Time Derivative of Rotation Matrices: A Tutorial
Shiyu Zhao
Abstract— The time derivative of a rotation matrix equals the
product of a skew-symmetric matrix and the rotation matrix
itself. This article gives a brief tutorial on the well-known result.
I. INTRODUCTION
The attitude of a ground or aerial robot is often represented
by a rotation matrix, whose time derivative is important
to characterize the rotational kinematics of the robot. It is
a well-known result that the time derivative of a rotation
matrix equals the product of a skew-symmetric matrix and
the rotation matrix itself. One classic method to derive this
result is as follows [1, Sec 4.1], [2, Sec 2.3.1], and [3,
Sec 4.2.2] (see [4] for other methods). Let R(t) ∈R3×3
with t ≥0 be a rotation matrix satisfying R(t)RT(t) = I
for all t where I is the identity matrix. Taking time derivative
on both sides of R(t)RT(t) = I gives
˙R(t)RT(t) + R(t) ˙RT(t) = 0,
which indicates that S(t) ≜˙R(t)RT(t) is a skew-symmetric
matrix satisfying S(t)+ST(t) = 0 for all t, and consequently
˙R(t) = S(t)R(t).
The above derivation is simple, but it is not straightforward
to see the precise physical meaning of S(t) (though S(t)
corresponds to an angular velocity vector, it is unclear
which reference frame this vector is expressed in). This
article gives another simple derivation, which is essentially
a reorganization of the derivation in [1]–[3], to clarify the
precise physical meanings of the quantities in the expression
of the time derivative of a rotation matrix.
Notation: For any vector w = [w1, w2, w3]T ∈R3, define
the skew-symmetric operator [·]× as
[w]× ≜


0
−w3
w2
w3
0
−w1
−w2
w1
0

∈R3×3.
(1)
The skew-symmetric operator is useful because it can convert
a cross product of two vectors into a matrix-vector product.
More specifically, for any w, x ∈R3, it can be easily verified
that w × x = [w]× x. Another useful property is that for
any w ∈R3 and any rotation matrix R ∈R3×3 satisfying
RRT = I and det(R) = 1 it holds that [Rw]× = R [w]× RT
[3, Section 4.2.1].
Shiyu Zhao is with the Department of Automatic Control and Systems En-
gineering, University of Sheffield, UK. szhao@sheffield.ac.uk
xA
yA
zA
xB
yB
zB
wB
Fig. 1: Frame A is fixed while frame B is rotating.
II. TIME DERIVATIVE OF ROTATION MATRICES
Consider two reference frames A and B in the three-
dimensional space (see Figure 1). Assume the origins of the
two frames collocate with each other. Suppose frame A is
fixed and frame B is rotating. In the area of robotics, frame A
usually corresponds to the world frame fixed on the ground,
and frame B usually corresponds to the body frame attached
to the body of a robot.
In the sequel, the time variables of all the matrices and
vectors are omitted for the sake of simplicity. Let the rotation
matrix RA
B ∈R3×3, which satisfies (RA
B)−1 = (RA
B)T and
det(RA
B) = 1, represent the rotational transformation from
frame B to frame A. For any point in the space, suppose
PA ∈R3 and PB ∈R3 are its coordinates expressed in
frames A and B, respectively, then
PA = RA
BPB.
Let RB
A = (RA
B)T be the rotation from frame A to frame B.
Suppose wB ∈R3 is the angular velocity of frame B
(relative to frame A) expressed in frame B. The vector wB
quantifies the rotational movement of frame B: ∥wB∥equals
to the angular rate, which quantifies how fast frame B is
rotating, and wB/∥wB∥indicates the axis of the rotational
movement. Since the angular velocity is a vector, it can also
be expressed in frame A as wA ∈R3, which satisfies
wA = RA
BwB.
The following is the main result on the relation between
rotational transformations and angular velocities.
Theorem 1 (Time Derivative of Rotation Matrices). The
time derivative of the rotational transformations RA
B and RB
A
are expressed as
˙RA
B = [wA]×RA
B
(2)
˙RA
B = RA
B[wB]×
(3)
˙RB
A = −RB
A[wA]×
(4)
˙RB
A = −[wB]×RB
A
(5)
arXiv:1609.06088v1  [cs.RO]  20 Sep 2016
Proof. We first prove (2). Consider an arbitrary point fixed
in frame B. If PA and PB are the coordinates of this point
in frames A and B, respectively, then PB is constant since
the point is fixed in frame B, and PA is time-varying since
frame B is rotating. As a result, we have ˙PB = 0. Taking
time derivative on both sides of PA = RA
BPB yields
˙PA = ˙RA
BPB.
(6)
On the other hand, by the relation between linear and angular
velocities, we have
˙PA = wA × PA = [wA]×PA.
(7)
Substituting (7) into (6) gives
˙RA
BPB = [wA]×PA = [wA]×RA
BPB.
(8)
Since PB may be arbitrarily chosen, equation (8) holds for
arbitrary PB ∈R3 and hence implies (2).
Equations (3)–(5) can be derived from (2). In particular,
by the property of the skew-symmetric operator, we have
[wA]× =

