example of quaternion conjugate

Quaternion Conjugate: Examples, Exercises, and Solutions

This entry is the self study companion to 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.

1 Formula summary

For

q = qw + qxi + qyj + qzk,

(1)

the conjugate is

q∗ = qw − qxi − qyj − qzk.

(2)

In scalar vector notation,

q = qw + q, q∗ = qw − q.

(3)

Conjugation is an involution:

(q∗)∗ = q.

(4)

It is real linear:

(p + q)∗ = p∗ + q∗,

(5)

(λq)∗ = λq∗, λ ∈ ℝ.

(6)

Most importantly, conjugation reverses product order:

(pq)∗ = q∗p∗.

(7)

The product of a quaternion with its conjugate is real:

qq∗ = q∗q = q2 + q2 + q2+ q2. w x y z

(8)

For a unit quaternion,

∗ − 1 q = q .

(9)

For the PhysicsLibrary passive attitude convention,

A B ∗ qB = ( qA) .

(10)

2 Exercises

Basic conjugation.
For

$q = 3 − 2i + 5j − 7k,$
compute q^∗ and the scalar first component columns of both q and q^∗.
Which component is unchanged?
Real and pure special cases.
Compute the conjugate of

$r = 5$
and of

$v = 2i − 3j + 4k.$
Show that every real quaternion satisfies

$∗ r = r$
and every pure quaternion satisfies

$v∗ = − v.$
Conjugation as an involution.
Starting from

$q = qw + q,$
prove directly that

$∗ ∗ (q ) = q.$
Then verify the result numerically for

$q = − 2 + i + 3j − 4k.$
Conjugate of a sum and real scalar multiple.
Let

$p = 2 + i − 3j$
and

$q = − 1 + 4j + 2k.$
Verify explicitly that

$(p + q)∗ = p∗ + q∗$
and

$(− 3p)∗ = − 3p ∗.$
Product reversal using basis quaternions.
Evaluate

$(ij)∗$
directly.
Then evaluate

$j∗i∗.$
Verify

$(ij)∗ = j∗i∗.$
Explain why the reversed factor order is essential.
Product reversal for general numerical quaternions.
Let

$p = 1 + 2i − j + 3k$
and

$q = 2 − i + 4j + k.$
Compute pq, then (pq)^∗.
Separately compute q^∗p^∗ and verify that the two results agree.
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.
Product with the conjugate.
For

$q = 2 − i + 2j + 2k,$
compute

$∗ qq$
and

$q∗q.$
Show explicitly that both are the same real quaternion.
Derive $qq∗$ in scalar vector form.
Let

$q = qw + q.$
Using the Hamilton scalar vector product, derive

$∗ 2 qq = qw + q ⋅ q.$
Why does the vector part vanish?
Conjugate and norm connection.
Suppose

$q = 1 − 2i + 2j − 4k.$
Compute qq^∗.
Without yet using the explicit inverse formula, identify the Euclidean norm squared of the four quaternion components.
Conjugate of a triple product.
Prove

$∗ ∗ ∗ ∗ (pqr) = r q p .$ (11)

Then state the corresponding pattern for

$(q1q2q3q4)∗.$
Conjugation of a pure quaternion product.
Let a and b be pure quaternions.
Starting from

$ab = − a ⋅ b + a × b,$
take the conjugate and show that

$(ab)∗ = − a ⋅ b − a × b.$ (12)

Then show that this equals

$ba.$
Unit quaternion and reverse passive frame map.
Frame B is obtained from frame A by a positive 90^∘ frame rotation about +z.
The PhysicsLibrary passive quaternion is

$B 1-−-k- qA = √ 2 .$
Compute

$AqB.$
State the physical meaning of the conjugate in frame-map language.
Conjugate sandwich reverses the coordinate map.
Let

$B q = qA$
be unit and suppose

$Bv = qAv q∗.$
Starting from this equation, solve algebraically for ^Av in terms of ^Bv and q.
Show that the reverse map is

$Av = q∗Bv q.$ (13)
$q$ versus $q∗$ is not the same as $q$ versus $− q$ .
For

$1 −-k- q = √2--,$
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.
Conjugate as a convention diagnostic.
A software library claims to store the passive frame quaternion ^Bq_A.
For a positive 90^∘ frame rotation about +z, it reports

$q = 1 +√-k. 2$
Under the PhysicsLibrary convention, determine whether this quaternion more naturally represents ^Bq_A or its reverse map ^Aq_B.
What simple conjugation operation converts it to the PhysicsLibrary A → B passive map?

3 Solutions

Solution 1: basic conjugation

The quaternion is

q = 3 − 2i + 5j − 7k.

