Sum of Quaternion Conjugates

Theorem

Let $\mathbf x, \mathbf y \in \mathbb H$ be quaternions.

Let $\overline {\mathbf x}$ be the conjugate of $\mathbf x$.

Then:

$\overline {\mathbf x + \mathbf y} = \overline {\mathbf x} + \overline {\mathbf y}$

Let:

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

$\mathbf y = e \mathbf 1 + f \mathbf i + g \mathbf j + h \mathbf k$

Then:

$\ds \overline {\mathbf x + \mathbf y}$

$=$

$\ds \overline {\paren {a \mathbf 1 + b \mathbf i + c \mathbf j + d \mathbf k} + \paren {e \mathbf 1 + f \mathbf i + g \mathbf j + h \mathbf k} }$

$\ds $

$=$

$\ds \overline {\paren {a + e} \mathbf 1 + \paren {b + f} \mathbf i + \paren {c + g} \mathbf j + \paren {d + h} \mathbf k}$

$\ds $

$=$

$\ds \paren {a + e} \mathbf 1 - \paren {b + f} \mathbf i - \paren {c + g} \mathbf j - \paren {d + h} \mathbf k$

$\ds $

$=$

$\ds \paren {a \mathbf 1 - b \mathbf i - c \mathbf j - d \mathbf k} + \paren {e \mathbf 1 - f \mathbf i - g \mathbf j - h \mathbf k}$

$\ds $

$=$

$\ds \overline {\mathbf x} + \overline {\mathbf y}$

$\blacksquare$