proof of Rodrigues’ rotation formula

proof of Rodrigues’ rotation formula

Let

[

𝐱

,

𝐲

,

𝐳

]

be a frame of

right-handed

orthonormal vectors

in

ℝ

3

,
and let

𝐯

=

a

⁢

𝐱

+

b

⁢

𝐲

+

c

⁢

𝐳

(with

a

,

b

,

c

∈

ℝ

) be any vector to be rotated on the

𝐳

axis,
by an angle

θ

counter-clockwise.

The image vector

𝐯

′

is the vector

𝐯

with its

component

in the

𝐱

,

𝐲

plane rotated, so we can write

𝐯

′

=

a

⁢

𝐱

′

+

b

⁢

𝐲

′

+

c

⁢

𝐳

,

where

𝐱

′

and

𝐲

′

are the

rotations

by angle

θ

of the

𝐱

and

𝐲

vectors in the

𝐱

,

𝐲

plane. By the

rotation formula

in two

dimensions

, we have

𝐱

′

=

cos

⁡

θ

⁢

𝐱

+

sin

⁡

θ

⁢

𝐲

,

𝐲

′

=

-

sin

⁡

θ

⁢

𝐱

+

cos

⁡

θ

⁢

𝐲

.

So

𝐯

′

=

cos

⁡

θ

⁢

(

a

⁢

𝐱

+

b

⁢

𝐲

)

+

sin

⁡

θ

⁢

(

a

⁢

𝐲

-

b

⁢

𝐱

)

+

c

⁢

𝐳

.

The vector

a

⁢

𝐱

+

b

⁢

𝐲

is the

projection

of

𝐯

onto the

𝐱

,

𝐲

plane, and

a

⁢

𝐲

-

b

⁢

𝐱

is its rotation by

90

∘

.
So these two vectors form an

orthogonal

frame in the

𝐱

,

𝐲

plane,
although they are not necessarily

unit vectors

.
Alternate expressions for these vectors are easily derived — especially with the help of the picture:

𝐯

-

(

𝐯

⋅

𝐳

)

⁢

𝐳

=

𝐯

-

c

⁢

𝐳

=

a

⁢

𝐱

+

b

⁢

𝐲

,

𝐳

×

𝐯

=

a

⁢

(

𝐳

×

𝐱

)

+

b

⁢

(

𝐳

×

𝐲

)

+

c

⁢

(

𝐳

×

𝐳

)

=

a

⁢

𝐲

-

b

⁢

𝐱

.

Substituting these into the expression for

𝐯

′

:

𝐯

′

=

cos

⁡

θ

⁢

(

𝐯

-

(

𝐯

⋅

𝐳

)

⁢

𝐳

)

+

sin

⁡

θ

⁢

(

𝐳

×

𝐯

)

+

c

⁢

𝐳

,

which could also have been derived directly
if we had first considered the frame

[

𝐯

-

(

𝐯

⋅

𝐳

)

,

𝐳

×

𝐯

]

instead of

[

𝐱

,

𝐲

]

.

We attempt to simplify further:

𝐯

′

=

𝐯

+

sin

⁡

θ

⁢

(

𝐳

×

𝐯

)

+

(

cos

⁡

θ

-

1

)

⁢

(

𝐯

-

(

𝐯

⋅

𝐳

)

⁢

𝐳

)

.

Since

𝐳

×

𝐯

is linear in

𝐯

, this

transformation

is represented by
a

linear operator

A

. Under a right-handed

orthonormal basis

,
the

matrix representation

of

A

is directly computed to be

A

⁢

𝐯

=

𝐳

×

𝐯

=

[

0

-

z

3

z

2

z

3

0

-

z

1

-

z

2

z

1

0

]

⁢

[

v

1

v

2

v

3

]

.

