Quaternion Group/Order 4 Matrices

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Representation of Quaternion Group

Let $\mathbf I, \mathbf J, \mathbf K, \mathbf L$ denote the following four elements of the matrix space $\map {\MM_\Z} 4$:

\(\ds \mathbf I\) \(=\) \(\ds \begin {bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end {bmatrix}\)
\(\ds \mathbf J\) \(=\) \(\ds \begin {bmatrix} 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end {bmatrix}\)
\(\ds \mathbf K\) \(=\) \(\ds \begin {bmatrix} 0 & -1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & -1 & 0 \end {bmatrix}\)
\(\ds \mathbf L\) \(=\) \(\ds \begin {bmatrix} 0 & 0 & 0 & -1 \\ 0 & 0 & 1 & 0 \\ 0 & -1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end {bmatrix}\)

where $\Z$ is the set of integers.


The set:

$\Dic 2 = \set {\mathbf I, -\mathbf I, \mathbf J, -\mathbf J, \mathbf K, -\mathbf K, \mathbf L, -\mathbf L}$

under the operation of conventional matrix multiplication, forms the quaternion group.

This can be generated by the $2$ elements $\mathbf J$ and $\mathbf K$.


Proof

\(\ds \mathbf J^2\) \(=\) \(\ds \begin {bmatrix} 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end {bmatrix} \begin {bmatrix} 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end {bmatrix}\)
\(\ds \) \(=\) \(\ds \begin {bmatrix} -1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end {bmatrix}\)
\(\ds \) \(=\) \(\ds -\mathbf I\)


\(\ds \mathbf K^2\) \(=\) \(\ds \begin {bmatrix} 0 & -1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & -1 & 0 \end {bmatrix} \begin {bmatrix} 0 & -1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & -1 & 0 \end {bmatrix}\)
\(\ds \) \(=\) \(\ds \begin {bmatrix} -1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end {bmatrix}\)
\(\ds \) \(=\) \(\ds -\mathbf I\) \(\ds = \mathbf J^2\)


\(\ds \mathbf J^4\) \(=\) \(\ds \paren {-\mathbf J^2}^2\)
\(\ds \) \(=\) \(\ds \begin {bmatrix} -1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end {bmatrix} \begin {bmatrix} -1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end {bmatrix}\)
\(\ds \) \(=\) \(\ds \begin {bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end {bmatrix}\)
\(\ds \) \(=\) \(\ds \mathbf I\)


\(\ds \mathbf J \mathbf K\) \(=\) \(\ds \begin {bmatrix} 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end {bmatrix} \begin {bmatrix} 0 & -1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & -1 & 0 \end {bmatrix}\)
\(\ds \) \(=\) \(\ds \begin {bmatrix} 0 & 0 & 0 & -1 \\ 0 & 0 & 1 & 0 \\ 0 & -1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end {bmatrix}\)
\(\ds \) \(=\) \(\ds \mathbf L\)


\(\ds \mathbf J^3\) \(=\) \(\ds \mathbf J \paren {\mathbf J^2}\)
\(\ds \) \(=\) \(\ds \begin {bmatrix} 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end {bmatrix} \begin {bmatrix} -1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end {bmatrix}\)
\(\ds \) \(=\) \(\ds \begin {bmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ -1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \end {bmatrix}\)
\(\ds \) \(=\) \(\ds -\mathbf J\)


\(\ds \mathbf K^3\) \(=\) \(\ds \mathbf K \paren {\mathbf K^2}\)
\(\ds \) \(=\) \(\ds \begin {bmatrix} 0 & -1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & -1 & 0 \end {bmatrix} \begin {bmatrix} -1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end {bmatrix}\)
\(\ds \) \(=\) \(\ds \begin {bmatrix} 0 & 1 & 0 & 0 \\ -1 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 \\ 0 & 0 & 1 & 0 \end {bmatrix}\)
\(\ds \) \(=\) \(\ds -\mathbf K\)


\(\ds \mathbf K \mathbf J\) \(=\) \(\ds \begin {bmatrix} 0 & -1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & -1 & 0 \end {bmatrix} \begin {bmatrix} 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end {bmatrix}\)
\(\ds \) \(=\) \(\ds \begin {bmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & -1 & 0 \\ 0 & 1 & 0 & 0 \\ -1 & 0 & 0 & 0 \end {bmatrix}\)
\(\ds \) \(=\) \(\ds -\mathbf L\)


We note that:

\(\ds \paren {\mathbf J \mathbf K}^2\) \(=\) \(\ds \mathbf L^2\)
\(\ds \) \(=\) \(\ds \begin {bmatrix} 0 & 0 & 0 & -1 \\ 0 & 0 & 1 & 0 \\ 0 & -1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end {bmatrix} \begin {bmatrix} 0 & 0 & 0 & -1 \\ 0 & 0 & 1 & 0 \\ 0 & -1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end {bmatrix}\)
\(\ds \) \(=\) \(\ds \begin {bmatrix} -1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end {bmatrix}\)
\(\ds \) \(=\) \(\ds -\mathbf I\)

Hence we can construct the Cayley table:

$\begin{array}{r|rrrrrrrr}
     &  \mathbf I &  \mathbf J & -\mathbf I & -\mathbf J &  \mathbf K &  \mathbf L & -\mathbf K & -\mathbf L \\

\hline 

\mathbf I &  \mathbf I &  \mathbf J & -\mathbf I & -\mathbf J &  \mathbf K &  \mathbf L & -\mathbf K & -\mathbf L \\
\mathbf J &  \mathbf J & -\mathbf I & -\mathbf J &  \mathbf I &  \mathbf L & -\mathbf K & -\mathbf L &  \mathbf K \\

-\mathbf I & -\mathbf I & -\mathbf J & \mathbf I & \mathbf J & -\mathbf K & -\mathbf L & \mathbf K & \mathbf L \\ -\mathbf J & -\mathbf J & \mathbf I & \mathbf J & -\mathbf I & -\mathbf L & \mathbf K & \mathbf L & -\mathbf K \\ 

\mathbf K &  \mathbf K & -\mathbf L & -\mathbf K &  \mathbf L & -\mathbf I &  \mathbf J &  \mathbf I & -\mathbf J \\
\mathbf L &  \mathbf L &  \mathbf K & -\mathbf L & -\mathbf K & -\mathbf J & -\mathbf I &  \mathbf J &  \mathbf I \\

-\mathbf K & -\mathbf K & \mathbf L & \mathbf K & -\mathbf L & \mathbf I & -\mathbf J & -\mathbf I & \mathbf J \\ -\mathbf L & -\mathbf L & -\mathbf K & \mathbf L & \mathbf K & \mathbf J & \mathbf I & -\mathbf J & -\mathbf I \end{array}$

and it can be seen that this is an instance of the quaternion group.

$\blacksquare$


Also see

In Matrix Form of Quaternion it is shown that a general element $\mathbf x$ of $\mathbb H$ has the form:

$\mathbf x = \begin {bmatrix} a + b i & c + d i \\ -c + d i & a - b i \end {bmatrix}$


Sources

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