example of quaternion inverse ExampleOfQuaternionInverse 2026-08-28 16:11:22 2026-08-28 16:11:22 Example quaternion inverse added keywords and updated Related % this is the default PhysicsLibrary preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics). % If you use \includegraphics{figure} with no file extension, PhysicsLibrary % will prefer figure.png before falling back to figure.eps. \usepackage{graphicx} \DeclareGraphicsExtensions{.png,.jpg,.jpeg,.pdf,.eps} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here \section*{Quaternion Inverse: Examples, Exercises, and Solutions} This entry is the self study companion to \emph{Quaternion Inverse}. The quaternion inverse combines the conjugate and norm into the quantity that undoes quaternion multiplication. For unit attitude quaternions the inverse reduces to the conjugate, but for a general nonzero quaternion the norm squared in the denominator is essential. All exercises are stated first. Complete worked solutions follow afterward. \section{Formula summary} For a nonzero quaternion \begin{equation} q = q_w + q_x\mathbf i + q_y\mathbf j + q_z\mathbf k, \end{equation} the inverse is \begin{equation} q^{-1} = \frac{q^*}{\lVert q\rVert^2}. \end{equation} Because \begin{equation} q q^* = q^*q = \lVert q\rVert^2, \end{equation} we have \begin{equation} q q^{-1} = q^{-1}q = 1. \end{equation} The inverse of a product reverses factor order: \begin{equation} (pq)^{-1} = q^{-1}p^{-1}. \end{equation} For a unit quaternion, \begin{equation} \lVert q\rVert=1 \end{equation} and therefore \begin{equation} q^{-1}=q^*. \end{equation} For the PhysicsLibrary passive attitude convention, \begin{equation} {}^{A}q_B = ({}^{B}q_A)^{-1} = ({}^{B}q_A)^*. \end{equation} \section{Exercises} \begin{enumerate} \item \textbf{Direct inverse calculation.} For \[ q = 2 - \mathbf i + 2\mathbf j + 2\mathbf k, \] compute \[ q^*, \qquad \lVert q\rVert^2, \qquad q^{-1}. \] \item \textbf{Verify both inverse identities.} Using the quaternion from Exercise 1, compute \[ q q^{-1} \] and \[ q^{-1}q. \] Verify that both equal the multiplicative identity. \item \textbf{Unit quaternion simplification.} Let \[ q = \frac12 \left( 1-\mathbf i-\mathbf j-\mathbf k \right). \] First verify that $q$ is unit. Then compute $q^{-1}$ and show that \[ q^{-1}=q^*. \] \item \textbf{Real and pure quaternion inverses.} Find the inverse of the real quaternion \[ r=-4 \] and the pure quaternion \[ v = 2\mathbf i-\mathbf j+2\mathbf k. \] Show that the inverse of a nonzero pure quaternion may be written \begin{equation} v^{-1} = -\frac{v}{\lVert v\rVert^2}. \end{equation} \item \textbf{Why the zero quaternion has no inverse.} Explain why \[ q=0 \] cannot have a multiplicative inverse. Relate the failure to the denominator in \[ q^{-1} = \frac{q^*}{\lVert q\rVert^2}. \] \item \textbf{Inverse of a product.} Prove \begin{equation} (pq)^{-1} = q^{-1}p^{-1} \end{equation} for nonzero quaternions $p$ and $q$. Your proof should verify both the left and right inverse identities. \item \textbf{Numerical product inverse.} Let \[ p = 1+2\mathbf i-\mathbf j+\mathbf k \] and \[ q = 2-\mathbf i+\mathbf j+3\mathbf k. \] Compute \[ pq, \qquad (pq)^{-1}, \qquad q^{-1}p^{-1}. \] Verify numerically that the last two quantities are equal. \item \textbf{Why \(p^{-1}q^{-1}\) is generally wrong.} Using the quaternions from Exercise 7, compute \[ p^{-1}q^{-1}. \] Compare it with \[ (pq)^{-1}. \] Explain the discrepancy. \item \textbf{Solve a left multiplication equation.