We also have

-

(

𝐯

-

(

𝐯

⋅

𝐳

)

⁢

𝐳

)

=

-

a

⁢

𝐱

-

b

⁢

𝐲

(rotate

a

⁢

𝐱

+

b

⁢

𝐲

by

180

∘

)

=

𝐳

×

(

a

⁢

𝐲

-

b

⁢

𝐱

)

(rotate

a

⁢

𝐲

-

b

⁢

𝐱

by

90

∘

)

=

𝐳

×

(

𝐳

×

(

a

⁢

𝐱

+

b

⁢

𝐲

+

c

⁢

𝐳

)

)

=

A

2

⁢

𝐯

.

So

𝐯

′

=

I

⁢

𝐯

+

sin

⁡

θ

⁢

A

⁢

𝐯

+

(

1

-

cos

⁡

θ

)

⁢

A

2

⁢

𝐯

,

proving Rodrigues’ rotation formula.

Relation with the matrix exponential

Here is a curious fact. Notice that the matrix

1

1

If we want to use coordinate-free , then in this

section

, “matrix” should be replaced by “linear operator” and

transposes

should be replaced by the

adjoint

operation.

A

is

skew-symmetric

.
This is not a coincidence — for any

skew-symmetric matrix

B

, we have

(

e

B

)

t

=

e

B

t

=

e

-

B

=

(

e

B

)

-

1

,
and

det

⁡

e

B

=

e

tr

⁡

B

=

e

0

=

1

,
so

e

B

is always a rotation. It is in fact the case that:

I

+

sin

⁡

θ

⁢

A

+

(

1

-

cos

⁡

θ

)

⁢

A

2

=

e

θ

⁢

A

for the matrix

A

we had above! To prove this, observe that powers of

A

cycle like so:

I

,

A

,

A

2

,

-

A

,

-

A

2

,

A

,

A

2

,

-

A

,

-

A

2

,

…

Then

sin

⁡

θ

⁢

A

=

∑

k

=

0

∞

(

-

1

)

k

⁢

θ

2

⁢

k

+

1

⁢

A

(

2

⁢

k

+

1

)

!

=

∑

k

=

0

∞

θ

2

⁢

k

+

1

⁢

A

2

⁢

k

+

1

(

2

⁢

k

+

1

)

!

=

∑

k

⁢

odd

(

θ

⁢

A

)

k

k

!

(

1

-

cos

⁡

θ

)

⁢

A

2

=

∑

k

=

1

∞

(

-

1

)

k

-

1

⁢

θ

2

⁢

k

⁢

A

2

(

2

⁢

k

)

!

=

∑

k

=

1

∞

θ

2

⁢

k

⁢

A

2

⁢

k

(

2

⁢

k

)

!

=

∑

k

≥

2

⁢

even

(

θ

⁢

A

)

k

k

!

.

Adding

sin

⁡

θ

⁢

A

,

(

1

-

cos

⁡

θ

)

⁢

A

2

and

I

together, we obtain the power series
for

e

θ

⁢

A

.

Second proof:

If we regard

θ

as time, and differentiate the

equation

𝐯

′

=

a

⁢

𝐱

′

+

b

⁢

𝐲

′

+

c

⁢

𝐳

with respect to

θ

, we obtain

d

⁢

𝐯

′

/

d

⁢

θ

=

a

⁢

𝐲

′

-

b

⁢

𝐱

′

=

𝐳

×

𝐯

′

=

A

⁢

𝐯

′

, whence the solution (to this linear ODE)
is

𝐯

′

=

e

θ

⁢

A

⁢

𝐯

.

Remark:

The

operator

e

θ

⁢

A

, as

θ

ranges over

ℝ

,
is a one-parameter subgroup of

SO

⁢

(

3

)

.
In higher dimensions

n

, every rotation in

SO

⁢

(

n

)

is of the form

e

A

for a skew-symmetric

A

, and the second proof above
can be modified to prove this more general fact.

Title

proof of Rodrigues’ rotation formula

Canonical name

ProofOfRodriguesRotationFormula

Date of creation

2013-03-22 15:23:25

Last modified on

2013-03-22 15:23:25

Owner

stevecheng (10074)

Last modified by

stevecheng (10074)

Numerical id

14

Author

stevecheng (10074)

Entry type

Proof

Classification

msc 51-00

Classification

msc 15-00

Related topic

DimensionOfTheSpecialOrthogonalGroup

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