Quaternion Conjugate

Quaternion conjugation reverses the sign of the vector part of a quaternion while leaving the scalar part unchanged. If

$\displaystyle q=q_w+q_x\mathbf i+q_y\mathbf j+q_z\mathbf k = q_w+\mathbf q,$

then the quaternion conjugate is

$\displaystyle q^* = q_w-q_x\mathbf i-q_y\mathbf j-q_z\mathbf k = q_w-\mathbf q.$

(1)

Conjugation is one of the central operations in quaternion algebra. It separates scalar and vector parts, produces a real quantity when a quaternion is multiplied by its conjugate, reverses the order of a product, and becomes the multiplicative inverse when the quaternion has unit norm.

In the later PhysicsLibrary attitude articles, unit quaternions represent passive coordinate transformations. Under that convention, conjugation has a direct frame meaning: it reverses the direction of the coordinate map.

Definition

For

$\displaystyle q=q_w+\mathbf q,$

define

$\displaystyle q^*=q_w-\mathbf q.$

(2)

In scalar first component form,

$\displaystyle [q^*]_{\mathrm{PL}} = \begin{bmatrix} q_w\ -q_x\ -q_y\ -q_z \end{bmatrix}.$

(3)

Thus conjugation changes the sign of the three imaginary coefficients and leaves the real coefficient unchanged.

The notation $q^*$ is used throughout the PhysicsLibrary quaternion series. Other sources may use $\overline q$ , $q^\dagger$ , or another symbol. A dagger is also widely used for conjugate transpose of matrices, so an unfamiliar source should always be checked before importing formulas.

Basic examples

$\displaystyle q=3-2\mathbf i+4\mathbf j-\mathbf k,$

then

$\displaystyle q^* = 3+2\mathbf i-4\mathbf j+\mathbf k.$

(4)

If $q$ is real,

$\displaystyle q=a,$

then

$\displaystyle q^*=q.$

(5)

If $q$ is pure,

$\displaystyle q=\mathbf q,$

then

$\displaystyle q^*=-q.$

(6)

The real quaternions are therefore exactly the quaternions fixed by conjugation, while pure quaternions reverse sign.

Conjugation is an involution

Applying conjugation twice returns the original quaternion:

$\displaystyle (q^*)^*=q.$

(7)

Indeed,

$\displaystyle q=q_w+\mathbf q$

gives

$\displaystyle q^*=q_w-\mathbf q,$

and a second conjugation gives

$\displaystyle (q^*)^*=q_w+\mathbf q=q.$

An operation that is its own inverse is called an involution.

Conjugation and addition

Let

$\displaystyle p=p_w+\mathbf p, \qquad q=q_w+\mathbf q.$

Then

$\displaystyle p+q=(p_w+q_w)+(\mathbf p+\mathbf q).$

Conjugating gives

$\displaystyle (p+q)^* = p^*+q^*.$

(8)

Similarly, for a real scalar $a$ ,

$\displaystyle (aq)^*=a q^*.$

(9)

Thus conjugation is linear over real scalars.

Conjugation reverses the order of multiplication

The most important structural identity is

$\displaystyle (pq)^* = q^*p^*.$

(10)

The factor order reverses. This is essential because quaternion multiplication is not commutative.

To derive the result, write

$\displaystyle p=(p_w,\mathbf p), \qquad q=(q_w,\mathbf q).$

The Hamilton scalar vector product is

$\displaystyle pq = \left( p_wq_w-\mathbf p\cdot\mathbf q,\; p_w\mathbf q+q_w\mathbf p+\mathbf p\times\mathbf q \right).$

Conjugating this product changes the sign of its vector part:

$\displaystyle (pq)^* = \left( p_wq_w-\mathbf p\cdot\mathbf q,\; -p_w\mathbf q-q_w\mathbf p-\mathbf p\times\mathbf q \right).$

Now

$\displaystyle q^*=(q_w,-\mathbf q), \qquad p^*=(p_w,-\mathbf p).$

Their product is

$\begin{equation*} \begin{aligned} q^*p^* ={}& \left( q_wp_w-(-\mathbf q)\cdot(-\... ...\mathbf q) +(-\mathbf q)\times(-\mathbf p) \right). \end{aligned}\end{equation*}$

Because the dot product is symmetric and

$\displaystyle \mathbf q\times\mathbf p = -\mathbf p\times\mathbf q,$

this becomes exactly

$\displaystyle q^*p^* = \left( p_wq_w-\mathbf p\cdot\mathbf q,\; -p_w\mathbf q-q_w\mathbf p-\mathbf p\times\mathbf q \right).$

Therefore

$\displaystyle (pq)^*=q^*p^*.$

Conjugation is consequently an anti homomorphism of quaternion multiplication: it preserves multiplication only after reversing factor order.

