# 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 $\left(\begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{matrix}\right)$ and $\left(\begin{matrix}0 & 0 & - 1 \\ 1 & 0 & 3 \\ 0 & 1 & 0\end{matrix}\right)$. 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 $\mathbb{Q}$ by adding the Real root will clearly not include the Complex roots. If it helps, the Real roots of ${x}^{3} - 3 x + 1 = 0$ are: ${x}_{1} = {\omega}^{\frac{1}{3}} + {\omega}^{- \frac{1}{3}}$ ${x}_{2} = {\omega}^{\frac{4}{3}} + {\omega}^{- \frac{4}{3}}$ ${x}_{3} = {\omega}^{\frac{7}{3}} + {\omega}^{- \frac{7}{3}}$ where $\omega = - \frac{1}{2} + \frac{\sqrt{3}}{2}$ is the primitive Complex cube root of $1$

Dec 19, 2016

We seem to need $6 \times 6$ 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 $6 \times 6$ matrices in order to be able to use rational coefficients.

Given:

${x}^{3} - 3 x + 1 = 0$

Use Cardano's method to solve the cubic:

Let $x = u + v$

Then:

${u}^{3} + {v}^{3} + 3 \left(u v - 1\right) \left(u + v\right) + 1 = 0$

Add the constraint $v = \frac{1}{u}$ to eliminate the term in $\left(u + v\right)$ and get:

${u}^{3} + \frac{1}{u} ^ 3 + 1 = 0$

Multiply through by ${u}^{3}$ to get the quadratic in ${u}^{3}$:

${\left({u}^{3}\right)}^{2} + \left({u}^{3}\right) + 1 = 0$

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

${u}^{3} = \omega \text{ }$ and ${u}^{3} = {\omega}^{2}$

where $\omega = - \frac{1}{2} + \frac{\sqrt{3}}{2}$ is the primitve Complex cube root of $1$

Hence the roots of our cubic are:

${x}_{1} = \sqrt[3]{\omega} + \frac{1}{\sqrt[3]{\omega}} = {\omega}^{\frac{1}{3}} + {\omega}^{\frac{8}{3}}$

${x}_{2} = \omega \sqrt[3]{\omega} + \frac{1}{\omega \sqrt[3]{\omega}} = {\omega}^{\frac{4}{3}} + {\omega}^{\frac{5}{3}}$

${x}_{3} = {\omega}^{2} \sqrt[3]{\omega} + \frac{1}{{\omega}^{2} \sqrt[3]{\omega}} = {\omega}^{\frac{7}{3}} + {\omega}^{\frac{2}{3}}$

Note that $\sqrt[3]{\omega}$ has minimum polynomial ${\Phi}_{9} = {x}^{6} + {x}^{3} + 1$

So to represent all three roots concurrently with rational matrices will almost certainly require $6 \times 6$ matrices.

First represent $\sqrt[3]{\omega}$ by the $3 \times 3$ matrix:

$\left(\begin{matrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ \omega & 0 & 0\end{matrix}\right)$

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

$\left(\begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{matrix}\right)$, $\left(\begin{matrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ \omega & 0 & 0\end{matrix}\right)$, $\left(\begin{matrix}0 & 0 & 1 \\ \omega & 0 & 0 \\ 0 & \omega & 0\end{matrix}\right)$

$\left(\begin{matrix}\omega & 0 & 0 \\ 0 & \omega & 0 \\ 0 & 0 & \omega\end{matrix}\right)$, $\left(\begin{matrix}0 & \omega & 0 \\ 0 & 0 & \omega \\ {\omega}^{2} & 0 & 0\end{matrix}\right)$, $\left(\begin{matrix}0 & 0 & \omega \\ {\omega}^{2} & 0 & 0 \\ 0 & {\omega}^{2} & 0\end{matrix}\right)$

$\left(\begin{matrix}{\omega}^{2} & 0 & 0 \\ 0 & {\omega}^{2} & 0 \\ 0 & 0 & {\omega}^{2}\end{matrix}\right)$, $\left(\begin{matrix}0 & {\omega}^{2} & 0 \\ 0 & 0 & {\omega}^{2} \\ 1 & 0 & 0\end{matrix}\right)$, $\left(\begin{matrix}0 & 0 & {\omega}^{2} \\ 1 & 0 & 0 \\ 0 & 1 & 0\end{matrix}\right)$

which makes the roots:

