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Physics 116A Fall 2019
Eigenvalues and eigenvectors of rotation matrices
These notes are a supplement to a previous class handout entitled, Rotation Matrices
in two, three and many dimensions. In these notes, we shall focus on the eigenvalues and
eigenvectors of proper and improper rotation matrices in two and three dimensions.
1. The eigenvalues and eigenvectors of proper and improper rotation matrices
in two dimensions
In the previous class handout cited above, we showed that the most general proper
rotation matrix in two-dimensions is of the form,
R(θ) = cosθ −sinθ ; where 0 ≤ θ < 2π; (1)
sinθ cosθ
which represents a proper counterclockwise rotation by an angle θ in the x–y plane.
Consider the eigenvalue problem,
~ ~
R(θ)v = λv: (2)
~
Since R(θ) rotates the vector v by an angle θ, we conclude that for θ 6= 0 (mod π), there
~
are no real eigenvectors v that are solutions to eq. (2). This can be easily checked by an
explicit calculation as follows.
det(R(θ)−λI) = 0 =⇒ detcosθ−λ −sinθ =0; (3)
sinθ cosθ −λ
which yields the characteristic equation,
(cosθ −λ)2 +sin2θ = 0: (4)
This equation simplifies to
λ2 −2λcosθ+1=0; (5)
which yields the eigenvalues,
√ 2 ±iθ
λ=cosθ± cos θ−1=cosθ±isinθ=e : (6)
Thus, we ave confirmed that the eigenvalues are not real if θ 6= 0 (mod π). For the special
cases of R = I and R = −I, corresponding to θ = 0 and π, respectively, we obtain real
eigenvalues as expected. In particular, the case of θ = π corresponds to a two dimension
~ ~
inversion x → −x, which implies that the eigenvalue of R(π) is doubly degenerate and
equal to −1.
1
The case of improper rotations in two dimensions is more interesting. In the previous
class handout cited above, we noted that the most general improper rotation matrix in
two-dimensions is of the form,
R(θ) = cosθ sinθ ; where 0 ≤ θ < 2π; (7)
sinθ −cosθ
which can be expressed as the product of a proper rotation and a reflection,
R= cosθ −sinθ 1 0 : (8)
sinθ cosθ 0 −1
However, it is easy to show that the action of R(θ) is equivalent to a pure reflection
through a line that passes through the origin. This can be seen by considering the
eigenvalue problem,
~ ~
R(θ)v = λv: (9)
Wecan again determine the eigenvalues of R(θ) by solving its characteristic equation,
det(R(θ)−λI) = 0 =⇒ detcosθ−λ sinθ =0; (10)
sinθ −cosθ−λ
which is equivalent to
(cosθ −λ)(−cosθ−λ)−sin2θ =0: (11)
This equation simplifies to
λ2 −1 = 0; (12)
which yields the eigenvalues, λ = ±1.
The interpretation of this result is immediate. The matrix R(θ) when operating on
~
a vector v represents a reflection of that vector through a line of reflection that passes
~ ~ ~
through the origin. In the case of λ = 1 we have R(θ)v = v, which means that v is a
vector that lies parallel to the line of reflection (and is thus unaffected by the reflection).
~ ~ ~
In the case of λ = −1 we have R(θ)v = −v, which means that v is a vector that is
~ ~
perpendicular to the line of reflection (and is thus is transformed, v → −v, by the
reflection).
One can therefore determine the line of reflection by computing the eigenvector that
corresponds to λ = 1,
cosθ sinθ x = x : (13)
sinθ −cosθ y y
If θ = 0 (mod 2π), then any vector of the form (x) is an eigenvector corresponding to the
0
eigenvalue λ = 1. This implies that the line of reflection is the x-axis, which corresponds
to the equation y = 0. In general (for any value of θ), the solution to eq. (13) is
xcosθ+ysinθ =x; (14)
2
or equivalently,
x(1−cosθ)−ysinθ =0: (15)
It is convenient to use trigonometric identities to rewrite eq. (15) as
2 1 1 1
2xsin 2θ −2ysin 2θ cos 2θ =0: (16)
1 1
If θ 6= 0 (mod 2π), then we can divide both sides of eq. (16) by sin 2θ to obtain
1 1
xsin 2θ −ycos 2θ = 0: (17)
We recognize eq. (17) as an equation for a straight line that passes through the origin
1
with a slope equal to tan 2θ . Thus, we have demonstrated that the most general
2×2orthogonal matrix with determinant equal to −1 given by R(θ) represents a pure
reflection through a straight line of slope tan 1θ that passes through the origin.
