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Show that each matrix in SO$(3)$ equals $e^X$ for some skew-symmetric $X$.

Here, SO(3) refers to the special orthogonal group that is the rotation group of $\mathbb{R}^3$. I am also supposed to use the following facts that I have derived in proving the above result.

  1. Given $B = \begin{pmatrix} 0 & -\theta & 0 \\ \theta & 0 & 0 \\ 0 &0 &0 \end{pmatrix}$, $e^B =\begin{pmatrix} \cos \theta & -\sin \theta & 0\\ \sin \theta & \cos \theta & 0\\ 0 & 0 & 1 \end{pmatrix}$, where $e^B$ denotes the exponential of $B$. (Note that this is what I calculated the exponential to be - is this correct?)

  2. For any orthogonal matrix $A$, we have $Ae^BA^T = e^{ABA^T}$. (I have already proven this result.)

Again, I need to use results 1 and 2 in the proof of my question, but I'm not seeing how these two results combine to show my desired result. Since $e^B$ is one of the 3-dimensional rotation matrices (assuming that I calculated my exponential correctly), do I need to make some sort of argument based on rotation? Or is there a simpler way?

  • The fact that needs to be proved is that every $SO(3)$ is an orthogonal change of basis away from a rotation about the $z$-axis. I'm sure there's a standard proof of this intuitively true fact, but I don't know it. – Charles Hudgins May 07 '22 at 01:22
  • The overkill solution is, of course, that ${\rm SO}(n)$ is a connected and compact Lie group, so its exponential map is surjective as a consequence of the Hopf-Rinow theorem from Riemannian geometry. Roughly: take a bi-invariant metric (by averaging any left-invariant metric with respect to the Haar measure of ${\rm SO}(n)$), so the Riemannian exponential agrees with the Lie exponential and geodesics starting at the identity matrix are one-parameter subgroups. Now H&R says that any matrix in ${\rm SO}(n)$ can be reached from the identity matrix with a minimizing geodesic. – Ivo Terek May 07 '22 at 03:51

1 Answers1

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Let $g\in SO(3,\mathbb{R})$ be any element.

(1) We first show that $1$ is an eigenvalue of $g$. Consider the characteristic polynomial $f_g$ of $g$. It is clear that $f_g\in \mathbb{R}[x]$ and $\deg(f_g)=3$. Thus $f_g$ has a real root, say $\lambda_1$. Let $\alpha_1$ be the unit eigenvalue of $\lambda_1$. Since $g$ preserve the length, we get $\lambda_1^2=\pm 1$. If $\lambda_1=-1$, let $\lambda_2, \lambda_3$ be the other complex eigenvalues. Then $\lambda_2\lambda_3=-1$. Since $\lambda_2,\lambda_3$ satisfies a monic real quadratic equation $x^2+ax-1,$ which has only real roots, we have $\{\lambda_2,\lambda_3\}=\{\pm 1 \}$. Thus $1$ is an eigenvalue of $g$.

(2) Let $\alpha_1$ be an unit eigenvector of $g$ and let $W=Span\{\alpha_1\}$. Consider $W^{\perp}$, which has dimension 2 and is invariant under $g$. Thus $g|_{W^{\perp}}\in SO(2,\mathbb{R})$. Thus there exists orthonormal basis $\alpha_2,\alpha_3$ and an angle $\theta$ such that $$g\begin{pmatrix}\alpha_2 \\ \alpha_3 \end{pmatrix}=\begin{pmatrix}\alpha_2 & \alpha_3 \end{pmatrix}\begin{pmatrix} \cos(\theta)&\sin(\theta)\\ -\sin(\theta)& \cos(\theta)\end{pmatrix}.$$ The rest is easy.

Q-Zhang
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