quaternion inverseQuaternionInverse2026-08-23 19:19:402026-08-23 19:24:34Definitionconverted to passive rotation conventionquaternioninversereciprocaldivisionunit quaternion% conjugatenormHamilton productnoncommutative algebra% this is the default PhysicsLibrary preamble. as your knowledge
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% define commands here\section*{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
\begin{equation}
\boxed{qq^{-1}=q^{-1}q=1.}
\end{equation}
The explicit formula is
\begin{equation}
\boxed{q^{-1}=\frac{q^*}{\lVert q\rVert^2}.}
\end{equation}
\section{Derivation from conjugate and norm}
The quaternion conjugate and quaternion norm satisfy
\begin{equation}
qq^*=q^*q=\lVert q\rVert^2.
\end{equation}
For $q\neq0$, the real scalar $\lVert q\rVert^2$ is nonzero. Dividing by it
gives
\begin{equation}
q\frac{q^*}{\lVert q\rVert^2}
=\frac{qq^*}{\lVert q\rVert^2}=1,
\end{equation}
and likewise
\begin{equation}
\frac{q^*}{\lVert q\rVert^2}q
=\frac{q^*q}{\lVert q\rVert^2}=1.
\end{equation}
Therefore equation (2) is both a left and a right inverse.
In components,
\begin{equation}
\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}.
}
\end{equation}
The zero quaternion has no multiplicative inverse.
\section{Uniqueness of the inverse}
Suppose $r$ and $s$ are both inverses of $q$. Then
\begin{equation}
r=r(qs)=(rq)s=s.
\end{equation}
Thus a nonzero quaternion has exactly one two-sided multiplicative inverse.
\section{Inverse of a product}
As with conjugation, the inverse of a product reverses the order of the factors:
\begin{equation}
\boxed{(pq)^{-1}=q^{-1}p^{-1}.}
\end{equation}
Indeed,
\begin{equation}
(pq)(q^{-1}p^{-1})=p(qq^{-1})p^{-1}=1,
\end{equation}
and
\begin{equation}
(q^{-1}p^{-1})(pq)=q^{-1}(p^{-1}p)q=1.
\end{equation}
For several factors,
\begin{equation}
\boxed{
(q_1q_2\cdots q_n)^{-1}
=q_n^{-1}\cdots q_2^{-1}q_1^{-1}.
}
\end{equation}
\section{Unit quaternions}
If $q$ is a unit quaternion,
\[
\lVert q\rVert=1,
\]
then the inverse formula simplifies to
\begin{equation}
\boxed{q^{-1}=q^*.}
\end{equation}
This identity is fundamental in rotation applications because a unit
quaternion can be inverted simply by changing the sign of its vector part.
\section{Pure quaternions}
For a pure quaternion $\mathbf v$,
\[
\mathbf v^*=-\mathbf v.
\]
Also,
\begin{equation}
\mathbf v^2=-\lVert\mathbf v\rVert^2.
\end{equation}
Therefore, for $\mathbf v\neq0$,
\begin{equation}
\boxed{
\mathbf v^{-1}
=-\frac{\mathbf v}{\lVert\mathbf v\rVert^2}.
}
\end{equation}
For a unit pure quaternion $\mathbf u$,
\begin{equation}
\boxed{\mathbf u^{-1}=-\mathbf u.}
\end{equation}
\section{Quaternion division is side-dependent}
A fraction such as $b/a$ is ambiguous for quaternions unless the side of
multiplication is defined.
If
\begin{equation}
a x=b,
\end{equation}
then multiply on the left by $a^{-1}$:
\begin{equation}
\boxed{x=a^{-1}b.}
\end{equation}
But if
\begin{equation}
x a=b,
\end{equation}
then multiply on the right:
\begin{equation}
\boxed{x=ba^{-1}.}
\end{equation}
In general,
\begin{equation}
\boxed{a^{-1}b\neq ba^{-1}.}
\end{equation}
For this reason, engineering derivations should normally write quaternion
inverses explicitly rather than use an undefined quotient notation.
\section{Worked examples}
\subsection*{Example 1: general inverse}
Let
\[
q=2+\mathbf i-2\mathbf j+2\mathbf k.
\]
Then
\[
q^*=2-\mathbf i+2\mathbf j-2\mathbf k,
\]
and
\[
\lVert q\rVert^2=13.
\]
Hence
\[
\boxed{
q^{-1}=\frac{1}{13}
\left(2-\mathbf i+2\mathbf j-2\mathbf k\right).
}
\]
The check is immediate:
\[
qq^{-1}=\frac{qq^*}{13}=1.
\]
\subsection*{Example 2: left and right equations differ}
Consider
\[
(1+\mathbf i)x=\mathbf j.
\]
Since
\[
(1+\mathbf i)^{-1}=\frac{1-\mathbf i}{2},
\]
\[
\begin{aligned}
x
&=\frac{1-\mathbf i}{2}\mathbf j\\
&=\frac{\mathbf j-\mathbf k}{2}.
\end{aligned}
\]
Now consider
\[
x(1+\mathbf i)=\mathbf j.
\]
Then
\[
\begin{aligned}
x
&=\mathbf j\frac{1-\mathbf i}{2}\\
&=\frac{\mathbf j+\mathbf k}{2},
\end{aligned}
\]
because $\mathbf j\mathbf i=-\mathbf k$.
\subsection*{Example 3: unit quaternion inverse}
For
\[
q=\frac{1}{\sqrt2}(1,1,0,0),
\]
we have $\lVert q\rVert=1$, so
\[
\boxed{
q^{-1}=q^*=\frac{1}{\sqrt2}(1,-1,0,0).
