# Eigenvectors of a Rotation Matrix

The eigenvector of the rotation matrix corresponding to eigenvalue 1 is the axis of rotation. The remaining eigenvalues are complex conjugates of each other and so are the corresponding eigenvectors. The two complex eigenvectors can be manipulated to determine a plane perpendicular to the first real eigen vector. I did this in matlab for many non-diagonal, non-identity matrices and found that eigenvectors were both unit and mutually orthonormal. I dont think this is a coincidence. Why is it so?

Section 3.1 of the link below: robotics.caltech.edu/~jwb/courses/ME115/handouts/rotation.pdf

• The assumption of the first question is false; any eigenvector of any matrix can be rescaled arbitrarily.
– mdp
Nov 3, 2014 at 11:04
• I'll modify my question. For a 3x3 rotation matrix, not the identity matrix, why must the eigen vectors be mutually orthogonal? Nov 3, 2014 at 11:59
• Even banning the identity matrix isn't enough; the matrix $$\begin{pmatrix}-1&0&0\\0&-1&0\\0&0&1\end{pmatrix}$$ is also a rotation matrix admitting non-orthogonal eigenvectors.
– mdp
Nov 3, 2014 at 12:08
• Alright, here is my actual doubt: The eigenvector of the rotation matrix corresponding to eigenvalue 1 is the axis of rotation. The remaining eigenvalues are complex conjugates of each other and so are the corresponding eigenvectors. The two complex eigenvectors can be manipulated to determine a plane perpendicular to the first real eigen vector. I did this in matlab for many non-diagonal, non-identity matrices and found that eigenvectors were both unit and mutually orthonormal. I dont think this is a coincidence. robotics.caltech.edu/~jwb/courses/ME115/handouts/rotation.pdf Nov 3, 2014 at 12:15
• The software reporting the eigenvectors must have been "normalizing" them to unit length. As I point out, this can be done with eigenvectors of any matrix. It is true that the in three dimensions, for a rotation matrix not the identity, there is a unique "axis of rotation", determined by an eigenvector for the eigenvalue 1. Since that "axis of rotation" is fixed, the rotation is indeed rotation of planes perpendicular to the axis of rotation, and this is a general fact about rotation in three dimensions. Nov 3, 2014 at 12:57

I will assume a real orthogonal matrix is involved. Further, for an orthogonal matrix to represent a "rotation" means that the determinant is 1.

1) Any (nonzero) multiple of an eigenvector is again an eigenvector, so it is not the case that eigenvectors of an orthogonal matrix must be unit vectors. However by the same token, any eigenvector can be scaled to be a vector of length one.

2) It is not generally the case that eigenvectors of a rotation matrix are mutually orthogonal, as a simple case illustrates. The trivial rotation corresponds to the identity matrix $I$, for which all nonzero vectors are eigenvectors corresponding to eigenvalue 1. Clearly we can pick two eigenvectors in this case which are not orthogonal.

More generally we can have an eigenspace for eigenvalue 1 of dimension greater than 1, and/or an eigenspace for eigenvalue -1 of even dimension greater than 0. In either of these cases we can also pick a pair of eigenvectors for the same eigenvalue that are not mutually orthogonal.

Typically a rotation matrix will have some complex eigenvalues, as for example in dimension two:

$$\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$$

whose characteristic polynomial is $\lambda^2 + 1$. In such cases there are no real eigenvectors corresponding to the complex eigenvalues.

A rotation preserves the length of a vector, so if real eigenvalues for a rotation matrix exist, they must have absolute value 1. Thus the only possible real eigenvalues are $\pm 1$.

The PDF linked from Comments on the Question describes $3\times 3$ orthogonal (real) matrices $A$, with special emphasis on rotations, i.e. where $\det A = 1$. Euler's Displacement Thm. is stated, to the effect that if the rotation is nonzero, there exists a unique axis of rotation, and the matrix $A$ itself can be expressed in terms of a unit vector $w$ directed along that axis and an angle of rotation $\phi$ (Rodriguez' Formula).

The immediate Question concerns conditions for which eigenvectors of $A$ will be orthogonal. Although the linked paper describes a real orthonormal basis for $\mathbb{R}^3$ in terms of unit eigenvectors, only one of the basis vectors is an eigenvector. The other two are linear combinations of complex eigenvectors.

