example of quaternion conjugate ExampleOfQuaternionConjugate 2026-08-28 15:46:03 2026-08-28 15:46:03 Example quaternion conjugate 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 Conjugate: Examples, Exercises, and Solutions} This entry is the self study companion to \emph{Quaternion Conjugate}. Quaternion conjugation is simple to define but unusually important. It reverses the sign of the vector part, reverses product order, produces the quadratic real quantity $qq^*$, and for unit attitude quaternions reverses the direction of a passive frame map. All exercises are stated first. Complete worked solutions follow afterward. \section{Formula summary} For \begin{equation} q = q_w + q_x\mathbf i + q_y\mathbf j + q_z\mathbf k, \end{equation} the conjugate is \begin{equation} q^* = q_w - q_x\mathbf i - q_y\mathbf j - q_z\mathbf k. \end{equation} In scalar vector notation, \begin{equation} q = q_w+\mathbf q, \qquad q^* = q_w-\mathbf q. \end{equation} Conjugation is an involution: \begin{equation} (q^*)^*=q. \end{equation} It is real linear: \begin{equation} (p+q)^* = p^*+q^*, \end{equation} \begin{equation} (\lambda q)^* = \lambda q^*, \qquad \lambda\in\mathbb R. \end{equation} Most importantly, conjugation reverses product order: \begin{equation} (pq)^* = q^*p^*. \end{equation} The product of a quaternion with its conjugate is real: \begin{equation} qq^* = q^*q = q_w^2+q_x^2+q_y^2+q_z^2. \end{equation} For a unit quaternion, \begin{equation} q^*=q^{-1}. \end{equation} For the PhysicsLibrary passive attitude convention, \begin{equation} {}^{A}q_B = ({}^{B}q_A)^*. \end{equation} \section{Exercises} \begin{enumerate} \item \textbf{Basic conjugation.} For \[ q = 3 - 2\mathbf i + 5\mathbf j - 7\mathbf k, \] compute $q^*$ and the scalar first component columns of both $q$ and $q^*$. Which component is unchanged? \item \textbf{Real and pure special cases.} Compute the conjugate of \[ r=5 \] and of \[ v = 2\mathbf i - 3\mathbf j + 4\mathbf k. \] Show that every real quaternion satisfies \[ r^*=r \] and every pure quaternion satisfies \[ v^*=-v. \] \item \textbf{Conjugation as an involution.} Starting from \[ q=q_w+\mathbf q, \] prove directly that \[ (q^*)^*=q. \] Then verify the result numerically for \[ q = -2 + \mathbf i + 3\mathbf j - 4\mathbf k. \] \item \textbf{Conjugate of a sum and real scalar multiple.} Let \[ p = 2+\mathbf i-3\mathbf j \] and \[ q = -1+4\mathbf j+2\mathbf k. \] Verify explicitly that \[ (p+q)^* = p^*+q^* \] and \[ (-3p)^* = -3p^*. \] \item \textbf{Product reversal using basis quaternions.} Evaluate \[ (\mathbf i\mathbf j)^* \] directly. Then evaluate \[ \mathbf j^*\mathbf i^*. \] Verify \[ (\mathbf i\mathbf j)^* = \mathbf j^*\mathbf i^*. \] Explain why the reversed factor order is essential. \item \textbf{Product reversal for general numerical quaternions.} Let \[ p = 1+2\mathbf i-\mathbf j+3\mathbf k \] and \[ q = 2-\mathbf i+4\mathbf j+\mathbf k. \] Compute $pq$, then $(pq)^*$. Separately compute $q^*p^*$ and verify that the two results agree. \item \textbf{Why \(p^*q^*\) is generally wrong.