Conjugation leaves the scalar coefficient unchanged and changes the sign of the vector part:

q∗ = 3 + 2i − 5j + 7k.

(14)

The scalar first component columns are

⌊ ⌋ ⌊ ⌋ 3 3 ||− 2|| ∗ || 2 || [q]PL = ⌈ 5 ⌉ , [q ]PL = ⌈ − 5⌉ . − 7 7

(15)

The scalar component is unchanged.

Solution 2: real and pure special cases

For the real quaternion

r = 5,

the vector part is zero, so

r∗ = 5 = r.

(16)

For the pure quaternion

v = 2i − 3j + 4k,

the scalar part is zero. Therefore

∗ v = − 2i + 3j − 4k = − v.

(17)

Thus real quaternions are fixed by conjugation, while pure quaternions change sign.

Solution 3: conjugation as an involution

Write

q = q + q. w

Then

∗ q = qw − q.

Conjugate again:

(q∗)∗ = qw + q = q.

Therefore

(q∗)∗ = q.

(18)

For

q = − 2 + i + 3j − 4k,

we have

q∗ = − 2 − i − 3j + 4k.

Conjugating again gives

∗ ∗ (q ) = − 2 + i + 3j − 4k = q.

Solution 4: conjugate of a sum and real scalar multiple

First,

p + q = 1 + i + j + 2k.

Therefore

(p + q)∗ = 1 − i − j − 2k.

Separately,

∗ p = 2 − i + 3j

and

∗ q = − 1 − 4j − 2k.

Hence

∗ ∗ p + q = 1 − i − j − 2k.

Thus

(p + q)∗ = p∗ + q∗.

(19)

Also,

− 3p = − 6 − 3i + 9j.

Therefore

∗ (− 3p) = − 6 + 3i − 9j.

Meanwhile,

∗ − 3p = − 3 (2 − i + 3j) = − 6 + 3i − 9j.

Thus

(− 3p)∗ = − 3p ∗.

(20)

Solution 5: product reversal using basis quaternions

Hamilton multiplication gives

ij = k.

Therefore

∗ ∗ (ij) = k = − k.

Now

i∗ = − i, j∗ = − j.

Hence

j∗i∗ = (− j)(− i) = ji = − k.

Thus

(ij)∗ = j∗i∗.

(21)

If the factor order were not reversed, one would obtain

i∗j∗ = ij = k,

which has the wrong sign.

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 − 10i − 3j + 14k.

Therefore

(pq )∗ = 5 + 10i + 3j − 14k.

(22)

Now

q∗ = 2 + i − 4j − k

and

∗ p = 1 − 2i + j − 3k.

Multiplying in the reversed order gives

q∗p∗ = 5 + 10i + 3j − 14k.

(23)

Hence

(pq)∗ = q∗p∗.

Solution 7: why $p∗q∗$ is generally wrong

Using the same conjugates,

p∗ = 1 − 2i + j − 3k

and

q∗ = 2 + i − 4j − k.

Direct multiplication gives

∗ ∗ p q = 5 − 16i − 7j.

(24)

But from Solution 6,

∗ (pq ) = 5 + 10i + 3j − 14k.

These are not equal.

The conjugation rule reverses factor order because quaternion multiplication is not generally commutative.

Solution 8: product with the conjugate

For

q = 2 − i + 2j + 2k,

the conjugate is

q∗ = 2 + i − 2j − 2k.

The scalar part of qq^∗ is

22 + (− 1)2 + 22 + 22 = 13.

The vector part cancels.

Therefore

qq∗ = 13.

(25)

Similarly,

q∗q = 13.

(26)

Both products are the same real quaternion.

Solution 9: derive $qq∗$ in scalar vector form

Let

q = qw + q

and

q∗ = qw − q.

Use the Hamilton scalar vector product.

The scalar part is

qwqw − q ⋅ (− q ) = q2w + q ⋅ q.

The vector part is

q (− q) + q q + q × (− q). w w

The first two terms cancel, and

q × q = 0.

Therefore the vector part vanishes.

Thus

qq∗ = q2w + q ⋅ q.

(27)

Solution 10: conjugate and norm connection

For

q = 1 − 2i + 2j − 4k,

the sum of the squared components is

12 + (− 2)2 + 22 + (− 4)2 = 1 + 4 + 4 + 16 = 25.

Therefore

qq∗ = 25.

(28)

This is the squared Euclidean norm of the four component column:

∥q∥2 = 25.

(29)

Hence

∥q∥ = 5.

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

∗ ∗ ∗ ∗ (pqr) = r q p .

(30)

Likewise,

∗ ∗ ∗ ∗ ∗ (q1q2q3q4) = q4q3q2q1.

(31)

Conjugation reverses the entire factor sequence.

