User:Phy1729/sandbox

a is assumed to be the identity matrix

we start with the 2x2 matrices that square to -1


 * $$\begin{bmatrix}

\pm i & 0 \\ 0 & \pm i  \end{bmatrix} \begin{bmatrix} 0 & 1 \\  -1 & 0   \end{bmatrix} \begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix}$$

let i be the first j the second and k the third thus


 * $$\begin{bmatrix}

a \pm bi & c + di \\ -c + di & a \pm bi  \end{bmatrix}$$

to find the sign of b multiply j and k to result in the matrix of i


 * $$\begin{bmatrix}

0 & 1 \\  -1 & 0   \end{bmatrix} \begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix} =\begin{bmatrix} i & 0 \\ 0 & -i \end{bmatrix}$$

from this we find that the correct 2x2 matrix is


 * $$\begin{bmatrix}

a + bi & c + di \\ -c + di & a - bi  \end{bmatrix}$$

converting to a 4x4 we have


 * $$\begin{bmatrix}

a & b & c & d \\ -b & a & -d & c \\ -c & d & a & -b \\ -d & -c & b & a \end{bmatrix}$$

which does not commute

on the other hand if one extrapolates the complex 2x2 matrix into a quaternion 4x4 the result is


 * $$\begin{bmatrix}

a & b \\ -b & a  \end{bmatrix}$$
 * $$\begin{bmatrix}

\begin{bmatrix} a & b \\ -b & a  \end{bmatrix} & \begin{bmatrix} a & b \\ -b & a  \end{bmatrix}  \\ -\begin{bmatrix} a & b \\ -b & a  \end{bmatrix} & \begin{bmatrix} a & b \\ -b & a  \end{bmatrix} \end{bmatrix}$$
 * $$\begin{bmatrix}

a & b & c & d \\ -b & a & -d & c \\ -c & -d & a & b \\ d & -c & -b & a  \end{bmatrix}$$

taking the k 4x4 matrix and squaring it results in


 * $$\begin{bmatrix}

0 & 0 & 0 & 1 \\  0 & 0 & -1 & 0  \\  0 & -1 & 0 & 0  \\  1 & 0 & 0 & 0   \end{bmatrix}^2$$
 * $$\begin{bmatrix}

1 & 0 & 0 & 0 \\  0 & 1 & 0 & 0  \\  0 & 0 & 1 & 0  \\  0 & 0 & 0 & 1   \end{bmatrix}$$

which lo and behold is the identity matrix which stands 1 which means that $$k^2 = 1$$ when the complex 2x2 matrix is extrapolated. This quaternion system does commute. If one were to form systems where $$i^2=k^2=-1$$ and $$j^2=1$$ or $$j^2=k^2=-1$$ and $$i^2=1$$ they would also commute