Multiplicative Inverse of Quaternion

Theorem
Let $$\mathbf x = a \mathbf 1 + b \mathbf i + c \mathbf j + d \mathbf k$$ be a quaternion such that $$\mathbf x \ne \mathbf 0$$.

Then $$\mathbf x$$ has an inverse $$\mathbf x^{-1}$$ under the operation of quaternion multiplication:


 * $$\mathbf x^{-1} = \lambda \overline{\mathbf x}$$

where:
 * $$\lambda = \frac 1 {a^2 + b^2 + c^2 + d^2} \mathbf 1$$

Proof
From Multiplicative Identity for Quaternions‎, we need to show that $$\lambda \overline{\mathbf x} \mathbf x = \mathbf 1 = \mathbf x \lambda \overline{\mathbf x}$$.

From Product of Quaternion with Conjugate we have that:


 * $$\overline{\mathbf x} \mathbf x = \left({a^2 + b^2 + c^2 + d^2}\right) \mathbf 1$$

Using the definition of $$\lambda$$ from above:

$$ $$ $$

Unless all of $$a, b, c, d = 0$$ we have that $$a^2 + b^2 + c^2 + d^2 \ne 0$$.

Therefore $$\lambda$$ is defined for all quaternions not equal to $$\mathbf 0$$.

It follows from Left Inverse for All is Right Inverse that $$\mathbf x \lambda \overline{\mathbf x} = \mathbf 1$$.

Hence the result.