} Using the same $p$ and $q$ from Exercise 6, compute \[ p^*q^* \] and compare it with $(pq)^*$. Relate the difference to noncommutativity. \item \textbf{Product with the conjugate.} For \[ q = 2-\mathbf i+2\mathbf j+2\mathbf k, \] compute \[ qq^* \] and \[ q^*q. \] Show explicitly that both are the same real quaternion. \item \textbf{Derive \(qq^*\) in scalar vector form.} Let \[ q=q_w+\mathbf q. \] Using the Hamilton scalar vector product, derive \[ qq^* = q_w^2+\mathbf q\cdot\mathbf q. \] Why does the vector part vanish? \item \textbf{Conjugate and norm connection.} Suppose \[ q = 1 - 2\mathbf i + 2\mathbf j - 4\mathbf k. \] Compute $qq^*$. Without yet using the explicit inverse formula, identify the Euclidean norm squared of the four quaternion components. \item \textbf{Conjugate of a triple product.} Prove \begin{equation} (pqr)^* = r^*q^*p^*. \end{equation} Then state the corresponding pattern for \[ (q_1q_2q_3q_4)^*. \] \item \textbf{Conjugation of a pure quaternion product.} Let $\mathbf a$ and $\mathbf b$ be pure quaternions. Starting from \[ \mathbf a\mathbf b = -\mathbf a\cdot\mathbf b + \mathbf a\times\mathbf b, \] take the conjugate and show that \begin{equation} (\mathbf a\mathbf b)^* = -\mathbf a\cdot\mathbf b - \mathbf a\times\mathbf b. \end{equation} Then show that this equals \[ \mathbf b\mathbf a. \] \item \textbf{Unit quaternion and reverse passive frame map.} Frame $B$ is obtained from frame $A$ by a positive $90^\circ$ frame rotation about $+\widehat{\mathbf z}$. The PhysicsLibrary passive quaternion is \[ {}^{B}q_A = \frac{1-\mathbf k}{\sqrt2}. \] Compute \[ {}^{A}q_B. \] State the physical meaning of the conjugate in frame-map language. \item \textbf{Conjugate sandwich reverses the coordinate map.} Let \[ q = {}^{B}q_A \] be unit and suppose \[ {}^{B}v = q\,{}^{A}v\,q^*. \] Starting from this equation, solve algebraically for ${}^{A}v$ in terms of ${}^{B}v$ and $q$. Show that the reverse map is \begin{equation} {}^{A}v = q^* \,{}^{B}v \,q. \end{equation} \item \textbf{\(q\) versus \(q^*\) is not the same as \(q\) versus \(-q\).} For \[ q = \frac{1-\mathbf k}{\sqrt2}, \] write $q^*$ and $-q$. Do $q$ and $q^*$ represent the same passive frame map? Do $q$ and $-q$ represent the same passive frame map? Explain the difference. \item \textbf{Conjugate as a convention diagnostic.} A software library claims to store the passive frame quaternion ${}^{B}q_A$. For a positive $90^\circ$ frame rotation about $+z$, it reports \[ q = \frac{1+\mathbf k}{\sqrt2}. \] Under the PhysicsLibrary convention, determine whether this quaternion more naturally represents ${}^{B}q_A$ or its reverse map ${}^{A}q_B$. What simple conjugation operation converts it to the PhysicsLibrary $A\rightarrow B$ passive map? \end{enumerate} \section{Solutions} \subsection*{Solution 1: basic conjugation} The quaternion is \[ q = 3 - 2\mathbf i + 5\mathbf j - 7\mathbf k. \] Conjugation leaves the scalar coefficient unchanged and changes the sign of the vector part: \begin{equation} q^* = 