2
Finally, it is worth noting that since R(θ) is both an orthogonal matrix, R(θ)R(θ)T= I,
and a symmetric matrix, R(θ)T = R(θ), it follows that
2
R(θ) =I; (18)
which is property that must be satisfied by a reflection matrix since two consecutive
reflections are equivalent to the identity operation when acting on a vector.
3. The eigenvalues and eigenvectors of proper rotation matrices in three
dimensions
The most general three-dimensional proper rotation matrix, which we henceforth
denote by R(nˆ;θ), can be specified by an axis of rotation pointing in the direction
of the unit vector nˆ, and a rotation angle θ. Conventionally, a positive rotation angle
corresponds to a counterclockwise rotation. The direction of the axis is determined by the
right hand rule. Namely, curl the fingers of your right hand around the axis of rotation,
where your fingers point in the θ direction. Then, your thumb points perpendicular to
the plane of rotation in the direction of nˆ. All possible proper rotations correspond to
0 ≤ θ ≤ π and the unit vector nˆ pointing in any direction.
To learn more about the properties of a general three-dimensional rotation, consider
the matrix representation R(nˆ;θ) with respect to the standard basis B = {i; j ; k}. We
s
can define a new coordinate system in which the unit vector nˆ points in the direction
of the new z-axis; the corresponding new basis will be denoted by B′. The matrix
representation of the rotation with respect to B′ is then given by
cosθ −sinθ 0
R(k;θ) ≡ sinθ cosθ 0; (19)
0 0 1
where the axis of rotation points in the z-direction (i.e., along the unit vector k).
1Note that for θ = 0 (mod 2π), eq. (17) reduces to y = 0 which is the equation for the x-axis, as
expected.
3
Using the formalism developed in the class handout, Vector coordinates, matrix ele-
ments, changes of basis, and matrix diagonalization, there exists an invertible matrix P
such that
R(nˆ;θ) = PR(k;θ)P−1; (20)
where R(k;θ) is given by eq. (19). In Appendix A, we will determine an explicit form
for the matrix P. However, the mere existence of the matrix P in eq. (20) is sufficient
to provide a simple algorithm for determining the rotation axis nˆ (up to an overall sign)
and the rotation angle θ that characterize a general three-dimensional rotation matrix.
To determine the rotation angle θ, we note that the properties of the trace imply that
Tr(PRP−1) = Tr(P−1PR) = TrR, since one can cyclically permute the matrices within
the trace without modifying its value. Hence, it immediately follows from eq. (20) that
Tr R(nˆ;θ) = Tr R(k;θ) = 2cosθ +1; (21)
after taking the trace of eq. (19). By convention, 0 ≤ θ ≤ π, which implies that sinθ ≥ 0.
Hence, the rotation angle is uniquely determined by eq. (21) To identify nˆ, we observe
that any vector that is parallel to the axis of rotation is unaffected by the rotation itself.
This last statement can be expressed as an eigenvalue equation,
R(nˆ;θ)nˆ = nˆ : (22)
Thus, nˆ is an eigenvector of R(nˆ;θ) corresponding to the eigenvalue 1. In particular, the
eigenvalue 1 is nondegenerate for any θ 6= 0, in which case nˆ can be determined up to an
overall sign by computing the eigenvalues and the normalized eigenvectors of R(nˆ;θ). A
simple proof of this result is given in Appendix B. Here, we shall establish this assertion
by noting that the eigenvalues of any matrix are invariant with respect to a similarity
transformation. In light of eq. (20), it follows that the eigenvalues of R(nˆ;θ) are identical
to the eigenvalues of R(k;θ). The latter can be obtained from the characteristic equation,
2 2
(1 −λ) (cosθ−λ) +sin θ = 0;
which simplifies to:
(1 −λ)(λ2 −2λcosθ+1) = 0;
after using sin2 θ + cos2 θ = 1. Using the results of eqs. (5) and (6), it follows that the
three eigenvalues of R(k;θ) are given by,
iθ −iθ
λ1 = 1; λ2 = e ; λ3 = e ; for 0 ≤ θ ≤ π:
There are three distinct cases:
Case 1: θ = 0 λ1 = λ2 = λ3 = 1, R(nˆ;0) = I,
Case 2: θ = π λ1 = 1; λ2 = λ3 = −1, R(nˆ;π),
iθ −iθ
Case 3: 0 < θ < π λ1 = 1;λ2 = e ; λ3 = e , R(nˆ;θ),
where the corresponding rotation matrix is indicated for each of the three cases.
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