RA
BwB

× = RA
B [wB]× (RA
B)T.
(9)
Substituting (9) into (2) yields
˙RA
B
=
[wA]× RA
B
=
RA
B [wB]× (RA
B)TRA
B = RA
B [wB]×, which is (3). Taking
transpose on both sides of (2) gives
˙RB
A = −RB
A [wA]×,
which is (4). By substituting (9) into (4) leads to (5).
As indicated by Theorem 1, the expression of the time
derivative depends on the definitions of the rotation transfor-
mation and angular velocity. One should be clear about their
physical meanings before applying (2)–(5).
III. PRACTICAL CONSIDERATION IN ROBOTIC MOTION
For a robot equipped with an inertial measurement unit
(IMU), the value of wB, which is the angular velocity of
the robot relative to the world frame expressed in its body
frame, can be directly measured. As a result, equations (3)
and (5), i.e.,
˙RA
B = RA
B[wB]×
˙RB
A = −[wB]×RB
A
are particularly useful in practice.
It must be noted that the origins of frames A and B are
assumed to collocate with each other in Theorem 1. This
assumption is, however, usually not satisfied for moving
robots because the body frame may translate in the space
(see Figure 2). Nevertheless, (3) and (5) still holds in this
case. To prove that, we may introduce an intermediate
frame A′ whose axes are parallel to those of frame A and
origin collocates with the origin of frame B. By considering
frames A′ and B, we have ˙RA′
B = RA′
B [wB]× and ˙RB
A′ =
−[wB]×RB
A′. Since the axes of frame A′ are parallel to those
of frame A, we always have RA′
B = RA
B and RB
A′ = RB
A
and consequently (3) and (5) still holds (note wB remain
the same). On the other hand, if the origins of frames A
and B do not collocate, equations (2) and (4) do not hold
any more because w′
A ̸= wA due to the nonzero translation
between frames A′ and A. With the above discussion, we
A
x
y
z
B
x
y
z
wB
RA
B
Fig. 2: The world frame A and body frame B for a robot moving in the
three-dimensional space.
A
xA
yA
B
xB
yB
α
Fig. 3: The world frame A and body frame B for a robot moving in the
plane.
know equations (3) and (5) are more useful than equations
(2) and (4) in practice.
If a robot is moving in the plane, the rotation (or orien-
tation) of the robot can be represented by a single angle α
(see Figure 3). Then, the rotation transformation from frame
B to frame A is
Rα =
 cos α
−sin α
sin α
cos α

.
In order to verify Rα, one may examine some specific points
in frame B such as e1 = [1, 0]T and e2 = [0, 1]T. Taking
time derivative on both sides of Rα gives
˙Rα =
 −sin α
−cos α
cos α
−sin α

˙α.
(10)
Expression (10) may also be obtained as a special case of (3).
In particular, we can consider the three-dimensional frame
with the z-axes pointing out of the paper in Figure 3. Then,
RA
B =