3 + 2\mathbf i - 5\mathbf j + 7\mathbf k. \end{equation} The scalar first component columns are \begin{equation} [q]_{\mathrm{PL}} = \begin{bmatrix} 3\\ -2\\ 5\\ -7 \end{bmatrix}, \qquad [q^*]_{\mathrm{PL}} = \begin{bmatrix} 3\\ 2\\ -5\\ 7 \end{bmatrix}. \end{equation} The scalar component is unchanged. \subsection*{Solution 2: real and pure special cases} For the real quaternion \[ r=5, \] the vector part is zero, so \begin{equation} r^*=5=r. \end{equation} For the pure quaternion \[ v = 2\mathbf i - 3\mathbf j + 4\mathbf k, \] the scalar part is zero. Therefore \begin{equation} v^* = -2\mathbf i + 3\mathbf j - 4\mathbf k = -v. \end{equation} Thus real quaternions are fixed by conjugation, while pure quaternions change sign. \subsection*{Solution 3: conjugation as an involution} Write \[ q=q_w+\mathbf q. \] Then \[ q^* = q_w-\mathbf q. \] Conjugate again: \[ (q^*)^* = q_w+\mathbf q = q. \] Therefore \begin{equation} (q^*)^*=q. \end{equation} For \[ q = -2+\mathbf i+3\mathbf j-4\mathbf k, \] we have \[ q^* = -2-\mathbf i-3\mathbf j+4\mathbf k. \] Conjugating again gives \[ (q^*)^* = -2+\mathbf i+3\mathbf j-4\mathbf k = q. \] \subsection*{Solution 4: conjugate of a sum and real scalar multiple} First, \[ p+q = 1+\mathbf i+\mathbf j+2\mathbf k. \] Therefore \[ (p+q)^* = 1-\mathbf i-\mathbf j-2\mathbf k. \] Separately, \[ p^* = 2-\mathbf i+3\mathbf j \] and \[ q^* = -1-4\mathbf j-2\mathbf k. \] Hence \[ p^*+q^* = 1-\mathbf i-\mathbf j-2\mathbf k. \] Thus \begin{equation} (p+q)^* = p^*+q^*. \end{equation} Also, \[ -3p = -6-3\mathbf i+9\mathbf j. \] Therefore \[ (-3p)^* = -6+3\mathbf i-9\mathbf j. \] Meanwhile, \[ -3p^* = -3(2-\mathbf i+3\mathbf j) = -6+3\mathbf i-9\mathbf j. \] Thus \begin{equation} (-3p)^* = -3p^*. \end{equation} \subsection*{Solution 5: product reversal using basis quaternions} Hamilton multiplication gives \[ \mathbf i\mathbf j=\mathbf k. \] Therefore \[ (\mathbf i\mathbf j)^* = \mathbf k^* = -\mathbf k. \] Now \[ \mathbf i^*=-\mathbf i, \qquad \mathbf j^*=-\mathbf j. \] Hence \[ \mathbf j^*\mathbf i^* = (-\mathbf j)(-\mathbf i) = \mathbf j\mathbf i = -\mathbf k. \] Thus \begin{equation} (\mathbf i\mathbf j)^* = \mathbf j^*\mathbf i^*. \end{equation} If the factor order were not reversed, one would obtain \[ \mathbf i^*\mathbf j^* = \mathbf i\mathbf j = \mathbf k, \] which has the wrong sign. \subsection*{Solution 6: product reversal for general numerical quaternions} Let \[ p=(1,[2,-1,3]) \] and \[ q=(2,[-1,4,1]). \] Using the Hamilton product, \[ pq = 5-10\mathbf i-3\mathbf j+14\mathbf k. \] Therefore \begin{equation} (pq)^* = 5+10\mathbf i+3\mathbf j-14\mathbf k. \end{equation} Now \[ q^* = 2+\mathbf i-4\mathbf j-\mathbf k \] and \[ p^* = 1-2\mathbf i+\mathbf j-3\mathbf k. \] Multiplying in the reversed order gives \begin{equation} q^*p^* = 5+10\mathbf i+3\mathbf j-14\mathbf k. \end{equation} Hence \[ (pq)^* = q^*p^*. \] \subsection*{Solution 7: why \(p^*q^*\) is generally wrong} Using the same conjugates, \[ p^* = 1-2\mathbf i+\mathbf j-3\mathbf k \] and \[ q^* = 2+\mathbf i-4\mathbf j-\mathbf k. \] Direct multiplication gives \begin{equation} p^*q^* = 5-16\mathbf i-7\mathbf j. \end{equation} But from Solution 6, \[ (pq)^* = 5+10\mathbf i+3\mathbf j-14\mathbf k. \] These are not equal. The conjugation rule reverses factor order because quaternion multiplication is not generally commutative. \subsection*{Solution 8: product with the conjugate} For \[ q = 2-\mathbf i+2\mathbf j+2\mathbf k, \] the conjugate is \[ q^* = 2+\mathbf i-2\mathbf j-2\mathbf k. \] The scalar part of $qq^*$ is \[ 2^2+(-1)^2+2^2+2^2 = 13. \] The vector part cancels. Therefore \begin{equation} qq^*=13. \end{equation} Similarly, \begin{equation} q^*q=13. \end{equation} Both products are the same real quaternion. \subsection*{Solution 9: derive \(qq^*\) in scalar vector form} Let \[ q=q_w+\mathbf q \] and \[ q^*=q_w-\mathbf q. \] Use the Hamilton scalar vector product. The scalar part is \[ q_wq_w - \mathbf q\cdot(-\mathbf q) = q_w^2+\mathbf q\cdot\mathbf q. \] The vector part is \[ q_w(-\mathbf q) + q_w\mathbf q + \mathbf q\times(-\mathbf q). \] The first two terms cancel, and \[ \mathbf q\times\mathbf q=0. \] Therefore the vector part vanishes. Thus \begin{equation} qq^* = q_w^2+\mathbf q\cdot\mathbf q. \end{equation} \subsection*{Solution 10: conjugate and norm connection} For \[ q = 1-2\mathbf i+2\mathbf j-4\mathbf k, \] the sum of the squared components is \[ 1^2+(-2)^2+2^2+(-4)^2 = 1+4+4+16 = 25. \] Therefore \begin{equation} qq^*=25. \end{equation} This is the squared Euclidean norm of the four component column: \begin{equation} \lVert q\rVert^2=25. \end{equation} Hence \[ \lVert q\rVert=5. \] \subsection*{Solution 11: conjugate of a triple product} Associativity permits us to write \[ pqr=(pq)r. \] Take the conjugate: \[ (pqr)^* = [(pq)r]^*. \] Using product reversal once, \[ [(pq)r]^* = r^*(pq)^*. \] Use it again: \[ (pq)^* = q^*p^*. \] Therefore \begin{equation} (pqr)^* = r^*q^*p^*. \end{equation} Likewise, \begin{equation} (q_1q_2q_3q_4)^* = q_4^*q_3^*q_2^*q_1^*. \end{equation} Conjugation reverses the entire factor sequence. \subsection*{Solution 12: conjugation of a pure quaternion product} For pure quaternions, \[ \mathbf a\mathbf b = -\mathbf a\cdot\mathbf b + \mathbf a\times\mathbf b. \] The dot product term is real and is unchanged by conjugation. The cross product term is pure and changes sign. Therefore \begin{equation} (\mathbf a\mathbf b)^* = -\mathbf a\cdot\mathbf b - \mathbf a\times\mathbf b. \end{equation} Now reverse the pure factors: \[ \mathbf b\mathbf a = -\mathbf b\cdot\mathbf a + \mathbf b\times\mathbf a. \] Because \[ \mathbf b\cdot\mathbf a = \mathbf a\cdot\mathbf b \] and \[ \mathbf b\times\mathbf a = -\mathbf a\times\mathbf b, \] we obtain \[ \mathbf b\mathbf a = -\mathbf a\cdot\mathbf b - \mathbf a\times\mathbf b. \] Hence \begin{equation} (\mathbf a\mathbf b)^* = \mathbf b\mathbf a. \end{equation} \subsection*{Solution 13: unit quaternion and reverse passive frame map} The forward passive frame quaternion is \[ {}^{B}q_A = \frac{1-\mathbf