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#

2 Answers
Dec 19, 2016

Answer:

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))#

Dec 19, 2016

Answer:

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))#