Specifically the characteristic polynomial of $A$ will be of degree 3, and if the angle of rotation $\phi \neq 0$, then the three eigenvalues (roots of the characteristic polynomial) will be as follows: $\lambda_1 = 1, \lambda_2 = \cos \phi + i \sin \phi, \lambda_3 = \cos \phi - i \sin \phi$.

Given the unit vector $w$ as above directed along the axis of rotation, this is an eigenvector corresponding to eigenvalue $\lambda_1 = 1$:

$$A w = w$$

Clearly the unique eigenvalue $\lambda_1 = 1$, being real, is distinct from the two complex (non-real) eigenvalues. Unless nonzero angle $\phi = \pi$ (which is the case Matt describes in one of his Comments on the Question), the eigenvalues $\lambda_2,\lambda_3$ are distinct (they are complex conjugates in any case).

Let $u,v$ be complex unit eigenvectors of $A$ corresponding to $\lambda_2,\lambda_3$ respectively. We first show that $w$ is orthogonal to both $u,v$.

Since $A^T=A^{-1}$ by definition of an orthogonal matrix, we have:

$$w^T A = (A^T w)^T = (A^{-1} w)^T = w^T$$

That is $w^T$ is a left eigenvector corresponding to eigenvalue $\lambda_1 = 1$, just as $w$ is a right eigenvector for $\lambda_1$. This implies:

$$\lambda_1 w^T u = w^T A u = \lambda_2 w^T u$$

Since $\lambda_2 \neq \lambda_1$, this tells us $w^T u = 0$. Note that this last is a scalar value, and regardless of the fact that $u$ is complex, it means that $w,u$ are orthogonal. Similarly $w,v$ are orthogonal.

Because $u,v$ are complex vectors, we want to clarify what it means for them to be "orthogonal".

Rather than defining the dot-product as simply $u^T v$, we need to take the complex conjugate of one vector in this product. It doesn't make a significant difference which one is conjugated; the effect of switching to conjugating the other only means the scalar dot-product is conjugated.

Thus $\overline{u}^T v = 0$ if and only if $u^T \overline{v} = 0$, so the notion of "orthgonality" is exactly the same, whichever vector in the definition gets conjugated.

Bearing in mind that conjugation does not alter a real value (matrix or vector), we can compute as follows:

$$\overline{u}^T A = \overline{u^T A} = \overline{A^T u}^T = \overline{A^{-1} u}^T$$

Since $Au = \lambda_2 u$, we have $A^{-1} u = \lambda_2^{-1} u$. Also since $|\lambda_2|=1$, $\lambda_2^{-1} = \overline{\lambda_2}$. Putting these observations together:

$$\overline{u}^T A = \overline{A^{-1} u}^T = \overline{\lambda_2^{-1} u^T} = \overline{\lambda_2^{-1}} \overline{u}^T = \lambda_2 \overline{u}^T$$

We can now compute the orthgonality of $u,v$ similarly to previously for $w,u$ and $w,v$:

$$\lambda_2 \overline{u}^T v = \overline{u}^T A v = \overline{u}^T (\lambda_3 v) = \lambda_3 \overline{u}^T v$$

That is, since $\lambda_2 \neq \lambda_3$, the above implies $\overline{u}^T v = 0$.

Therefore $\{u,v,w\}$ form an orthonormal basis that diagonalizes $A$, i.e. the unitarily similar matrix is diagonal:

$$\begin{pmatrix} \lambda_2 & 0 & 0 \\ 0 & \lambda_3 & 0 \\ 0 & 0 & 1 \end{pmatrix}$$

However it is more convenient to work with an orthonormal basis of real vectors (the paper outlines how to do this) $\{c_2,c_3,w\}$ with respect to which the representation is only "block" diagonal:

$$\begin{pmatrix} \cos \phi & \sin \phi & 0 \\ -\sin \phi & \cos \phi & 0 \\ 0 & 0 & 1 \end{pmatrix}$$

Bear in mind that although this is an orthonormal basis, the first two real basis elements are not eigenvectors, but rather linear combinations of $u,v$ chosen to give us real vectors. The third basis element $w$ is the same real unit vector, in the direction of the axis of rotation, as before.