Conjugating a product of many factors

Repeated use of equation (10) gives

$\displaystyle (q_1q_2\cdots q_n)^* = q_n^*\cdots q_2^*q_1^*.$

(11)

Every factor is conjugated and the entire order is reversed.

For three factors,

$\displaystyle (abc)^*=c^*b^*a^*.$

This identity becomes especially useful in frame composition and in simplifying quaternion coordinate transformation expressions.

Scalar and vector parts from conjugation

Conjugation gives convenient formulas for extracting the scalar and vector parts of a quaternion.

Add $q$ and $q^*$ :

$\displaystyle q+q^* = (q_w+\mathbf q)+(q_w-\mathbf q) = 2q_w.$

Therefore

$\displaystyle \operatorname{Sc}(q) = \frac{q+q^*}{2}.$

(12)

Subtract:

$\displaystyle q-q^* = (q_w+\mathbf q)-(q_w-\mathbf q) = 2\mathbf q.$

Therefore

$\displaystyle \operatorname{Vec}(q) = \frac{q-q^*}{2}.$

(13)

These formulas characterize the decomposition of a quaternion into parts that are unchanged and sign reversed by conjugation.

Product with the conjugate

Let

$\displaystyle q=q_w+\mathbf q.$

Using the scalar vector product,

$\displaystyle qq^* = (q_w,\mathbf q)(q_w,-\mathbf q).$

The scalar part is

$\displaystyle q_w^2-\mathbf q\cdot(-\mathbf q) = q_w^2+\mathbf q\cdot\mathbf q,$

and the vector part is

$\displaystyle q_w(-\mathbf q) + q_w\mathbf q + \mathbf q\times(-\mathbf q) = \mathbf 0.$

Hence

$\displaystyle qq^* = q^*q = q_w^2+q_x^2+q_y^2+q_z^2.$

(14)

The result is always a nonnegative real number.

This identity is the basis for the quaternion norm. It also leads directly to the inverse of every nonzero quaternion.

Relation to norm and inverse

The quaternion norm is defined by

$\displaystyle \lVert q\rVert = \sqrt{qq^*}.$

Therefore

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

(15)

If $q\neq0$ , divide by $\lVert q\rVert^2$ to obtain

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

(16)

For a unit quaternion,

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

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

(17)

This special case is fundamental in attitude mechanics because orientation quaternions are maintained at unit norm.

Conjugation of a passive frame quaternion

The algebraic operation above does not depend on any rotation convention. Its physical interpretation does.

PhysicsLibrary uses a unit quaternion

$\displaystyle {}^{B}q_A$

to map coordinates from frame $A$ into frame $B$ :

$\displaystyle {}^{B}v = {}^{B}q_A\, {}^{A}v\, ({}^{B}q_A)^*.$

(18)

Because the quaternion is unit,

$\displaystyle ({}^{B}q_A)^{-1} = ({}^{B}q_A)^*.$

Therefore the inverse coordinate map is

$\displaystyle {}^{A}q_B = ({}^{B}q_A)^*.$

(19)

It follows that

$\displaystyle {}^{A}v = {}^{A}q_B\, {}^{B}v\, ({}^{A}q_B)^*.$

(20)

Conjugating a unit frame quaternion therefore reverses the map direction:

$\displaystyle A\longrightarrow B$

becomes

$\displaystyle B\longrightarrow A.$

This is the quaternion counterpart of transposing a passive direction cosine matrix.

Passive axis angle interpretation

Suppose frame $B$ is obtained from frame $A$ by a positive right hand rule frame rotation through angle $\theta$ about unit axis $\widehat{\mathbf u}$ . Under the PhysicsLibrary passive convention,

$\displaystyle {}^{B}q_A = \cos\frac{\theta}{2} - \widehat{\mathbf u}\sin\frac{\theta}{2}.$

(21)

Conjugation gives

$\displaystyle {}^{A}q_B = \cos\frac{\theta}{2} + \widehat{\mathbf u}\sin\frac{\theta}{2}.$

(22)

Thus conjugation reverses the passive frame transformation. The same numerical quaternion on the right side of equation (23) is also the familiar positive active rotor for the corresponding physical vector rotation in a fixed frame.

