What are the Complex (Non-Real) Eigenvectors of $3\times 3$ Rotation Matrices?

Question

A $3\times 3$ rotation matrix $R$ that rotates $\mathbb{R}^3$ around the unit vector $v\in\mathbb{R}^3$ by angle $\theta$ (as defined by Rodrigues' rotation formula) satisfies the eigendecomposition $$ R=W\Sigma W^\mathrm{*} \enspace, $$ where $$ W=\left(\begin{matrix}v \;|\; x \;|\; y \end{matrix}\right) $$ is a unitary matrix of eigenvectors, and $$ \Sigma=\mathrm{diag}\left(1,e^{i\theta},e^{-i\theta}\right) $$ is the matrix of the corresponding eigenvalues. What expressions define the non-real eigenvectors $x$ and $y$?

Answer 1

Theorem

$x$ and $y$ are non-real eigenvectors of $R$ if and only if \begin{align} x &= a+ib\\ y &= \pm(a-ib)\\ b &= a\times v \enspace, \end{align} where $a\in\mathbb{R}^3$ is an arbitrary vector satisfying $$ ||a||^2=\frac{1}{2} \;,\; v^\mathrm{T}a=0 \enspace, $$ which implies that $W$ has 1 d.o.f.

Proof of Necessity

We will first of all show that if $w=c+id$ and $u$ are non-real eigenvectors of $R$ with eigenvalues $e^{i\theta}$ and $e^{-i\theta}$ respectively, for some $c$ and $d$ in $\mathbb{R}^3$, then $$ v^\mathrm{T}c=0 \;,\; \; d = c\times v \;,\; ||c||^2=\frac{1}{2} \;,\quad \mbox{and}\quad u=\pm\overline{w}=\pm(c-id) $$ We will only prove that $\overline{w}$ is a possible value of eigenvector $u$. Clearly if this is the case then $u=\pm\overline{w}$.

Proof that $v^\mathrm{T}c=0$

Since $W$ is unitary, $w$ must satisfy $$ v^\mathrm{T}w \;=\; v^\mathrm{T}c+iv^\mathrm{T}d \;=\; 0 \enspace.$$ As $v$ is real, this condition is only satisfied when $v$ is orthogonal to both $c$ and $d$.

Proof that $d = c\times v$

By the eigenvector equation, \begin{align} Rw &= e^{i\theta}w \\ &= (\cos\theta+i\sin\theta)w \\ &= -d\sin\theta+c\cos\theta+i(c\sin\theta+d\cos\theta) \enspace. \end{align} Furthermore, by Rodrigues' rotation formula, \begin{align} Rw &= w\cos\theta + (v\times w)\sin\theta+v(v^\mathrm{T}w)(1-\cos\theta) \\ &= w\cos\theta + (v\times w)\sin\theta \quad,\quad\mbox{as $v^\mathrm{T}w=0$}\\ &= (v\times c)\sin\theta+c\cos\theta+ i\left((v\times d)\sin\theta+d\cos\theta\right) \end{align} The result follows by comparing similar terms in the real and imaginary parts of these two expressions for $Rw$.

Proof that $||c||^2=\frac{1}{2}$

The fact that the eigenvectors have unit norms implies that \begin{align} 1 &= w^*w \\ &= (c^\mathrm{T}-id^\mathrm{T})(c+id) \\ &= c^\mathrm{T}c+d^\mathrm{T}d \\ &= c^\mathrm{T}c+(c\times v)^\mathrm{T}(c\times v) \\ &= c^\mathrm{T}c+(c^\mathrm{T}c)v^\mathrm{T}v-(c^\mathrm{T}v)(v^\mathrm{T}c) \quad,\quad \mbox{by cross-product properties}\\ &= 2||c||^2 \\ \Rightarrow ||c||^2 &= \frac{1}{2} \enspace. \end{align}

Proof that $\overline{w}$ is an Eigenvector

As $W$ is unitary, $\overline{w}$ must satisfy $$ v^\mathrm{T}\overline{w} \;=\; \overline{w}^*w \;=\; 0 \quad\mbox{and}\quad ||\overline{w}||^2=1 \enspace. $$ The first 3 lines of the proof that $||c||^2=\frac{1}{2}$ show that $||\overline{w}||^2=1$, so it suffices to show that $\overline{w}$ has the two orthogonality properties and that $e^{-i\theta}$ is $\overline{w}$'s eigenvalue.

