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Problem

Given the skew-symmetric matrix:

$S(t)={\left(\begin{array}{ccc}0,-t_3,t_2\\t_3,0,-t_1\\-t_2,t_1,0\end{array}\right)}$,

with $t{\in}S^2$, i.e. a normalized three dimensional vector. Find a rotation matrix $R{\in}SO(3)$ such that the product $S(t)R$ is also skew-symmetric.

Proposed answer

Using the Rodrigues' rotation formula:

$R(a,\theta) = I\cos\theta + S(a)\sin\theta + aa\!^\mathsf{T}(1{-}\cos\theta)$

we have:

$S(t)R(a,\theta)=S(t)\cos\theta + S(t)S(a)\sin\theta + S(t)aa\!^\mathsf{T}(1-\cos\theta)$

By simple inspection two solutions arise:

  • $R{=}I$, which is trivial, and
  • $R{=}R(t,(2k+1)\pi)$, $k{=}0{,}1{,}{\ldots}$, giving $S(t)R{=}{-}S(t)$.

It is possible to find other solutions?

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Let $0 \neq S \in \mathfrak{so}(3)$ be a non-zero skew-symmetric matrix, and let $R \in SO(3)$. The claim is that, indeed, $SR$ is skew-symmetric if and only if $R = I_3$ or $R$ is the rotation by $\pi$ about the axis $\ker(S)$; since $SI = S$ is necessarily skew-symmetric, we only need to consider the case where $R \neq I_3$. Observe that $$ (SR)^T+SR = R^TS^T+SR = -R^TS+SR =-R^T(S-RSR), $$ so that $SR$ is skew-symmetric if and only if $S=RSR$.

Before continuing, let me check that $\ker(S)$ is indeed $1$-dimensional. Observe that, since $S$ is skew-symmetric, that $\ker(S)$ and $\ker(S)^\perp$ are both $S$-invariant. Now, since $S \neq 0$, fix a unit vector $0 \neq x \in \ker(S)^\perp$. Then, since $S$ is skew-symmetric, $x$ is orthogonal to $0 \neq Sx \in \ker(S)^\perp$, so that $\ker(S)^\perp$ is at least two-dimensional and hence $\ker(S)$ is at most $1$-dimensional. However, $S$ cannot be invertible, for $\operatorname{set}(S) = \operatorname{det}(S^T) = \operatorname{det}(-S) = -\operatorname{det}(S)$, so that, indeed, $\operatorname{det}(S) = 0$. Thus, $\ker(S)$ is, in fact, $1$-dimensional.

For future convenience, observe that if $0 \neq x \in \ker(S)^\perp$ is a unit vector, then for $y := \tfrac{1}{\|Sx\|}Sx$ and $0 \neq z \in \ker(S)$ any unit vector, $\{x,y\}$ is an orthonormal basis for $\ker(S)^\perp$, $\ker(S) = \mathbb{R}z$, and $\beta = \{x,y,z\}$ is an orthonormal basis for $\mathbb{R}^3$. Indeed, let’s see how $S$ looks in terms of this basis. On the one hand, we already have that $Sz = 0$ and that $Sx = \alpha y$ for $\alpha = \|Sx\|$. On the other hand, by the same argument as above, $y$ is orthogonal to $0 \neq Sy \in \ker(S)^\perp$, so that $$ Sy = \langle x, Sy \rangle = -\langle Sx, y \rangle = -\langle \alpha y, y \rangle = -\alpha. $$ Thus, $$ Sx = \alpha y, \quad Sy = -\alpha x, \quad Sz = 0, $$ for $\alpha := \|Sx\| \neq 0$.

Suppose, on the one hand, that $SR$ is skew-symmetric, and hence that $S=RSR$. We first need to show that $\ker(S) = \ker(R-I_3)$, so assume otherwise. Let $z \in \ker(S)$ be a unit vector. Fix a unit vector $x \in \ker(R-I_3)$, so that $\ker(R-I_3) = \mathbb{R}x$. Then, $$ z = z_0 + cx $$ for $z_0 \in \ker(R-I_3)^\perp$ and $c = \langle x,z\rangle$; by assumption, $z_0 \neq 0$. Then $$ 0 = Sz = RSRz = RSR(z_0 + cx) = RS(Rz_0 + cx), $$ so that $Rz_0 + cx \in \ker(S)$, i.e., $Rz_0 + cx = az$ for some $a \in \mathbb{R}$. But then, $$ Rz_0 + cx = az = az_0 + acx, $$ so that $Rz_0 = az_0$ and $cx = acx$. Since $z_0$ is non-zero, it is therefore a real eigenvector for $R$, forcing $a = -1$ and hence $c = 0$. Thus, $R$ is the rotation by $\pi$ about the axis $\ker(R-I_3) = \mathbb{R}x$, and $\ker(S) \subset \ker(R-I_3)^\perp = \ker(R+I_3)$.

