Is this matrix decomposition possible?

Given a $2\times2$ matrix $S$ with entries in $\mathbb{Z}$ or $\mathbb{Q}$ , when is it possible to write $S=\frac{1}{3}(ABC+CAB+BCA)$ such that $A+B+C=0$, where $A, B, C$ are matrices over the same ground ring as S. Always? How would I find $A, B, C$?

Solution 1:

To challenge this knot, we have no choice but to replace any specific matrices. As one of the approaches, The rotation matrix is a part of correct answer. Because the matrix has special characteristics. As reasons, variables and equations can be reduced and commutativity is allowed under any two dimension. For example, the following figure shows that even if rotating operations are exchanged, eventual angles are corresponded as same direction. Therefore, this property can be applied to supply the better solution.

Initially, let replace $\{A,B,C\}$ with rotation matrices $\{R_{a},R_{b},R_{c}\}$. Further, rotation matrix can be handled to be identity with Euler formula:

$$ R_{a}=R(a)= \begin{bmatrix} \cos{(a)} & -\sin{(a)} \\ \sin{(a)} & \cos{(a)} \end{bmatrix} \equiv e^{ia} $$

Then the supplied top equation can be expressed by follows:

$$ \begin{aligned} 3S =& ABC+BCA+CAB \\ =& R_{a}R_{b}R_{c}+R_{b}R_{c}R_{a}+R_{c}R_{a}R_{b} \\ S=& R_{a}R_{b}R_{c} \\ \Leftrightarrow e^{i\phi}=& e^{ia} e^{ib} e^{ic} =e^{i(a+b+c)} \\ \end{aligned} \\ $$

where, $S\equiv e^{i\phi}$ and $\phi$ is phase degree. Next, the bottom constraint equation is:

$$ O=R_{a}+R_{b}+R_{c} \\ \Leftrightarrow 0=e^{ia} + e^{ib} + e^{ic} $$

Hence, two equations can be expressed by the following simultaneous equation:

$$ \begin{cases} a+b+c=\phi \\ e^{ia}+e^{ib}+e^{ic}=0 \end{cases} $$

We can solve to replace as $\delta_{1}=c-a$ and $\delta_{2}=c-b$. Eventually, the solutions are below (However, these parameters are limited on range of $[0 \ \ 2\pi + \phi]$):

$$ (a,b,c) =\biggl\{ \cfrac{\phi}{3}, \cfrac{\phi+2\pi}{3}, \cfrac{\phi+4\pi}{3} \biggr\} $$

Therefore, the top equation can be calculated as follows:

$$ \begin{aligned} S =& R_{a}R_{b}R_{c} \\ =& R_{\phi /3}(R_{\phi /3} R_{120^\circ})(R_{\phi /3} R_{240^\circ}) \\ =& R_{\phi /3}^3R_{(120+240)^\circ} \\ R_{\phi}=& R_{\phi/3}^3 \end{aligned} $$

This means that end phase is exactly three times of initial phase. This notion can be drawn on unit ring like following figure. Of course, these operations can be exchangeable as above explanation.

$\hspace{3cm}$

Next, we have to determine the matrix $S$. Then, its internal elements must be rational numbers or integers owing to sentence. To estimate these factors, the Pythagorean triple gives good clue. Following descriptions are initial four triples (Of course inverted ordering or sort sequence are also true. However except unit matrix):

$$ \cfrac{1}{5} \begin{bmatrix} 3 & -4 \\ 4 & 3 \end{bmatrix} ,\cfrac{1}{13} \begin{bmatrix} 5 & -12 \\ 12 & 5 \end{bmatrix} ,\cfrac{1}{17} \begin{bmatrix} 8 & -15 \\ 15 & 8 \end{bmatrix} ,\cdots $$

To satisfy these conditions, each matrices are applied:

$$ \begin{array} AA= & \cfrac{1}{5} \begin{bmatrix} 3 & -4 \\ 4 & 5 \end{bmatrix} ,& \cfrac{1}{13} \begin{bmatrix} 5 & -12 \\ 12 & 5 \end{bmatrix} ,&\cdots \\ B= & \cfrac{1}{10} \begin{bmatrix} -(4\sqrt{3}+3) & -(3\sqrt{3}-4) \\ -(3\sqrt{3}-4) & -(4\sqrt{3}+3) \end{bmatrix} ,& \cfrac{1}{26} \begin{bmatrix} -(12\sqrt{3}+5) & -(5\sqrt{3}-12) \\ 5\sqrt{3}-12 & -(12\sqrt{3}+5) \end{bmatrix} ,&\cdots \\ C= & \cfrac{1}{10} \begin{bmatrix} -(4\sqrt{3}-3) & -(3\sqrt{3}+4) \\ -(3\sqrt{3}+4) & -(4\sqrt{3}-3) \end{bmatrix} ,& \cfrac{1}{26} \begin{bmatrix} 12\sqrt{3}-5 & 5\sqrt{3}+12 \\ -(5\sqrt{3}+12) & 12\sqrt{3}-5 \end{bmatrix} ,&\cdots \\ S= & \cfrac{1}{125} \begin{bmatrix} -117 & -44 \\ 44 & -117 \end{bmatrix} ,& \cfrac{1}{2197} \begin{bmatrix} -2035 & 828 \\ -828 & -2035 \end{bmatrix} ,&\cdots \end{array} $$

As well known, Pythagorean triples exist endlessly so that these commutative matrices (and whole solutions) also exist infinity.