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?

Question

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#

Liam Brennan · Accepted Answer

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

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))#

Anna Allen · Answer

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

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))#

Lily Anderson · Answer

undefined To find the other two roots of the polynomial $x^3 - 3x + 1 = 0$ given one root as a rational companion matrix, we can use the fact that the roots of a polynomial correspond to the eigenvalues of its companion matrix.

Given the rational companion matrix: 
$$M = \begin{pmatrix} 
0 & 0 & -1 \
1 & 0 & 3 \
0 & 1 & 0 \
\end{pmatrix}$$

First, we need to find the eigenvalues of this matrix. The eigenvalues are the roots of the characteristic polynomial of the matrix.

Next, once we find the other two eigenvalues, we can construct the rational matrices for each of them.

Let's find the eigenvalues of matrix $M$:

The characteristic polynomial is given by:
$$ \text{det}(M - \lambda I) = 0 $$

$$ \text{det}\left( \begin{pmatrix} 
-\lambda & 0 & -1 \
1 & -\lambda & 3 \
0 & 1 & -\lambda \
\end{pmatrix} \right) = 0 $$

$$ (-\lambda)((-\lambda)(-\lambda) - (1)(1)) - (0)((-\lambda)(3) - (0)(1)) + (-1)((1)(3) - (0)(1)) = 0 $$

$$ -\lambda^3 + \lambda + 3 = 0 $$

Now, we solve this cubic equation to find the other two roots.

After obtaining the other two roots, we construct the rational matrices representing each of them.

Thus, the rational matrices representing the other two roots can be found once we solve the cubic equation and obtain the eigenvalues of the given rational companion matrix.