, 0, 0, 0, 0, -1),(1, 0, 0, 0, 0, 0), (0, 1, 0, 0, 0, 0),(0, 0, 1, 0, 0, -1),(0, 0, 0, 1, 0, 0),(0, 0, 0, 0, 1, 0))`

Then:

`A^2 = ((0, 0, 0, 0, -1, 0),(0, 0, 0, 0, 0, -1),(1, 0, 0, 0, 0, 0),(0, 1, 0, 0, -1, 0),(0, 0, 1, 0, 0, -1),(0, 0, 0, 1, 0, 0))`

`A^3 = ((0, 0, 0, -1, 0, 0),(0, 0, 0, 0, -1, 0), (0, 0, 0, 0, 0, -1), (1, 0, 0, -1, 0, 0), (0, 1, 0, 0, -1, 0), (0, 0, 1, 0, 0, -1))`

`A^4 = ((0, 0, -1, 0, 0, 1), (0, 0, 0, -1, 0, 0), (0, 0, 0, 0, -1, 0), (0, 0, -1, 0, 0, 0), (1, 0, 0, -1, 0, 0), (0, 1, 0, 0, -1, 0))`

`A^5 = ((0, -1, 0, 0, 1, 0), (0, 0, -1, 0, 0, 1), (0, 0, 0, -1, 0, 0), (0, -1, 0, 0, 0, 0), (0, 0, -1, 0, 0, 0), (1, 0, 0, -1, 0, 0))`

`A^6 = ((-1, 0, 0, 1, 0, 0), (0, -1, 0, 0, 1, 0), (0, 0, -1, 0, 0, 1), (-1, 0, 0, 0, 0, 0), (0, -1, 0, 0, 0, 0), (0, 0, -1, 0, 0, 0))`

`A^7 = ((0, 0, 1, 0, 0, 0), (-1, 0, 0, 1, 0, 0), (0, -1, 0, 0, 1, 0), (0, 0, 0, 0, 0, 1), (-1, 0, 0, 0, 0, 0), (0, -1, 0, 0, 0, 0))`

`A^8 = ((0, 1, 0, 0, 0, 0), (0, 0, 1, 0, 0, 0), (-1, 0, 0, 1, 0, 0), (0, 0, 0, 0, 1, 0), (0, 0, 0, 0, 0, 1), (-1, 0, 0, 0, 0, 0))`

Then the roots of `x^3-3x+1=0` are:

`x_1 = A^1+A^8 = ((0, 1, 0, 0, 0, -1), (1, 0, 1, 0, 0, 0), (-1, 1, 0, 1, 0, 0), (0, 0, 1, 0, 1, -1), (0, 0, 0, 1, 0, 1), (-1, 0, 0, 0, 1, 0))`

`x_2 = A^4+A^5 = ((0, -1, -1, 0, 1, 1), (0, 0, -1, -1, 0, 1), (0, 0, 0, -1, -1, 0), (0, -1, -1, 0, 0, 0), (1, 0, -1, -1, 0, 0), (1, 1, 0, -1, -1, 0))`

`x_3 = A^7+A^2 = ((0, 0, 1, 0, -1, 0), (-1, 0, 0, 1, 0, -1), (1, -1, 0, 0, 1, 0), (0, 1, 0, 0, -1, 1), (-1, 0, 1, 0, 0, -1), (0, -1, 0, 1, 0, 0))`