cos α
−sin α
0
sin α
cos α
0
0
0
1

.
The angular velocity of the robot can be expressed as wA =
wB = ˙αe3 where e3 = [0, 0, 1]T. By applying (2) or (3), it
is straightforward to obtain (10).
REFERENCES
[1] R. M. Murray, Z. Li, and S. S. Sastry, A Mathematical Introduction to
Robotic Manipulation. CRC Press, 1994.
[2] Y. Ma, S. Soatto, J. Kosecka, and S. Sastry, An Invitation to 3D Vision.
New York: Springer, 2004.
[3] M. W. Spong, S. Hutchinson, and M. Vidyasagar, Robot Modeling and
Control. John Wiley & Sons, Inc., 2006.
[4] F. Hamano, “Derivative of rotation matrix - direct matrix derivation
of well-known formula,” in Proceedings of Green Energy and Systems
Conference, (Long Beach, CA), 2013.
Time Derivative of Rotation Matrices: A Tutorial
Shiyu Zhao
Abstract— The time derivative of a rotation matrix equals the
product of a skew-symmetric matrix and the rotation matrix
itself. This article gives a brief tutorial on the well-known result.
I. INTRODUCTION
The attitude of a ground or aerial robot is often represented
by a rotation matrix, whose time derivative is important
to characterize the rotational kinematics of the robot. It is
a well-known result that the time derivative of a rotation
matrix equals the product of a skew-symmetric matrix and
the rotation matrix itself. One classic method to derive this
result is as follows [1, Sec 4.1], [2, Sec 2.3.1], and [3,
Sec 4.2.2] (see [4] for other methods). Let R(t) ∈R3×3
with t ≥0 be a rotation matrix satisfying R(t)RT(t) = I
for all t where I is the identity matrix. Taking time derivative
on both sides of R(t)RT(t) = I gives
˙R(t)RT(t) + R(t) ˙RT(t) = 0,
which indicates that S(t) ≜˙R(t)RT(t) is a skew-symmetric
matrix satisfying S(t)+ST(t) = 0 for all t, and consequently
˙R(t) = S(t)R(t).
The above derivation is simple, but it is not straightforward
to see the precise physical meaning of S(t) (though S(t)
corresponds to an angular velocity vector, it is unclear
which reference frame this vector is expressed in). This
article gives another simple derivation, which is essentially
a reorganization of the derivation in [1]–[3], to clarify the
precise physical meanings of the quantities in the expression
of the time derivative of a rotation matrix.
Notation: For any vector w = [w1, w2, w3]T ∈R3, define
the skew-symmetric operator [·]× as
[w]× ≜


0
−w3
w2
w3
0
−w1
−w2
w1
0

∈R3×3.
(1)
The skew-symmetric operator is useful because it can convert
a cross product of two vectors into a matrix-vector product.
More specifically, for any w, x ∈R3, it can be easily verified
that w × x = [w]× x. Another useful property is that for
any w ∈R3 and any rotation matrix R ∈R3×3 satisfying
RRT = I and det(R) = 1 it holds that [Rw]× = R [w]× RT
[3, Section 4.2.1].
Shiyu Zhao is with the Department of Automatic Control and Systems En-
gineering, University of Sheffield, UK. szhao@sheffield.ac.uk
xA
yA
zA
xB
yB
zB
wB
Fig. 1: Frame A is fixed while frame B is rotating.
II. TIME DERIVATIVE OF ROTATION MATRICES
Consider two reference frames A and B in the three-
dimensional space (see Figure 1). Assume the origins of the
two frames collocate with each other. Suppose frame A is
fixed and frame B is rotating. In the area of robotics, frame A
usually corresponds to the world frame fixed on the ground,
and frame B usually corresponds to the body frame attached
to the body of a robot.
In the sequel, the time variables of all the matrices and
vectors are omitted for the sake of simplicity. Let the rotation
matrix RA
B ∈R3×3, which satisfies (RA
B)−1 = (RA
B)T and
det(RA
B) = 1, represent the rotational transformation from
frame B to frame A. For any point in the space, suppose
PA ∈R3 and PB ∈R3 are its coordinates expressed in
frames A and B, respectively, then
PA = RA
BPB.
Let RB
A = (RA
B)T be the rotation from frame A to frame B.
Suppose wB ∈R3 is the angular velocity of frame B
(relative to frame A) expressed in frame B. The vector wB
quantifies the rotational movement of frame B: ∥wB∥equals
to the angular rate, which quantifies how fast frame B is
rotating, and wB/∥wB∥indicates the axis of the rotational
movement. Since the angular velocity is a vector, it can also
be expressed in frame A as wA ∈R3, which satisfies
wA = RA
BwB.
The following is the main result on the relation between
rotational transformations and angular velocities.
Theorem 1 (Time Derivative of Rotation Matrices). The
time derivative of the rotational transformations RA
B and RB
A
are expressed as
˙RA
B = [wA]×RA
B
(2)
˙RA
B = RA
B[wB]×
(3)
˙RB
A = −RB
A[wA]×
(4)
˙RB
A = −[wB]×RB
A
(5)
arXiv:1609.06088v1  [cs.RO]  20 Sep 2016
Proof. We first prove (2). Consider an arbitrary point fixed
in frame B. If PA and PB are the coordinates of this point
in frames A and B, respectively, then PB is constant since
the point is fixed in frame B, and PA is time-varying since
frame B is rotating. As a result, we have ˙PB = 0. Taking
time derivative on both sides of PA = RA
BPB yields
˙PA = ˙RA
BPB.
(6)
On the other hand, by the relation between linear and angular
velocities, we have
˙PA = wA × PA = [wA]×PA.
(7)
Substituting (7) into (6) gives
˙RA
BPB = [wA]×PA = [wA]×RA
BPB.
(8)
Since PB may be arbitrarily chosen, equation (8) holds for
arbitrary PB ∈R3 and hence implies (2).
Equations (3)–(5) can be derived from (2). In particular,
by the property of the skew-symmetric operator, we have
[wA]× =