} Let \[ p x=r, \] where \[ p=1+\mathbf i \] and \[ r=2+\mathbf j. \] Solve for $x$ by multiplying by the appropriate inverse on the correct side. Verify your answer by direct substitution. \item \textbf{Solve a right multiplication equation.} Let \[ x p=r, \] with the same \[ p=1+\mathbf i \] and \[ r=2+\mathbf j. \] Solve for $x$. Compare the result with Exercise 9 and explain why the two answers differ. \item \textbf{Inverse of a power.} Prove for every positive integer $n$ that \begin{equation} (q^n)^{-1} = (q^{-1})^n. \end{equation} Does factor reversal create any complication when all factors are the same quaternion? \item \textbf{Inverse of a real scalar multiple.} Let $\lambda\neq0$ be real and $q\neq0$. Prove \begin{equation} (\lambda q)^{-1} = \frac{1}{\lambda} q^{-1}. \end{equation} Then verify the identity with \[ \lambda=-3, \qquad q=1+\mathbf j. \] \item \textbf{Passive frame map reversal.} Frame $B$ is obtained from frame $A$ by a positive $90^\circ$ frame rotation about $+\widehat{\mathbf z}$. PhysicsLibrary uses \[ {}^{B}q_A = \frac{1-\mathbf k}{\sqrt2}. \] Compute \[ ({}^{B}q_A)^{-1} \] and identify it as a frame labeled quaternion. \item \textbf{Undo a passive coordinate transformation.} Suppose \begin{equation} {}^{B}v = {}^{B}q_A \,{}^{A}v \,({}^{B}q_A)^* \end{equation} with a unit attitude quaternion. Use the inverse to solve for ${}^{A}v$. Write the reverse transformation using both inverse notation and conjugate notation. \item \textbf{Inverse of \(q\) versus inverse of \(-q\).} For nonzero $q$, prove \begin{equation} (-q)^{-1} = -q^{-1}. \end{equation} If $q$ is a unit attitude quaternion, explain why $q^{-1}$ and $-q^{-1}$ still represent the same reverse physical orientation. \item \textbf{Software diagnostic: conjugate is not always the inverse.} A software routine defines \[ \operatorname{inv}(q)=q^* \] for every quaternion. Test the routine on \[ q = 2+\mathbf i. \] Compute the routine's result, the true inverse, and \[ q q^*. \] Under what condition is the software routine correct? \end{enumerate} \section{Solutions} \subsection*{Solution 1: direct inverse calculation} For \[ q = 2-\mathbf i+2\mathbf j+2\mathbf k, \] the conjugate is \begin{equation} q^* = 2+\mathbf i-2\mathbf j-2\mathbf k. \end{equation} The squared norm is \[ \begin{aligned} \lVert q\rVert^2 &= 2^2+(-1)^2+2^2+2^2\\ &= 4+1+4+4\\ &= 13. \end{aligned} \] Therefore \begin{equation} q^{-1} = \frac{ 2+\mathbf i-2\mathbf j-2\mathbf k }{ 13 }. \end{equation} \subsection*{Solution 2: verify both inverse identities} From Exercise 1, \[ q^{-1} = \frac{q^*}{13}. \] Therefore \[ q q^{-1} = q\frac{q^*}{13} = \frac{q q^*}{13}. \] Since \[ q q^*=13, \] we obtain \begin{equation} q q^{-1}=1. \end{equation} Likewise, \[ q^{-1}q = \frac{q^*q}{13}. \] Because \[ q^*q=13, \] we obtain \begin{equation} q^{-1}q=1. \end{equation} Thus the same quaternion is both the left and right multiplicative inverse. \subsection*{Solution 3: unit quaternion simplification} The quaternion is \[ q = \frac12 (1-\mathbf i-\mathbf j-\mathbf k). \] Its squared norm is \[ \lVert q\rVert^2 = \frac14 (1+1+1+1) = 1. \] Thus \begin{equation} \lVert q\rVert=1. \end{equation} The conjugate is \[ q^* = \frac12 (1+\mathbf i+\mathbf j+\mathbf k). \] Since the norm squared is one, \[ q^{-1} = \frac{q^*}{1}. \] Therefore \begin{equation} q^{-1} = q^* = \frac12 (1+\mathbf i+\mathbf j+\mathbf k). \end{equation} \subsection*{Solution 4: real and pure quaternion inverses} For \[ r=-4, \] the conjugate is also $-4$, and \[ \lVert r\rVert^2=16. \] Therefore \begin{equation} r^{-1} = -\frac14. \end{equation} Now consider \[ v = 2\mathbf i-\mathbf j+2\mathbf k. \] Because $v$ is pure, \[ v^*=-v. \] Its squared norm is \[ \lVert v\rVert^2 = 4+1+4 = 9. \] Therefore \begin{equation} v^{-1} = \frac{v^*}{\lVert v\rVert^2} = -\frac{v}{9}. \end{equation} Explicitly, \begin{equation} v^{-1} = -\frac29\mathbf i + \frac19\mathbf j - \frac29\mathbf k. \end{equation} \subsection*{Solution 5: why the zero quaternion has no inverse} An inverse $q^{-1}$ would have to satisfy \[ q q^{-1}=1. \] If \[ q=0, \] then for every quaternion $x$, \[ 0x=0. \] Therefore no quaternion can satisfy \[ 0x=1. \] The formula \[ q^{-1} = \frac{q^*}{\lVert q\rVert^2} \] shows the same obstruction algebraically. For $q=0$, \[ \lVert q\rVert^2=0, \] so the expression would require division by zero. Thus \begin{equation} 0^{-1} \quad\hbox{does not exist}. \end{equation} \subsection*{Solution 6: inverse of a product} Consider \[ q^{-1}p^{-1}. \] Multiply it on the right of $pq$: \[ \begin{aligned} (pq)(q^{-1}p^{-1}) &= p(qq^{-1})p^{-1}\\ &= p(1)p^{-1}\\ &= pp^{-1}\\ &= 1. \end{aligned} \] Thus $q^{-1}p^{-1}$ is a right inverse of $pq$. Now multiply it on the left: \[ \begin{aligned} (q^{-1}p^{-1})(pq) &= q^{-1}(p^{-1}p)q\\ &= q^{-1}(1)q\\ &= q^{-1}q\\ &= 1. \end{aligned} \] Thus it is also a left inverse. Therefore \begin{equation} (pq)^{-1} = q^{-1}p^{-1}. \end{equation} The factor order reverses. \subsection*{Solution 7: numerical product inverse} Let \[ p = 1+2\mathbf i-\mathbf j+\mathbf k \] and \[ q = 2-\mathbf i+\mathbf j+3\mathbf k. \] Hamilton multiplication gives \begin{equation} pq = 2-\mathbf i-8\mathbf j+6\mathbf k. \end{equation} Its squared norm is \[ \lVert pq\rVert^2 = 2^2+(-1)^2+(-8)^2+6^2 = 105. \] Therefore \begin{equation} (pq)^{-1} = \frac{ 2+\mathbf i+8\mathbf j-6\mathbf k }{ 105 }. \end{equation} Now \[ \lVert p\rVert^2 = 1+4+1+1 = 7, \] so \begin{equation} p^{-1} = \frac{ 1-2\mathbf i+\mathbf j-\mathbf k }{ 7 }. \end{equation} Also, \[ \lVert q\rVert^2 = 4+1+1+9 = 15, \] so \begin{equation} q^{-1} = \frac{ 2+\mathbf i-\mathbf j-3\mathbf k }{ 15 }. \end{equation} Multiplying in the reversed order, \[ q^{-1}p^{-1} = \frac{1}{105} (2+\mathbf i-\mathbf j-3\mathbf k) (1-2\mathbf i+\mathbf j-\mathbf k). \] The numerator product is \[ 2+\mathbf i+8\mathbf j-6\mathbf k. \] Hence \begin{equation} q^{-1}p^{-1} = \frac{ 2+\mathbf i+8\mathbf j-6\mathbf k }{ 105 } = (pq)^{-1}. \end{equation} \subsection*{Solution 8: why \(p^{-1}q^{-1}\) is generally wrong} Using the inverses from Solution 7, \[ p^{-1}q^{-1} = \frac{1}{105} (1-2\mathbf i+\mathbf j-\mathbf k) (2+\mathbf i-\mathbf j-3\mathbf k). \] The numerator product is \[ 2-7\mathbf i-4\mathbf j-8\mathbf k. \] Therefore \begin{equation} p^{-1}q^{-1} = \frac{ 2-7\mathbf i-4\mathbf j-8\mathbf k }{ 105 }. \end{equation} But \[ (pq)^{-1} = \frac{ 2+\mathbf i+8\mathbf j-6\mathbf k }{ 105 }. \] They differ because quaternion multiplication is not generally commutative. The inverse must undo the last factor first, so product inversion reverses factor order. \subsection*{Solution 9: solve a left multiplication equation} We are given \[ p x=r. \] Left multiply by $p^{-1}$: \[ p^{-1}px = p^{-1}r. \] Thus \begin{equation} x = p^{-1}r. \end{equation} For \[ p=1+\mathbf i, \] the squared norm is \[ 2, \] so \[ p^{-1} = \frac{1-\mathbf i}{2}. \] Therefore \[ \begin{aligned} x &= \frac12 (1-\mathbf i)(2+\mathbf j)\\ &= \frac12 (2+\mathbf j-2\mathbf i-\mathbf i\mathbf j)\\ &= \frac12 (2-2\mathbf i+\mathbf j-\mathbf k). \end{aligned} \] Hence \begin{equation} x = 1-\mathbf i + \frac12\mathbf j - \frac12\mathbf k. \end{equation} Direct substitution gives \[ (1+\mathbf i)x = 2+\mathbf j = r. \] \subsection*{Solution 10: solve a right multiplication equation} Now \[ x p=r. \] Right multiply by $p^{-1}$: \[ xpp^{-1} = rp^{-1}. \] Thus \begin{equation} x = rp^{-1}. \end{equation} Using \[ p^{-1} = \frac{1-\mathbf i}{2}, \] we obtain \[ \begin{aligned} x &= \frac12 (2+\mathbf j)(1-\mathbf i)\\ &= \frac12 (2-2\mathbf i+\mathbf j-\mathbf j\mathbf i). \end{aligned} \] Because \[ \mathbf j\mathbf i=-\mathbf k, \] we get \[ x = \frac12 (2-2\mathbf i+\mathbf j+\mathbf k). \] Therefore \begin{equation} x = 1-\mathbf i + \frac12\mathbf j + \frac12\mathbf k. \end{equation} This differs from Solution 9 in the sign of the $\mathbf k$ component. The difference arises because left and right division are distinct operations in a noncommutative algebra. \subsection*{Solution 11: inverse of a power} For \[ q^n = \underbrace{ q q\cdots q }_{n\ \mathrm{factors}}, \] repeated application of product inversion gives \[ (q^n)^{-1} = \underbrace{ q^{-1}q^{-1}\cdots q^{-1} }_{n\ \mathrm{factors}}. \] Hence \begin{equation} (q^n)^{-1} = (q^{-1})^n. \end{equation} Factor reversal creates no visible change because all factors are identical. \subsection*{Solution 12: inverse of a real scalar multiple} Let \[ p=\lambda q, \qquad \lambda\neq0. \] Since $\lambda$ is real, it commutes with every quaternion. Consider \[ \frac{1}{\lambda}q^{-1}. \] Then \[ \begin{aligned} (\lambda q) \left( \frac{1}{\lambda}q^{-1} \right) &= \lambda \frac{1}{\lambda} qq^{-1}\\ &= 1. \end{aligned} \] The product in the other order also equals one. Therefore \begin{equation} (\lambda q)^{-1} = \frac{1}{\lambda}q^{-1}. \end{equation} For \[ q=1+\mathbf j, \] \[ q^{-1} = \frac{1-\mathbf j}{2}. \] With \[ \lambda=-3, \] we obtain \begin{equation} (-3q)^{-1} = -\frac16 (1-\mathbf j). \end{equation} \subsection*{Solution 13: passive frame map reversal} The forward passive map is \[ {}^{B}q_A = \frac{1-\mathbf k}{\sqrt2}. \] It is unit, so inverse equals conjugate: \[ ({}^{B}q_A)^{-1} = ({}^{B}q_A)^*. \] Therefore \begin{equation} ({}^{B}q_A)^{-1} = \frac{1+\mathbf k}{\sqrt2}. \end{equation} The inverse reverses the frame map, so \begin{equation} {}^{A}q_B = \frac{1+\mathbf k}{\sqrt2}. \end{equation} \subsection*{Solution 14: undo a passive coordinate transformation} We begin with \[ {}^{B}v = {}^{B}q_A \,{}^{A}v \,({}^{B}q_A)^*. \] Let \[ q = {}^{B}q_A. \] Because $q$ is unit, \[ q^{-1}=q^*. \] Left multiply by $q^{-1}$ and right multiply by $(q^*)^{-1}$: \[ {}^{A}v = q^{-1} \,{}^{B}v \,(q^*)^{-1}. \] Since \[ (q^*)^{-1}=q, \] we obtain \begin{equation} {}^{A}v = q^{-1} \,{}^{B}v \,q. \end{equation} Using the unit relation $q^{-1}=q^*$, \begin{equation} {}^{A}v = q^* \,{}^{B}v \,q. \end{equation} Equivalently, with frame labels, \begin{equation} {}^{A}v = {}^{A}q_B \,{}^{B}v \,({}^{A}q_B)^*. \end{equation} \subsection*{Solution 15: inverse of \(q\) versus inverse of \(-q\)} Use \[ (-q)^* = -q^* \] and \[ \lVert-q\rVert^2 = \lVert q\rVert^2. \] Then \[ \begin{aligned} (-q)^{-1} &= \frac{ (-q)^* }{ \lVert-q\rVert^2 }\\ &= \frac{ -q^* }{ \lVert q\rVert^2 }\\ &= -q^{-1}. \end{aligned} \] Therefore \begin{equation} (-q)^{-1} = -q^{-1}. \end{equation} For unit attitude quaternions, $q^{-1}$ and $-q^{-1}$ are antipodal unit quaternions representing the same reverse physical orientation. \subsection*{Solution 16: software diagnostic: conjugate is not always the inverse} The software routine returns \[ q^* = 2-\mathbf i. \] But \[ \lVert q\rVert^2 = 2^2+1^2 = 5. \] Therefore the true inverse is \begin{equation} q^{-1} = \frac{ 2-\mathbf i }{ 5 }. \end{equation} The product with the conjugate is \[ \begin{aligned} q q^* &= (2+\mathbf i)(2-\mathbf i)\\ &= 4-\mathbf i^2\\ &= 5. \end{aligned} \] Thus \begin{equation} q q^*=5, \end{equation} not $1$. The routine \[ \operatorname{inv}(q)=q^* \] is correct only when \begin{equation} \lVert q\rVert=1. \end{equation} For general nonzero quaternions, division by the norm squared is required. \section{Compact review} The central inverse identities reinforced by this companion are \begin{equation} q^{-1} = \frac{q^*}{\lVert q\rVert^2}, \qquad q\neq0, \end{equation} \begin{equation} q q^{-1} = q^{-1}q = 1, \end{equation} \begin{equation} (pq)^{-1} = q^{-1}p^{-1}, \end{equation} and, for unit quaternions, \begin{equation} q^{-1}=q^*. \end{equation} For passive PhysicsLibrary attitude quaternions, \begin{equation} {}^{A}q_B = ({}^{B}q_A)^{-1} = ({}^{B}q_A)^*. \end{equation} Thus algebraic inversion and reversal of a unit passive frame map are the same operation. \section{Sources and exercise provenance} The exercises and solutions in this companion are newly written or expanded for PhysicsLibrary from the algebra developed in \emph{Quaternion Inverse}. Hamilton is the foundational source for quaternion division and reciprocals. Joly and Kelland--Tait provide classical systematic treatments of quaternion inverse operations. Sommer and coauthors provide a modern engineering discussion of unit quaternion conventions and frame transformations. \begin{thebibliography}{9} \bibitem{Hamilton1899} W. R. Hamilton, \emph{Elements of Quaternions}, 2nd ed., edited by C. J. Joly, Longmans, Green, and Co., 1899. Public domain historical source. \PMlinkexternal{Internet Archive scan} {https://archive.org/details/elementsofquater01hamiuoft} \bibitem{Joly1905} C. J. Joly, \emph{A Manual of Quaternions}, Macmillan and Co., London, 1905. Public domain historical source. \PMlinkexternal{Internet Archive search} {https://archive.org/search?query=A+Manual+of+Quaternions+Joly} \bibitem{KellandTait1882} P. Kelland and P. G. Tait, \emph{Introduction to Quaternions, with Numerous Examples}, 2nd ed., Macmillan and Co., London, 1882. Public domain historical source. \PMlinkexternal{Internet Archive search} {https://archive.org/search?query=Introduction+to+Quaternions+Kelland+Tait} \bibitem{Sommer2018} H. Sommer, I. Gilitschenski, M. Bloesch, S. Weiss, R. Siegwart, and J. Nieto, ``Why and How to Avoid the Flipped Quaternion Multiplication,'' \emph{Aerospace}, vol. 5, no. 3, article 72, 2018. Published under CC BY 4.0. \PMlinkexternal{Publisher article} {https://www.mdpi.com/2226-4310/5/3/72} \end{thebibliography} \section*{License} Unless otherwise noted, this PhysicsLibrary entry is intended for release under the Creative Commons Attribution ShareAlike 4.0 International license.