Matrix Form of Quaternion

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Theorem

Let $\mathbf x$ be a quaternion such that:

$\mathbf x = a \mathbf 1 + b \mathbf i + c \mathbf j + d \mathbf k$

When the quaternion basis is expressed in the form of matrices:

$\mathbf 1 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \qquad \mathbf i = \begin{bmatrix} i & 0 \\ 0 & -i \end{bmatrix} \qquad \mathbf j = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \qquad \mathbf k = \begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix}$

the general quaternion $\mathbf x$ has the form:

$\mathbf x = \begin{bmatrix} a + bi & c + di \\ -c + di & a - bi \end{bmatrix}$

or:

$\mathbf x = \begin{bmatrix} w & z \\ -\overline z & \overline w \end{bmatrix}$

where :

$w$ and $z$ are complex numbers
$\overline z$ is the complex conjugate of $z$.


Proof

\(\ds \mathbf x\) \(=\) \(\ds a \mathbf 1 + b \mathbf i + c \mathbf j + d \mathbf k\)
\(\ds \) \(=\) \(\ds a \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + b \begin{bmatrix} i & 0 \\ 0 & -i \end{bmatrix} + c \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} + d \begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix}\)
\(\ds \) \(=\) \(\ds \begin{bmatrix} a & 0 \\ 0 & a \end{bmatrix} + \begin{bmatrix} b i & 0 \\ 0 & - b i \end{bmatrix} + \begin{bmatrix} 0 & c \\ -c & 0 \end{bmatrix} + \begin{bmatrix} 0 & d i \\ d i & 0 \end{bmatrix}\)
\(\ds \) \(=\) \(\ds \begin{bmatrix} a + b i & c + d i \\ -c + d i & a - b i \end{bmatrix}\)

$\blacksquare$


Sources

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