}
\]
\section{Common pitfalls}
\begin{enumerate}
\item \textbf{Using $q^{-1}=q^*$ for a nonunit quaternion.}
The general formula requires division by $\lVert q\rVert^2$.
\item \textbf{Forgetting that inverse order reverses.}
\[
(pq)^{-1}=q^{-1}p^{-1}.
\]
\item \textbf{Treating quaternion division like scalar division.}
The equations $ax=b$ and $xa=b$ generally have different solutions.
\item \textbf{Attempting to invert the zero quaternion.}
Its norm is zero, so the inverse formula is undefined.
\item \textbf{Confusing storage layout with inverse convention.}
Scalar-first versus scalar-last storage does not change the abstract inverse.
\end{enumerate}
\section{Exercises}
\begin{enumerate}
\item Find the inverse of
\[
q=3-2\mathbf i+\mathbf j+4\mathbf k.
\]
\item Find the inverse of
\[
q=1+\mathbf i+\mathbf j+\mathbf k
\]
and verify both $qq^{-1}=1$ and $q^{-1}q=1$.
\item Prove
\[
(pq)^{-1}=q^{-1}p^{-1}.
\]
\item Prove
\[
(pqr)^{-1}=r^{-1}q^{-1}p^{-1}.
\]
\item Solve both
\[
(1+\mathbf i)x=\mathbf j,
\qquad
x(1+\mathbf i)=\mathbf j.
\]
\item Let
\[
\mathbf v=2\mathbf i-\mathbf j+2\mathbf k.
\]
Find $\mathbf v^{-1}$ and verify $\mathbf v\mathbf v^{-1}=1$.
\item A student claims that the inverse of $q=2+\mathbf i$ is
$q^*=2-\mathbf i$. Identify the error and compute the correct inverse.
\item If $q$ is a unit quaternion, prove that $q^{-1}=q^*$.
\end{enumerate}
\section{Solutions}
\subsection*{1. General inverse}
\[
q^*=3+2\mathbf i-\mathbf j-4\mathbf k,
\qquad
\lVert q\rVert^2=30.
\]
Therefore
\[
\boxed{
q^{-1}=\frac{3+2\mathbf i-\mathbf j-4\mathbf k}{30}.
}
\]
\subsection*{2. Inverse of $1+\mathbf i+\mathbf j+\mathbf k$}
\[
q^*=1-\mathbf i-\mathbf j-\mathbf k,
\qquad
\lVert q\rVert^2=4.
\]
Thus
\[
\boxed{q^{-1}=\frac14(1-\mathbf i-\mathbf j-\mathbf k).}
\]
Since $qq^*=q^*q=4$,
\[
qq^{-1}=1,
\qquad
q^{-1}q=1.
\]
\subsection*{3. Inverse of a product}
\[
(pq)(q^{-1}p^{-1})=p(qq^{-1})p^{-1}=1,
\]
and similarly on the left. Hence
\[
\boxed{(pq)^{-1}=q^{-1}p^{-1}.}
\]
\subsection*{4. Three factors}
Apply the two-factor rule twice:
\[
(pqr)^{-1}=r^{-1}(pq)^{-1}=r^{-1}q^{-1}p^{-1}.
\]
\subsection*{5. Left and right equations}
Since
\[
(1+\mathbf i)^{-1}=\frac{1-\mathbf i}{2},
\]
the first equation gives
\[
\boxed{x=\frac{\mathbf j-\mathbf k}{2}},
\]
whereas the second gives
\[
\boxed{x=\frac{\mathbf j+\mathbf k}{2}}.
\]
The difference comes from noncommutativity.
\subsection*{6. Pure quaternion inverse}
\[
\lVert\mathbf v\rVert^2=4+1+4=9.
\]
Hence
\[
\boxed{
\mathbf v^{-1}=-\frac{2\mathbf i-\mathbf j+2\mathbf k}{9}.
}
\]
Since $\mathbf v^2=-9$,
\[
\mathbf v\mathbf v^{-1}=1.
\]
\subsection*{7. Conjugate is not generally the inverse}
For $q=2+\mathbf i$,
\[
\lVert q\rVert^2=5.
\]
Therefore
\[
\boxed{q^{-1}=\frac{2-\mathbf i}{5}.}
\]
The student's expression omitted the squared-norm denominator.
\subsection*{8. Unit quaternion}
If $\lVert q\rVert=1$, then
\[
q^{-1}=\frac{q^*}{\lVert q\rVert^2}=q^*.
\]
\section{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.
\section{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.
\begin{thebibliography}{9}
\bibitem{Hamilton1899}
W. R. Hamilton,
\emph{Elements of Quaternions}, 2nd ed., Vol. I,
edited by C. J. Joly,
Longmans, Green, and Co., London, 1899.
Public-domain historical source.
\PMlinkexternal{Internet Archive copy}{https://archive.org/details/elementsofquater01hamiuoft}
\bibitem{Tait1890}
P. G. Tait,
\emph{An Elementary Treatise on Quaternions}, 3rd ed.,
Cambridge University Press, Cambridge, 1890.
Public-domain historical source.
\bibitem{Joly1905}
C. J. Joly,
\emph{A Manual of Quaternions},
Macmillan and Co., London, 1905.
Public-domain historical source.
\bibitem{Hathaway1896}
A. S. Hathaway,
\emph{A Primer of Quaternions}, 1896.
Public-domain text; Project Gutenberg edition and \LaTeX{} source available.
\PMlinkexternal{Project Gutenberg edition}{https://www.gutenberg.org/ebooks/9934}
\end{thebibliography}