$v^\mathrm{T}\overline{w}=0$ follows from $v$'s orthogonality to $c$ and $d$, which we established above. To prove that $\overline{w}^*w=0$: \begin{align} \overline{w}^*w &= w^\mathrm{T}w \\ &= (c+id)^\mathrm{T}(c+id) \\ &= c^\mathrm{T}c+2ic^\mathrm{T}d-d^\mathrm{T}d \\ &= ||c||^2+0-||c\times v||^2 \\ &= ||c||^2-||c||^2 \quad,\quad\mbox{as $v^\mathrm{T}c=0$ and $||v||=1$}\\ &= 0 \enspace. \\ \end{align}

To show that $e^{-i\theta}$ is $\overline{w}$'s eigenvalue, note that our proof that $d = c\times v$ shows that \begin{align} Rc &= -d\sin\theta+c\cos\theta \\ Rd &= c\sin\theta+d\cos\theta \enspace. \end{align} Substituting these results into $R\overline{w}$ and rearranging gives \begin{align} R\overline{w} &= Rc-iRd \\ &= (c-id)\cos\theta-(ic+d)\sin\theta \\ &= (c-id)\cos\theta-i(c-id)\sin\theta \\ &= (\cos\theta-i\sin\theta)(c-id) \\ &= e^{-i\theta}\overline{w} \enspace. \end{align}

Proof of Sufficiency

We will now prove that $x=a+ib$ and $y=a-ib$ are a valid pair of non-real eigenvectors of $R$. Again, it is obvious that if $y$ is an eigenvector, then $-y$ is also an eigenvector corresponding to the same eigenvalue.

Proof that $W^*W=I$

For $W$ to be unitary, $x$ and $y$ must satisfy $$ v^\mathrm{T}x \;=\; v^\mathrm{T}y \;=\; y^*x \;=\; 0 \enspace, $$ and $$ x^*x=y^*y=1\enspace. $$ It is obviously true that $v^\mathrm{T}x \;=\; v^\mathrm{T}y \;=\; 0$, since $v^\mathrm{T}a=0$ by definition and $v^\mathrm{T}(a\times v)=0$. The third orthogonality condition follows from our proof above that $\overline{w}$ is an eigenvector. Also, as $y=\overline{x}$, the unit norm conditions follow from the first 3 lines of our proof above that $||c||^2=\frac{1}{2}$ — just substitute $x$ for $w$.

Proof that $x$ and $y$ are Eigenvectors

Next we must prove that $x$ satisfies the eigenvector equation $$ Rx \;=\; e^{i\theta}x \enspace. $$ Combining Rodrigues' rotation formula with the fact that $v^\mathrm{T}x=0$ gives $$ Rx = x\cos\theta + (v\times x)\sin\theta \enspace. $$ As \begin{align} v\times x &= v\times(a+i(a\times v)) \\ &= v\times a + iv\times(a\times v) \\ &= v\times a + ia \quad,\quad\mbox{as $v^\mathrm{T}a=0$ and $||v||=1$}\\ &= i(i(a\times v) + a) \\ &= ix \enspace, \end{align} it follows by substitution that \begin{align} Rx &= x\cos\theta + ix\sin\theta \\ &= (\cos\theta + i\sin\theta)x \\ &= e^{i\theta}x \enspace. \end{align}

To show that $y$ is also an eigenvector with eigenvalue $e^{-i\theta}$, we can once again use Rodrigues' rotation formula and the fact that $v^\mathrm{T}y=0$ to obtain \begin{align} Ry &= y\cos\theta + (v\times y)\sin\theta \\ &= y\cos\theta-iy\sin\theta \enspace, \end{align} where the result $$ v\times y = -iy $$ follows from a slight modification of the proof that $v\times x = ix$. Thus \begin{align} Ry &= (\cos\theta-i\sin\theta)y \\ &= e^{-i\theta}y \enspace. \end{align}

QED.