This distinction is why a sign change in the vector part must not be described without also stating the intended frame map.

Relationship to direction cosine matrices

Let

$\displaystyle {}^{B}C_A = C({}^{B}q_A)$

be the passive direction cosine matrix corresponding to the unit quaternion ${}^{B}q_A$ .

The inverse frame map satisfies

$\displaystyle {}^{A}C_B = ({}^{B}C_A)^{-1}.$

Since a direction cosine matrix is orthogonal,

$\displaystyle ({}^{B}C_A)^{-1} = ({}^{B}C_A)^T.$

Quaternion conjugation gives the same result:

$\displaystyle C(q^*) = C(q)^T.$

(23)

Thus, for unit frame quaternions,

$\displaystyle C(({}^{B}q_A)^*) = C({}^{A}q_B) = {}^{A}C_B = ({}^{B}C_A)^T.$

(24)

This provides a useful implementation check between quaternion and matrix orientation code.

Conjugation and frame composition

Suppose

$\displaystyle {}^{B}q_A$

maps $A$ coordinates into $B$ , and

$\displaystyle {}^{C}q_B$

maps $B$ coordinates into $C$ .

The direct passive frame map is

$\displaystyle {}^{C}q_A = {}^{C}q_B\,{}^{B}q_A.$

(25)

Conjugate both sides:

$\displaystyle ({}^{C}q_A)^* = ({}^{B}q_A)^* ({}^{C}q_B)^*.$

Using the inverse frame labels,

$\displaystyle {}^{A}q_C = {}^{A}q_B\,{}^{B}q_C.$

(26)

The reversed product order produced by conjugation is exactly what is required when the entire frame chain is traversed in the opposite direction.

Worked example 1: numerical conjugate

Let

$\displaystyle q=2-3\mathbf i+4\mathbf j+5\mathbf k.$

Then

$\displaystyle q^* = 2+3\mathbf i-4\mathbf j-5\mathbf k.$

(27)

Applying conjugation again returns

$\displaystyle (q^*)^*=q.$

Worked example 2: product with the conjugate

For

$\displaystyle q=2-3\mathbf i+4\mathbf j+5\mathbf k,$

equation (15) gives

$\displaystyle qq^* = 2^2+(-3)^2+4^2+5^2.$

Therefore

$\displaystyle qq^*=54.$

(28)

The result is purely real.

The norm is consequently

$\displaystyle \lVert q\rVert=\sqrt{54}=3\sqrt6,$

and the inverse is

$\displaystyle q^{-1} = \frac{1}{54} \left( 2+3\mathbf i-4\mathbf j-5\mathbf k \right).$

Worked example 3: order reversal

Let

$\displaystyle p=1+\mathbf i, \qquad q=1+\mathbf j.$

First,

$\displaystyle pq = 1+\mathbf i+\mathbf j+\mathbf k.$

Therefore

$\displaystyle (pq)^* = 1-\mathbf i-\mathbf j-\mathbf k.$

Now compute the reversed conjugate product:

$\displaystyle q^*p^* = (1-\mathbf j)(1-\mathbf i).$

Expanding,

$\begin{equation*} \begin{aligned} q^*p^* &= 1-\mathbf i-\mathbf j+\mathbf j\mathbf i\ &= 1-\mathbf i-\mathbf j-\mathbf k. \end{aligned}\end{equation*}$

Hence

$\displaystyle (pq)^*=q^*p^*.$

If one incorrectly used $p^*q^*$ , the sign of the $\mathbf k$ term would be wrong.

Worked example 4: reversing a passive frame rotation

Let frame $B$ be obtained from frame $A$ by a positive $90^\circ$ rotation about $+z$ . The PhysicsLibrary passive quaternion is

$\displaystyle {}^{B}q_A = \frac{\sqrt2}{2} - \frac{\sqrt2}{2}\mathbf k.$

Its conjugate is

$\displaystyle {}^{A}q_B = ({}^{B}q_A)^* = \frac{\sqrt2}{2} + \frac{\sqrt2}{2}\mathbf k.$

The first quaternion maps coordinate components from $A$ to $B$ . The conjugate maps them from $B$ back to $A$ .