RA
BwB

× = RA
B [wB]× (RA
B)T.
(9)
Substituting (9) into (2) yields
˙RA
B
=
[wA]× RA
B
=
RA
B [wB]× (RA
B)TRA
B = RA
B [wB]×, which is (3). Taking
transpose on both sides of (2) gives
˙RB
A = −RB
A [wA]×,
which is (4). By substituting (9) into (4) leads to (5).
As indicated by Theorem 1, the expression of the time
derivative depends on the definitions of the rotation transfor-
mation and angular velocity. One should be clear about their
physical meanings before applying (2)–(5).
III. PRACTICAL CONSIDERATION IN ROBOTIC MOTION
For a robot equipped with an inertial measurement unit
(IMU), the value of wB, which is the angular velocity of
the robot relative to the world frame expressed in its body
frame, can be directly measured. As a result, equations (3)
and (5), i.e.,
˙RA
B = RA
B[wB]×
˙RB
A = −[wB]×RB
A
are particularly useful in practice.
It must be noted that the origins of frames A and B are
assumed to collocate with each other in Theorem 1. This
assumption is, however, usually not satisfied for moving
robots because the body frame may translate in the space
(see Figure 2). Nevertheless, (3) and (5) still holds in this
case. To prove that, we may introduce an intermediate
frame A′ whose axes are parallel to those of frame A and
origin collocates with the origin of frame B. By considering
frames A′ and B, we have ˙RA′
B = RA′
B [wB]× and ˙RB
A′ =
−[wB]×RB
A′. Since the axes of frame A′ are parallel to those
of frame A, we always have RA′
B = RA
B and RB
A′ = RB
A
and consequently (3) and (5) still holds (note wB remain
the same). On the other hand, if the origins of frames A
and B do not collocate, equations (2) and (4) do not hold
any more because w′
A ̸= wA due to the nonzero translation
between frames A′ and A. With the above discussion, we
A
x
y
z
B
x
y
z
wB
RA
B
Fig. 2: The world frame A and body frame B for a robot moving in the
three-dimensional space.
A
xA
yA
B
xB
yB
α
Fig. 3: The world frame A and body frame B for a robot moving in the
plane.
know equations (3) and (5) are more useful than equations
(2) and (4) in practice.
If a robot is moving in the plane, the rotation (or orien-
tation) of the robot can be represented by a single angle α
(see Figure 3). Then, the rotation transformation from frame
B to frame A is
Rα =
 cos α
−sin α
sin α
cos α

.
In order to verify Rα, one may examine some specific points
in frame B such as e1 = [1, 0]T and e2 = [0, 1]T. Taking
time derivative on both sides of Rα gives
˙Rα =
 −sin α
−cos α
cos α
−sin α

˙α.
(10)
Expression (10) may also be obtained as a special case of (3).
In particular, we can consider the three-dimensional frame
with the z-axes pointing out of the paper in Figure 3. Then,
RA
B =


cos α
−sin α
0
sin α
cos α
0
0
0
1

.
The angular velocity of the robot can be expressed as wA =
wB = ˙αe3 where e3 = [0, 0, 1]T. By applying (2) or (3), it
is straightforward to obtain (10).
REFERENCES
[1] R. M. Murray, Z. Li, and S. S. Sastry, A Mathematical Introduction to
Robotic Manipulation. CRC Press, 1994.
[2] Y. Ma, S. Soatto, J. Kosecka, and S. Sastry, An Invitation to 3D Vision.
New York: Springer, 2004.
[3] M. W. Spong, S. Hutchinson, and M. Vidyasagar, Robot Modeling and
Control. John Wiley & Sons, Inc., 2006.
[4] F. Hamano, “Derivative of rotation matrix - direct matrix derivation
of well-known formula,” in Proceedings of Green Energy and Systems
Conference, (Long Beach, CA), 2013.