${x}_{1} = \left(\begin{matrix}0 & 1 & {\omega}^{2} \\ 1 & 0 & 1 \\ \omega & 1 & 0\end{matrix}\right)$

${x}_{2} = \left(\begin{matrix}0 & \omega & \omega \\ {\omega}^{2} & 0 & \omega \\ {\omega}^{2} & {\omega}^{2} & 0\end{matrix}\right)$

${x}_{3} = \left(\begin{matrix}0 & {\omega}^{2} & 1 \\ \omega & 0 & {\omega}^{2} \\ 1 & \omega & 0\end{matrix}\right)$

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

$\omega = \left(\begin{matrix}0 & - 1 \\ 1 & - 1\end{matrix}\right)$

${\omega}^{2} = \left(\begin{matrix}- 1 & 1 \\ - 1 & 0\end{matrix}\right)$

Along with $0 = \left(\begin{matrix}0 & 0 \\ 0 & 0\end{matrix}\right)$ and $1 = \left(\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right)$ we can substitute these values into our $3 \times 3$ matrices to get:

${x}_{1} = \left(\begin{matrix}\left(\begin{matrix}0 & 0 \\ 0 & 0\end{matrix}\right) \text{ & ((10)(01))"" & ((-1 1)(-10))"" \\ ((10)(01))"" & ((0 0)(00))"" & ((10)(01))"" \\ ((0 -1)(1 -1))"" & ((10)(01))"" & ((0 0)(00))}\end{matrix}\right) = \left(\begin{matrix}0 & 0 & 1 & 0 & - 1 & 1 \\ 0 & 0 & 0 & 1 & - 1 & 0 \\ 1 & 0 & 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 & 0 & 1 \\ 0 & - 1 & 1 & 0 & 0 & 0 \\ 1 & - 1 & 0 & 1 & 0 & 0\end{matrix}\right)$

${x}_{2} = \left(\begin{matrix}\left(\begin{matrix}0 & 0 \\ 0 & 0\end{matrix}\right) \text{ & ((0 -1)(1 -1))"" & ((0 -1)(1 -1))"" \\ ((-1 1)(-10))"" & ((0 0)(00))"" & ((0 -1)(1 -1))"" \\ ((-1 1)(-10))"" & ((-1 1)(-10))"" & ((0 0)(00))}\end{matrix}\right) = \left(\begin{matrix}0 & 0 & 0 & - 1 & 0 & - 1 \\ 0 & 0 & 1 & - 1 & 1 & - 1 \\ - 1 & 1 & 0 & 0 & 0 & - 1 \\ - 1 & 0 & 0 & 0 & 1 & - 1 \\ - 1 & 1 & - 1 & 1 & 0 & 0 \\ - 1 & 0 & - 1 & 0 & 0 & 0\end{matrix}\right)$

${x}_{3} = \left(\begin{matrix}\left(\begin{matrix}0 & 0 \\ 0 & 0\end{matrix}\right) \text{ & ((-1 1)(-10))"" & ((10)(01))"" \\ ((0 -1)(1 -1))"" & ((0 0)(00))"" & ((-1 1)(-10))"" \\ ((10)(01))"" & ((0 -1)(1 -1))"" & ((00)(00))}\end{matrix}\right) = \left(\begin{matrix}0 & 0 & - 1 & 1 & 1 & 0 \\ 0 & 0 & - 1 & 0 & 0 & 1 \\ 0 & - 1 & 0 & 0 & - 1 & 1 \\ 1 & - 1 & 0 & 0 & - 1 & 0 \\ 1 & 0 & 0 & - 1 & 0 & 0 \\ 0 & 1 & 1 & - 1 & 0 & 0\end{matrix}\right)$

Dec 19, 2016

Here's another way to construct rational $6 \times 6$ matrices for the $3$ roots...