If a fixed physical vector has

$\displaystyle {}^{A}\mathbf v = \begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix},$

then the positive frame rotation produces

$\displaystyle {}^{B}\mathbf v = \begin{bmatrix} 0\\ -1\\ 0 \end{bmatrix}.$

Applying the conjugate frame map returns

$\displaystyle {}^{A}\mathbf v = \begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix}.$

Common pitfalls

Changing the scalar part. Quaternion conjugation leaves unchanged and reverses only the vector part.
Forgetting product order reversal. The correct identity is

$\displaystyle (pq)^*=q^*p^*,$
not .
Assuming $q^*=q^{-1}$ for every quaternion. This is true only for unit quaternions. In general,

$\displaystyle q^{-1}=q^*/\lVert q\rVert^2.$
Confusing quaternion conjugation with component storage order. Scalar first versus scalar last storage does not change the definition of the conjugate.
Confusing quaternion conjugation with complex conjugation of a matrix. The operations are related in some representations but are not the same object.
Conjugating a passive frame quaternion without reversing its frame meaning. For unit frame quaternions,

$\displaystyle {}^{A}q_B=({}^{B}q_A)^*.$
Changing Hamilton multiplication because the attitude interpretation is passive. Conjugation and passive frame interpretation do not change the Hamilton product.

Exercises

The exercises are stated first so the article can be used for self study. Complete solutions follow afterward.

Direct conjugation. For

$\displaystyle q=4-2\mathbf i+7\mathbf j-3\mathbf k,$
find .
Pure and real cases. Find the conjugate of

$\displaystyle q=5$
and of

$\displaystyle p=2\mathbf i-\mathbf j+4\mathbf k.$
Double conjugation. Prove directly that

$\displaystyle (q^*)^*=q.$
Scalar and vector extraction. Starting from $q=q_w+\mathbf q$ , derive

$\displaystyle \operatorname{Sc}(q)=\frac{q+q^*}{2}$
and

$\displaystyle \operatorname{Vec}(q)=\frac{q-q^*}{2}.$
Product with the conjugate. For

$\displaystyle q=1+2\mathbf i-2\mathbf j+\mathbf k,$
compute .
Order reversal. Let

$\displaystyle p=1+\mathbf i, \qquad q=2+\mathbf j.$
Compute and and verify that they agree.
Incorrect order. For the same and , compute and show that it differs from .
Unit quaternion inverse. If $\lVert q\rVert=1$ , prove that $q^*=q^{-1}$ .
Passive frame inverse. If ${}^{B}q_A$ is a unit quaternion mapping coordinates into , write the quaternion that maps coordinates back into .
Frame chain reversal. Given

$\displaystyle {}^{C}q_A = {}^{C}q_B\,{}^{B}q_A,$
conjugate the equation and derive the inverse frame chain.

Solutions

1. Direct conjugation

Starting from

$\displaystyle q=4-2\mathbf i+7\mathbf j-3\mathbf k,$

reverse all three imaginary coefficients:

$\displaystyle q^* = 4+2\mathbf i-7\mathbf j+3\mathbf k.$

2. Pure and real cases

A real quaternion is unchanged:

$\displaystyle 5^*=5.$

For the pure quaternion

$\displaystyle p=2\mathbf i-\mathbf j+4\mathbf k,$

the conjugate is

$\displaystyle p^* = -2\mathbf i+\mathbf j-4\mathbf k = -p.$

3. Double conjugation

$\displaystyle q=q_w+\mathbf q,$

then

$\displaystyle q^*=q_w-\mathbf q.$

Conjugating again gives

$\displaystyle (q^*)^* = q_w+\mathbf q = q.$

4. Scalar and vector extraction

Add:

$\displaystyle q+q^* = (q_w+\mathbf q)+(q_w-\mathbf q) = 2q_w.$

Therefore

$\displaystyle \operatorname{Sc}(q) = \frac{q+q^*}{2}.$

Subtract:

