Quaternion Inverse

Every nonzero quaternion has a unique multiplicative inverse. The inverse is built directly from the quaternion conjugate and quaternion norm and is the quaternion analogue of a reciprocal. Because quaternion multiplication is noncommutative, the inverse also makes clear why left and right division must be distinguished.

For a nonzero quaternion $q$ , the inverse $q^{-1}$ is defined by

$\displaystyle \boxed{qq^{-1}=q^{-1}q=1.}$

(1)

The explicit formula is

$\displaystyle \boxed{q^{-1}=\frac{q^*}{\lVert q\rVert^2}.}$

(2)

Derivation from conjugate and norm

The quaternion conjugate and quaternion norm satisfy

$\displaystyle qq^*=q^*q=\lVert q\rVert^2.$

(3)

For $q\neq0$ , the real scalar $\lVert q\rVert^2$ is nonzero. Dividing by it gives

$\displaystyle q\frac{q^*}{\lVert q\rVert^2} =\frac{qq^*}{\lVert q\rVert^2}=1,$

(4)

and likewise

$\displaystyle \frac{q^*}{\lVert q\rVert^2}q =\frac{q^*q}{\lVert q\rVert^2}=1.$

(5)

Therefore equation (2) is both a left and a right inverse.

In components,

$\displaystyle \boxed{ q^{-1} = \frac{q_w-q_x\mathbf i-q_y\mathbf j-q_z\mathbf k} {q_w^2+q_x^2+q_y^2+q_z^2}. }$

(6)

The zero quaternion has no multiplicative inverse.

Uniqueness of the inverse

Suppose $r$ and $s$ are both inverses of $q$ . Then

$\displaystyle r=r(qs)=(rq)s=s.$

(7)

Thus a nonzero quaternion has exactly one two-sided multiplicative inverse.

Inverse of a product

As with conjugation, the inverse of a product reverses the order of the factors:

$\displaystyle \boxed{(pq)^{-1}=q^{-1}p^{-1}.}$

(8)

Indeed,

$\displaystyle (pq)(q^{-1}p^{-1})=p(qq^{-1})p^{-1}=1,$

(9)

and

$\displaystyle (q^{-1}p^{-1})(pq)=q^{-1}(p^{-1}p)q=1.$

(10)

For several factors,

$\displaystyle \boxed{ (q_1q_2\cdots q_n)^{-1} =q_n^{-1}\cdots q_2^{-1}q_1^{-1}. }$

(11)

Unit quaternions

If $q$ is a unit quaternion,

$\displaystyle \lVert q\rVert=1,$

then the inverse formula simplifies to

$\displaystyle \boxed{q^{-1}=q^*.}$

(12)

This identity is fundamental in rotation applications because a unit quaternion can be inverted simply by changing the sign of its vector part.

Pure quaternions

For a pure quaternion $\mathbf v$ ,

$\displaystyle \mathbf v^*=-\mathbf v.$

Also,

$\displaystyle \mathbf v^2=-\lVert\mathbf v\rVert^2.$

(13)

Therefore, for $\mathbf v\neq0$ ,

$\displaystyle \boxed{ \mathbf v^{-1} =-\frac{\mathbf v}{\lVert\mathbf v\rVert^2}. }$

(14)

For a unit pure quaternion $\mathbf u$ ,

$\displaystyle \boxed{\mathbf u^{-1}=-\mathbf u.}$

(15)

Quaternion division is side-dependent

A fraction such as $b/a$ is ambiguous for quaternions unless the side of multiplication is defined.

$\displaystyle a x=b,$

(16)

then multiply on the left by $a^{-1}$ :

$\displaystyle \boxed{x=a^{-1}b.}$

(17)

But if

$\displaystyle x a=b,$

(18)

then multiply on the right:

$\displaystyle \boxed{x=ba^{-1}.}$

(19)

In general,

$\displaystyle \boxed{a^{-1}b\neq ba^{-1}.}$

(20)

For this reason, engineering derivations should normally write quaternion inverses explicitly rather than use an undefined quotient notation.

Worked examples

Example 1: general inverse

Let

$\displaystyle q=2+\mathbf i-2\mathbf j+2\mathbf k.$

Then

$\displaystyle q^*=2-\mathbf i+2\mathbf j-2\mathbf k,$

and

$\displaystyle \lVert q\rVert^2=13.$

Hence

$\displaystyle \boxed{ q^{-1}=\frac{1}{13} \left(2-\mathbf i+2\mathbf j-2\mathbf k\right). }$

The check is immediate:

$\displaystyle qq^{-1}=\frac{qq^*}{13}=1.$

Example 2: left and right equations differ

Consider

$\displaystyle (1+\mathbf i)x=\mathbf j.$

Since

$\displaystyle (1+\mathbf i)^{-1}=\frac{1-\mathbf i}{2},$

$\begin{equation*} \begin{aligned} x &=\frac{1-\mathbf i}{2}\mathbf j\ &=\frac{\mathbf j-\mathbf k}{2}. \end{aligned}\end{equation*}$

Now consider

$\displaystyle x(1+\mathbf i)=\mathbf j.$

Then

$\begin{equation*} \begin{aligned} x &=\mathbf j\frac{1-\mathbf i}{2}\ &=\frac{\mathbf j+\mathbf k}{2}, \end{aligned}\end{equation*}$

because $\mathbf j\mathbf i=-\mathbf k$ .