Solution 12: conjugation of a pure quaternion product

For pure quaternions,

ab = − a ⋅ b + a × b.

The dot product term is real and is unchanged by conjugation.

The cross product term is pure and changes sign.

Therefore

∗ (ab) = − a ⋅ b − a × b.

(32)

Now reverse the pure factors:

ba = − b ⋅ a + b × a.

Because

b ⋅ a = a ⋅ b

and

b × a = − a × b,

we obtain

ba = − a ⋅ b − a × b.

Hence

(ab)∗ = ba.

(33)

Solution 13: unit quaternion and reverse passive frame map

The forward passive frame quaternion is

B 1 − k qA = -√----. 2

Its conjugate is

B ∗ 1-+-k- ( qA) = √2-- .

For a unit attitude quaternion,

A B ∗ qB = ( qA) .

Therefore

1 + k AqB = -√---. 2

(34)

The conjugate reverses the passive coordinate map: it maps B coordinates back into A coordinates.

Solution 14: conjugate sandwich reverses the coordinate map

Start from

Bv = qAv q∗.

Left multiply by q^∗:

q∗Bv = q∗qAv q∗.

Because q is unit,

q∗q = 1.

Thus

q∗Bv = Av q∗.

Right multiply by q:

∗B A ∗ q v q = v q q.

Again,

q∗q = 1.

Therefore

Av = q∗Bv q.

(35)

The reverse coordinate transformation uses the conjugate quaternion in the opposite sandwich order.

Solution 15: $q$ versus $∗ q$ is not the same as $q$ versus $− q$

Given

q = 1 −√-k, 2

the conjugate is

∗ 1-+-k- q = √2-- .

(36)

The negative quaternion is

−-1 +-k- − q = √2-- .

(37)

The pair q and q^∗ generally represents opposite passive frame maps:

B A qA versus qB.

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.

Solution 16: conjugate as a convention diagnostic

PhysicsLibrary assigns a positive 90^∘ frame rotation about +z the passive A → B quaternion

B 1-−-k- qA = √2-- .

The software reports

1 + k q = -√---. 2

This is exactly the conjugate of the PhysicsLibrary A → B map.

Therefore it more naturally corresponds to

1 + k AqB = -√---. 2

(38)

To convert it to the PhysicsLibrary A → B passive map, take the conjugate:

1 − k qPL = q∗ = -√----. 2

4 Compact review

The core identities reinforced by this companion are

∗ q = qw − q,

(39)

(q∗)∗ = q,

(40)

(pq)∗ = q∗p∗,

(41)

qq∗ = q∗q = ∥q∥2,

(42)

and, for unit passive attitude quaternions,

AqB = (BqA )∗.

(43)

Conjugation is therefore both an algebraic operation and, for unit attitude quaternions, the operation that reverses a frame to frame coordinate map.

5 Sources and exercise provenance

The exercises and solutions in this companion are newly written or expanded for PhysicsLibrary from the algebra developed in 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.

References

[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

[2] C. J. Joly, A Manual of Quaternions, Macmillan and Co., London, 1905. Public domain historical source. Internet Archive search

[3] P. Kelland and P. G. Tait, Introduction to Quaternions, with Numerous Examples, 2nd ed., Macmillan and Co., London, 1882. Public domain historical source. Internet Archive search

[4] H. Sommer, I. Gilitschenski, M. Bloesch, S. Weiss, R. Siegwart, and J. Nieto, “Why and How to Avoid the Flipped Quaternion Multiplication,” Aerospace, vol. 5, no. 3, article 72, 2018. Published under CC BY 4.0. Publisher article

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Quaternion Conjugate: Examples, Exercises, and Solutions

1 Formula summary

2 Exercises

3 Solutions

Solution 1: basic conjugation

Solution 2: real and pure special cases

Solution 3: conjugation as an involution

Solution 4: conjugate of a sum and real scalar multiple

Solution 5: product reversal using basis quaternions

Solution 6: product reversal for general numerical quaternions

Solution 7: why is generally wrong

Solution 8: product with the conjugate

Solution 9: derive in scalar vector form

Solution 10: conjugate and norm connection

Solution 11: conjugate of a triple product

Solution 12: conjugation of a pure quaternion product

Solution 13: unit quaternion and reverse passive frame map

Solution 14: conjugate sandwich reverses the coordinate map

Solution 15: versus is not the same as versus

Solution 16: conjugate as a convention diagnostic

4 Compact review

5 Sources and exercise provenance

References

License

Solution 7: why $p∗q∗$ is generally wrong

Solution 9: derive $qq∗$ in scalar vector form

Solution 15: $q$ versus $∗ q$ is not the same as $q$ versus $− q$