k}{\sqrt2}. \] Its conjugate is \[ ({}^{B}q_A)^* = \frac{1+\mathbf k}{\sqrt2}. \] For a unit attitude quaternion, \[ {}^{A}q_B = ({}^{B}q_A)^*. \] Therefore \begin{equation} {}^{A}q_B = \frac{1+\mathbf k}{\sqrt2}. \end{equation} The conjugate reverses the passive coordinate map: it maps $B$ coordinates back into $A$ coordinates. \subsection*{Solution 14: conjugate sandwich reverses the coordinate map} Start from \[ {}^{B}v = q\,{}^{A}v\,q^*. \] Left multiply by $q^*$: \[ q^*{}^{B}v = q^*q\,{}^{A}v\,q^*. \] Because $q$ is unit, \[ q^*q=1. \] Thus \[ q^*{}^{B}v = {}^{A}v\,q^*. \] Right multiply by $q$: \[ q^*{}^{B}v\,q = {}^{A}v\,q^*q. \] Again, \[ q^*q=1. \] Therefore \begin{equation} {}^{A}v = q^* \,{}^{B}v \,q. \end{equation} The reverse coordinate transformation uses the conjugate quaternion in the opposite sandwich order. \subsection*{Solution 15: \(q\) versus \(q^*\) is not the same as \(q\) versus \(-q\)} Given \[ q = \frac{1-\mathbf k}{\sqrt2}, \] the conjugate is \begin{equation} q^* = \frac{1+\mathbf k}{\sqrt2}. \end{equation} The negative quaternion is \begin{equation} -q = \frac{-1+\mathbf k}{\sqrt2}. \end{equation} The pair $q$ and $q^*$ generally represents opposite passive frame maps: \[ {}^{B}q_A \quad\hbox{versus}\quad {}^{A}q_B. \] They are inverse transformations. By contrast, $q$ and $-q$ represent the same physical orientation and the same passive frame map. Thus conjugation changes map direction, while an overall sign change does not change orientation. \subsection*{Solution 16: conjugate as a convention diagnostic} PhysicsLibrary assigns a positive $90^\circ$ frame rotation about $+z$ the passive $A\rightarrow B$ quaternion \[ {}^{B}q_A = \frac{1-\mathbf k}{\sqrt2}. \] The software reports \[ q = \frac{1+\mathbf k}{\sqrt2}. \] This is exactly the conjugate of the PhysicsLibrary $A\rightarrow B$ map. Therefore it more naturally corresponds to \begin{equation} {}^{A}q_B = \frac{1+\mathbf k}{\sqrt2}. \end{equation} To convert it to the PhysicsLibrary $A\rightarrow B$ passive map, take the conjugate: \[ q_{\mathrm{PL}} = q^* = \frac{1-\mathbf k}{\sqrt2}. \] \section{Compact review} The core identities reinforced by this companion are \begin{equation} q^* = q_w-\mathbf q, \end{equation} \begin{equation} (q^*)^*=q, \end{equation} \begin{equation} (pq)^*=q^*p^*, \end{equation} \begin{equation} qq^*=q^*q=\lVert q\rVert^2, \end{equation} and, for unit passive attitude quaternions, \begin{equation} {}^{A}q_B = ({}^{B}q_A)^*. \end{equation} Conjugation is therefore both an algebraic operation and, for unit attitude quaternions, the operation that reverses a frame to frame coordinate map. \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 Conjugate}. Hamilton is the foundational source for quaternion conjugation. Joly and Kelland--Tait provide classical systematic treatments of conjugates, products, and norms. Sommer and coauthors provide a modern engineering discussion of quaternion convention management and passive 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.