#### Explanation:

Another way to construct $6 \times 6$ rational solutions is to start with the companion matrix of ${\Phi}_{9} = {x}^{6} + {x}^{3} + 1$ which is the minimal polynomial of $\sqrt[3]{\omega}$

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

${x}_{1} = {\omega}^{\frac{1}{3}} + {\omega}^{\frac{8}{3}}$

${x}_{2} = {\omega}^{\frac{4}{3}} + {\omega}^{\frac{5}{3}}$

${x}_{3} = {\omega}^{\frac{7}{3}} + {\omega}^{\frac{2}{3}}$

we will be able to construct rational matrix solutions from a matrix which behaves like $\sqrt[3]{\omega}$

The companion matrix of ${\Phi}_{9}$ is:

$A = \left(\begin{matrix}0 & 0 & 0 & 0 & 0 & - 1 \\ 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & - 1 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0\end{matrix}\right)$

Then:

${A}^{2} = \left(\begin{matrix}0 & 0 & 0 & 0 & - 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & - 1 \\ 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & - 1 & 0 \\ 0 & 0 & 1 & 0 & 0 & - 1 \\ 0 & 0 & 0 & 1 & 0 & 0\end{matrix}\right)$

${A}^{3} = \left(\begin{matrix}0 & 0 & 0 & - 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & - 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & - 1 \\ 1 & 0 & 0 & - 1 & 0 & 0 \\ 0 & 1 & 0 & 0 & - 1 & 0 \\ 0 & 0 & 1 & 0 & 0 & - 1\end{matrix}\right)$

${A}^{4} = \left(\begin{matrix}0 & 0 & - 1 & 0 & 0 & 1 \\ 0 & 0 & 0 & - 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & - 1 & 0 \\ 0 & 0 & - 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & - 1 & 0 & 0 \\ 0 & 1 & 0 & 0 & - 1 & 0\end{matrix}\right)$

${A}^{5} = \left(\begin{matrix}0 & - 1 & 0 & 0 & 1 & 0 \\ 0 & 0 & - 1 & 0 & 0 & 1 \\ 0 & 0 & 0 & - 1 & 0 & 0 \\ 0 & - 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & - 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & - 1 & 0 & 0\end{matrix}\right)$

${A}^{6} = \left(\begin{matrix}- 1 & 0 & 0 & 1 & 0 & 0 \\ 0 & - 1 & 0 & 0 & 1 & 0 \\ 0 & 0 & - 1 & 0 & 0 & 1 \\ - 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & - 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & - 1 & 0 & 0 & 0\end{matrix}\right)$

${A}^{7} = \left(\begin{matrix}0 & 0 & 1 & 0 & 0 & 0 \\ - 1 & 0 & 0 & 1 & 0 & 0 \\ 0 & - 1 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \\ - 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & - 1 & 0 & 0 & 0 & 0\end{matrix}\right)$

${A}^{8} = \left(\begin{matrix}0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ - 1 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \\ - 1 & 0 & 0 & 0 & 0 & 0\end{matrix}\right)$

Then the roots of ${x}^{3} - 3 x + 1 = 0$ are:

${x}_{1} = {A}^{1} + {A}^{8} = \left(\begin{matrix}0 & 1 & 0 & 0 & 0 & - 1 \\ 1 & 0 & 1 & 0 & 0 & 0 \\ - 1 & 1 & 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 & 1 & - 1 \\ 0 & 0 & 0 & 1 & 0 & 1 \\ - 1 & 0 & 0 & 0 & 1 & 0\end{matrix}\right)$

${x}_{2} = {A}^{4} + {A}^{5} = \left(\begin{matrix}0 & - 1 & - 1 & 0 & 1 & 1 \\ 0 & 0 & - 1 & - 1 & 0 & 1 \\ 0 & 0 & 0 & - 1 & - 1 & 0 \\ 0 & - 1 & - 1 & 0 & 0 & 0 \\ 1 & 0 & - 1 & - 1 & 0 & 0 \\ 1 & 1 & 0 & - 1 & - 1 & 0\end{matrix}\right)$

${x}_{3} = {A}^{7} + {A}^{2} = \left(\begin{matrix}0 & 0 & 1 & 0 & - 1 & 0 \\ - 1 & 0 & 0 & 1 & 0 & - 1 \\ 1 & - 1 & 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 & - 1 & 1 \\ - 1 & 0 & 1 & 0 & 0 & - 1 \\ 0 & - 1 & 0 & 1 & 0 & 0\end{matrix}\right)$