# How do you find a 2xx2 matrix A with rational coefficients such that A^2+A+((1,0),(0,1)) = ((0,0),(0,0)) ?

Dec 10, 2016

$A = \left(\begin{matrix}- \frac{1}{2} - \frac{i \sqrt{3}}{2} & 0 \\ 0 & - \frac{1}{2} + \frac{i \sqrt{3}}{2}\end{matrix}\right)$

#### Explanation:

By Cayley-Hamilton, the $A$ characteristic polynomial is given by

$p \left(\lambda\right) = {\lambda}^{2} + \lambda + 1 = 0$ and the eigenvalues are

$\lambda = - \frac{1}{2} \pm \frac{i \sqrt{3}}{2}$ so the matrix

$A = \left(\begin{matrix}- \frac{1}{2} - \frac{i \sqrt{3}}{2} & 0 \\ 0 & - \frac{1}{2} + \frac{i \sqrt{3}}{2}\end{matrix}\right)$ is a solution.

Dec 10, 2016

I get $A = \left(\begin{matrix}- \frac{1}{2} & b \\ c & - \frac{1}{2}\end{matrix}\right)$ with $b , c \in \mathbb{Q}$ and $b c = - \frac{3}{4}$

#### Explanation:

I used $A = \left(\begin{matrix}a & b \\ c & d\end{matrix}\right)$

I get the system (equations numbered in the obvious way)

${a}^{2} + b c + a = - 1$
$a b + b d + b = 0$
$a c + c d + c = 0$
${d}^{2} + b c + d = - 1$

Subtracting equation 4 from equation 1 gets us

${a}^{2} + a - \left({d}^{2} + d\right) = 0$

${a}^{2} + a + \frac{1}{4} - \left({d}^{2} + d + \frac{1}{4}\right) = 0$

${\left(a + \frac{1}{2}\right)}^{2} - {\left(d + \frac{1}{2}\right)}^{2} = 0$

The only real solution is $a = d = - \frac{1}{2}$.

From equation 1 (0r eq 4) we now get $b c = - \frac{3}{4}$

At this point, I can find no other restrictions on $b$ and $c$.

So, I tried $b = 1$ and $c = - \frac{3}{4}$

Unless I made some silly error,

$A = \left(\begin{matrix}- \frac{1}{2} & 1 \\ - \frac{3}{4} & - \frac{1}{2}\end{matrix}\right)$ is a solution.

More generally,

If

$A = \left(\begin{matrix}- \frac{1}{2} & b \\ c & - \frac{1}{2}\end{matrix}\right)$ with $b c = - \frac{3}{4}$,

then

${A}^{2} = \left(\begin{matrix}- \frac{1}{2} & b \\ c & - \frac{1}{2}\end{matrix}\right) \left(\begin{matrix}- \frac{1}{2} & b \\ c & - \frac{1}{2}\end{matrix}\right) = \left(\begin{matrix}- \frac{1}{2} & - b \\ - c & - \frac{1}{2}\end{matrix}\right)$.

Now if we add $A$, we get

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

Dec 10, 2016

A couple of possible solutions:

$A = \left(\begin{matrix}- \frac{1}{2} & \frac{1}{2} \\ - \frac{3}{2} & - \frac{1}{2}\end{matrix}\right) \text{ }$ or $\text{ } A = \left(\begin{matrix}- \frac{1}{2} & - \frac{1}{2} \\ \frac{3}{2} & - \frac{1}{2}\end{matrix}\right)$

#### Explanation:

Just in case you have not encountered the notation, the set of all rational numbers is denoted by $\mathbb{Q}$.

First consider matrices of the form $\left(\begin{matrix}a & 0 \\ 0 & a\end{matrix}\right)$ where $a \in \mathbb{Q}$

If you add, subtract, multiply or divide (i.e. multiply by the multiplicative inverse), such matrices then you end up with a matrix of the same form.

Matrices of the form $\left(\begin{matrix}a & 0 \\ 0 & a\end{matrix}\right)$ where $a \in \mathbb{Q}$ are said to be closed under addition, multiplication, subtraction and multiplicative inverse (of non-zero elements). There is also an additive identity $\left(\begin{matrix}0 & 0 \\ 0 & 0\end{matrix}\right)$ and multiplicative identity $\left(\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right)$. Both addition and multiplication are commutative.

