Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $y$?

share|improve this question
    
Please note: I've already derived the solution and written up the proof. I haven't been able to find the solution anywhere else online, but I currently have to wait 8 hours before stackexchange will let me post my answer. –  Ose Pedro Mar 14 '13 at 3:32
    
Nice work and thank you for posting the details! –  Amzoti Mar 14 '13 at 4:42

1 Answer 1

up vote 1 down vote accepted

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.