$\displaystyle q-q^* = (q_w+\mathbf q)-(q_w-\mathbf q) = 2\mathbf q.$

Therefore

$\displaystyle \operatorname{Vec}(q) = \frac{q-q^*}{2}.$

5. Product with the conjugate

For

$\displaystyle q=1+2\mathbf i-2\mathbf j+\mathbf k,$

the squared norm is

$\displaystyle 1^2+2^2+(-2)^2+1^2=10.$

Hence

$\displaystyle qq^*=10.$

6. Order reversal

First,

$\begin{equation*} \begin{aligned} pq &=(1+\mathbf i)(2+\mathbf j)\ &=2+\mathbf j+2\mathbf i+\mathbf k. \end{aligned}\end{equation*}$

Thus

$\displaystyle (pq)^* = 2-2\mathbf i-\mathbf j-\mathbf k.$

Now

$\displaystyle q^*=2-\mathbf j, \qquad p^*=1-\mathbf i.$

Then

$\begin{equation*} \begin{aligned} q^*p^* &=(2-\mathbf j)(1-\mathbf i)\ &=2-2\m... ...f j\mathbf i\ &=2-2\mathbf i-\mathbf j-\mathbf k. \end{aligned}\end{equation*}$

Therefore

$\displaystyle (pq)^*=q^*p^*.$

7. Incorrect order

Compute

$\begin{equation*} \begin{aligned} p^*q^* &=(1-\mathbf i)(2-\mathbf j)\ &=2-\ma... ...f i\mathbf j\ &=2-2\mathbf i-\mathbf j+\mathbf k. \end{aligned}\end{equation*}$

This differs from

$\displaystyle (pq)^* = 2-2\mathbf i-\mathbf j-\mathbf k.$

The factor order must reverse.

8. Unit quaternion inverse

From

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

and $\lVert q\rVert=1$ ,

$\displaystyle qq^*=1.$

Similarly,

$\displaystyle q^*q=1.$

Therefore $q^*$ is both the left and right inverse of $q$ :

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

9. Passive frame inverse

The inverse coordinate map is represented by the conjugate:

$\displaystyle {}^{A}q_B = ({}^{B}q_A)^*.$

10. Frame chain reversal

Begin with

$\displaystyle {}^{C}q_A = {}^{C}q_B\,{}^{B}q_A.$

Conjugating gives

$\displaystyle ({}^{C}q_A)^* = ({}^{B}q_A)^* ({}^{C}q_B)^*.$

Relabeling each conjugate as the inverse frame map,

$\displaystyle {}^{A}q_C = {}^{A}q_B\,{}^{B}q_C.$

Thus the conjugated chain traverses the same sequence of frames in the reverse direction.

What comes next

Quaternion conjugation immediately produces the real quantity

$\displaystyle qq^*=q^*q.$

The next PhysicsLibrary article develops the quaternion norm from this identity and proves its important multiplicative properties. The following inverse article then uses conjugation and norm together to obtain

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

Later orientation articles use the unit quaternion specialization $q^{-1}=q^*$ to reverse passive frame transformations, invert direction cosine matrices, and derive coordinate transformation and attitude kinematic formulas.

Sources and historical notes

Quaternion conjugation appears naturally in Hamilton's algebra and in later quaternion texts under several historical notations. Modern engineering notation often writes $q^*$ or $\overline q$ and interprets conjugation as reversing the vector part.

Hamilton's Elements of Quaternions and Joly's A Manual of Quaternions provide public domain historical treatments. The modern PhysicsLibrary presentation emphasizes the anti order property, the relationship to norm and inverse, and the passive frame interpretation needed in mechanics and navigation.

Bibliography

1: W. R. Hamilton, Elements of Quaternions, 2nd ed., edited by C. J. Joly, Longmans, Green, and Co., 1899. Public domain historical source. Internet Archive scan https://archive.org/details/elementsofquater01hamiuoft
2: C. J. Joly, A Manual of Quaternions, Macmillan and Co., London, 1905. Public domain historical source. Internet Archive scan https://archive.org/details/manualofquaterni00jolyrich
3: A. S. Hathaway, A Primer of Quaternions, 1896. Public domain historical source. Project Gutenberg edition https://www.gutenberg.org/ebooks/9934
4: Wikibooks contributors, “Abstract Algebra/Quaternions.” Openly licensed instructional source. Wikibooks quaternion article https://en.wikibooks.org/wiki/Abstract_Algebra/Quaternions

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