In other words, such matrices form a field. This field is isomorphic to $\mathbb{Q}$, that is there is an exact correspondence between $\mathbb{Q}$ and this field of matrices, given by the mapping:

$a \to \left(\begin{matrix}a & 0 \\ 0 & a\end{matrix}\right)$

which preserves the arithmetical structure.

Now consider the matrix:

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

See what happens if we square this matrix:

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

So $\left(\begin{matrix}0 & 1 \\ - 3 & 0\end{matrix}\right)$ is a "square root of $- 3$"

If we add this to our field of matrices, and add all other matrices required to make it closed under arithmetic operations, then we have the set of matrices of the form:

$\left(\begin{matrix}a & b \\ - 3 b & a\end{matrix}\right) \text{ }$ where $a , b \in \mathbb{Q}$

This corresponds to the Complex number $a + b \sqrt{- 3}$, behaving arithmetically just like it.

Now:

${x}^{2} + x + 1 = {\left(x + \frac{1}{2}\right)}^{2} - {\left(\frac{\sqrt{- 3}}{2}\right)}^{2}$

$\textcolor{w h i t e}{{x}^{2} + x + 1} = \left(x + \frac{1}{2} - \frac{\sqrt{- 3}}{2}\right) \left(x + \frac{1}{2} + \frac{\sqrt{- 3}}{2}\right)$

Hence zeros:

$x = - \frac{1}{2} \pm \frac{\sqrt{- 3}}{2}$

which correspond to the matrices:

$\left(\begin{matrix}- \frac{1}{2} & \frac{1}{2} \\ - \frac{3}{2} & - \frac{1}{2}\end{matrix}\right) \text{ }$ and $\text{ } \left(\begin{matrix}- \frac{1}{2} & - \frac{1}{2} \\ \frac{3}{2} & - \frac{1}{2}\end{matrix}\right)$

$\textcolor{w h i t e}{}$
Footnote

$\left(\begin{matrix}0 & 1 \\ - 3 & 0\end{matrix}\right)$ is only one of a family of matrices with rational coefficients with square $\left(\begin{matrix}- 3 & 0 \\ 0 & - 3\end{matrix}\right)$.

In fact any matrix of the following form will work:

$\left(\begin{matrix}0 & k \\ - \frac{3}{k} & 0\end{matrix}\right)$

and hence there are a family of solutions to the original problem.

Dec 11, 2016

One solution is the companion matrix $\left(\begin{matrix}0 & - 1 \\ 1 & - 1\end{matrix}\right)$

#### Explanation:

I think I was taught this about 35 years ago and had forgotten it...

$\textcolor{w h i t e}{}$
Companion matrix

Given a monic polynomial:

${x}^{n} + {a}_{n - 1} {x}^{n - 1} + {a}_{n - 2} {x}^{n - 2} + \ldots + {a}_{1} x + {a}_{0}$

The companion matrix is an $n \times n$ matrix of the form:

$C = \left(\begin{matrix}0 & 0 & \ldots & 0 & - {a}_{0} \\ 1 & 0 & \ldots & 0 & - {a}_{1} \\ 0 & 1 & \ldots & 0 & - {a}_{2} \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \ldots & 1 & - {a}_{n - 1}\end{matrix}\right)$

Then $C$ satisfies:

${C}^{n} + {a}_{n - 1} {C}^{n - 1} + {a}_{n - 2} {C}^{n - 2} + \ldots + {a}_{1} C + {a}_{0} = 0$

${x}^{2} + b x + c$

The companion matrix is:

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

So for our example:

${x}^{2} + x + 1$

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

$\textcolor{w h i t e}{}$
Footnote

I was interested in more complicated examples such as:

${x}^{5} + 4 x + 2 = 0$

The companion matrix for this example is:

$\left(\begin{matrix}0 & 0 & 0 & 0 & - 2 \\ 1 & 0 & 0 & 0 & - 4 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0\end{matrix}\right)$