Example 3: unit quaternion inverse

For

$\displaystyle q=\frac{1}{\sqrt2}(1,1,0,0),$

we have $\lVert q\rVert=1$ , so

$\displaystyle \boxed{ q^{-1}=q^*=\frac{1}{\sqrt2}(1,-1,0,0). }$

Common pitfalls

Using $q^{-1}=q^*$ for a nonunit quaternion. The general formula requires division by $\lVert q\rVert^2$ .
Forgetting that inverse order reverses.

$\displaystyle (pq)^{-1}=q^{-1}p^{-1}.$
Treating quaternion division like scalar division. The equations and generally have different solutions.
Attempting to invert the zero quaternion. Its norm is zero, so the inverse formula is undefined.
Confusing storage layout with inverse convention. Scalar-first versus scalar-last storage does not change the abstract inverse.

Exercises

Find the inverse of

$\displaystyle q=3-2\mathbf i+\mathbf j+4\mathbf k.$
Find the inverse of

$\displaystyle q=1+\mathbf i+\mathbf j+\mathbf k$
and verify both $qq^{-1}=1$ and $q^{-1}q=1$ .
Prove

$\displaystyle (pq)^{-1}=q^{-1}p^{-1}.$
Prove

$\displaystyle (pqr)^{-1}=r^{-1}q^{-1}p^{-1}.$
Solve both

$\displaystyle (1+\mathbf i)x=\mathbf j, \qquad x(1+\mathbf i)=\mathbf j.$
Let

$\displaystyle \mathbf v=2\mathbf i-\mathbf j+2\mathbf k.$
Find $\mathbf v^{-1}$ and verify $\mathbf v\mathbf v^{-1}=1$ .
A student claims that the inverse of $q=2+\mathbf i$ is $q^*=2-\mathbf i$ . Identify the error and compute the correct inverse.
If is a unit quaternion, prove that $q^{-1}=q^*$ .

Solutions

1. General inverse

$\displaystyle q^*=3+2\mathbf i-\mathbf j-4\mathbf k, \qquad \lVert q\rVert^2=30.$

Therefore

$\displaystyle \boxed{ q^{-1}=\frac{3+2\mathbf i-\mathbf j-4\mathbf k}{30}. }$

2. Inverse of $1+\mathbf i+\mathbf j+\mathbf k$

$\displaystyle q^*=1-\mathbf i-\mathbf j-\mathbf k, \qquad \lVert q\rVert^2=4.$

Thus

$\displaystyle \boxed{q^{-1}=\frac14(1-\mathbf i-\mathbf j-\mathbf k).}$

Since

$\displaystyle qq^{-1}=1, \qquad q^{-1}q=1.$

3. Inverse of a product

$\displaystyle (pq)(q^{-1}p^{-1})=p(qq^{-1})p^{-1}=1,$

and similarly on the left. Hence

$\displaystyle \boxed{(pq)^{-1}=q^{-1}p^{-1}.}$

4. Three factors

Apply the two-factor rule twice:

$\displaystyle (pqr)^{-1}=r^{-1}(pq)^{-1}=r^{-1}q^{-1}p^{-1}.$

5. Left and right equations

Since

$\displaystyle (1+\mathbf i)^{-1}=\frac{1-\mathbf i}{2},$

the first equation gives

$\displaystyle \boxed{x=\frac{\mathbf j-\mathbf k}{2}},$

whereas the second gives

$\displaystyle \boxed{x=\frac{\mathbf j+\mathbf k}{2}}.$

The difference comes from noncommutativity.

6. Pure quaternion inverse

$\displaystyle \lVert\mathbf v\rVert^2=4+1+4=9.$

Hence

$\displaystyle \boxed{ \mathbf v^{-1}=-\frac{2\mathbf i-\mathbf j+2\mathbf k}{9}. }$

Since $\mathbf v^2=-9$ ,

$\displaystyle \mathbf v\mathbf v^{-1}=1.$

7. Conjugate is not generally the inverse

For $q=2+\mathbf i$ ,

$\displaystyle \lVert q\rVert^2=5.$

Therefore

$\displaystyle \boxed{q^{-1}=\frac{2-\mathbf i}{5}.}$

The student's expression omitted the squared-norm denominator.

8. Unit quaternion

If $\lVert q\rVert=1$ , then

$\displaystyle q^{-1}=\frac{q^*}{\lVert q\rVert^2}=q^*.$

What comes next

Conjugate, norm, and inverse are now available as separate linked definitions. The next PhysicsLibrary article develops axis–angle representation and unit quaternions, including the half-angle form used for rigid-body rotations.

Sources and historical notes

Hamilton, Tait, Joly, and Hathaway all derive reciprocal or inverse operations from the conjugate and magnitude. Historical texts also emphasize that the order of factors reverses under reciprocation. The notation, engineering interpretation, worked examples, exercises, and solutions here are newly written.

Bibliography

1: W. R. Hamilton, Elements of Quaternions, 2nd ed., Vol. I, edited by C. J. Joly, Longmans, Green, and Co., London, 1899. Public-domain historical source. Internet Archive copy
2: P. G. Tait, An Elementary Treatise on Quaternions, 3rd ed., Cambridge University Press, Cambridge, 1890. Public-domain historical source.
3: C. J. Joly, A Manual of Quaternions, Macmillan and Co., London, 1905. Public-domain historical source.
4: A. S. Hathaway, A Primer of Quaternions, 1896. Public-domain text; Project Gutenberg edition and L^ATEX source available. Project Gutenberg edition

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