But now, starting with the unit vectors $x \in \ker(R-I_3) \subset \ker(S)^\perp$ and $z \in \ker(S)$, we can construct, as above, an orthonormal basis $\{x,y,z\}$ of $\mathbb{R}^3$, where $y = \tfrac{1}{\|Sx\|}Sx$, such that $$ Sx = \alpha y, \quad Sy = -\alpha x, \quad Sz = 0 $$ for $\alpha = \|Sx\| \neq 0$. But then, in particular, since $y \in \ker(R-I_3)^\perp = \ker(R+I_3)$, $$ -\alpha x = Sy = RSRy = -RSy = \alpha R x = \alpha x, $$ contradicting $\alpha \neq 0$.

So, as we have seen, $\ker(S) = \ker(R-I_3)$, so again, choose a unit vector $x \in \ker(S)^\perp = \ker(R-I_3)^\perp$ and construct an orthonormal basis $\{x,y,z\}$ of $\mathbb{R}^3$ such that $$ Sx = \alpha y, \quad Sy = -\alpha x, \quad Sz = 0, $$ for $\alpha = \|Sx\| \neq 0$. Since $S$ is skew-symmetric, $\ker(S)^\perp = \ker(R-I_3)^\perp$ is $R$-invariant, so that, in particular, $Rx \in \ker(S)^\perp$. But then, since $SR$ is skew-symmetric, $$ 0 = \langle x, SR x \rangle = -\langle Sx, Rx \rangle = -\alpha \langle y, Rx \rangle, $$ so that $Rx \in \ker(S)^\perp$ is orthogonal to $y$, and hence $Rx = cx$ for some $c \in \mathbb{R}$, i.e., $x \in \ker(S)^\perp = \ker(R-I_3)^\perp$ is a non-zero real eigenvector for $R$. Thus, necessarily, $c = -1$, so that, indeed, $R$ is the rotation by $\pi$ about the axis $\ker(R-I_3) = \ker(S)$.

Suppose, on the other hand, that $R$ is the rotation by $\pi$ about the axis $\ker(S) = \ker(R-I_3)$. Let $z \in \mathbb{R}^3$, so that $z = z_0 + z_1$ for $z_0 \in \ker(S) = \ker(R-I_3)$ and $z_1 \in \ker(S)^\perp = \ker(R+I_3)$. Then, since $S$ is skew-symmetric, $\ker(S)^\perp = \ker(R+I_3)$ is $S$-invariant, and hence $$ RSRv = RSR(v_0 + v_1) = RS(v_0-v_1) = -RSv_1 = Sv_1 = Sv_0 + Sv_1 = Sv, $$ as required.

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  • $\begingroup$ Maybe there is a typo error: $$Rw_0{+}cv=aw=a(w_0{+}cv)=aw_0{+}acv \implies Rw_0=aw_0+c(a{-}1)v$$ However, it is not clear to me why the only possible alternatives are $a{=}0$ and $c{=}1$. $\endgroup$ – manuel Jun 1 '14 at 9:32
  • $\begingroup$ @manuel You're absolutely right. I've overhauled the answer, which should now be complete and correct. $\endgroup$ – Branimir Ćaćić Jun 1 '14 at 19:05
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First of all, remind that exists an orthonormal basis $\alpha$ of $\mathbb{R}^3$ such that $[S]_{\alpha}$ is $\left(\begin{array}{ccc} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right)$ and exists an orthonormal basis $\beta$ of $\mathbb{R}^3$ such that $[R]_{\beta}$ is $\left(\begin{array}{ccc} cos(a) & -sen(a) & 0 \\ sen(a) & cos(a) & 0 \\ 0 & 0 & 1 \end{array} \right)$, for some $a$.

Let us prove the following proposition:


Proposition: If $RS$ is skew-symmetric then we can find $\alpha=\beta$ above.


So, you only need to figure out what values of $a$, the matrix $[R]_{\alpha}[S]_{\alpha}$ is skew-symmetric, which is a simple task.

Proof of Proposition:

If $R=Id$ the proposition is obvious. Suppose $R\neq Id$. Therefore the eigenspace of $R$, associated to $1$, has dimension $1$.

First, if $RS$ is skew-symmetric then $-RS=S^tR^t=-SR^t$. Now, $RSR=S$, since $RR^t=Id$.

Let $v\in \mathbb{R}^3$ such that $Rv=v$ and $\|v\|=1$. So $RSRv=Sv$ and $RSv=Sv$.

Now, we have two possibilities: $Sv=0 $ and $Sv\neq 0$.

1) Suppose $Sv=0$. So $Rv=v$ and $Sv=0$. If $\{r,w,v\}$ is an orthonormal ordered basis then or $\alpha=\{r,w,v\}$ or $\alpha=\{w,r,v\}$ satisfies $[S]_{\alpha}=\left(\begin{array}{ccc} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right)$. Now for both basis $[R]_{\alpha}=\left(\begin{array}{ccc} cos(a) & -sen(a) & 0 \\ sen(a) & cos(a) & 0 \\ 0 & 0 & 1 \end{array} \right)$, for some $a$.

2) Suppose $Sv\neq 0$. Since the eigenspace of $R$, associated to $1$, has dimension $1$ then $Sv=\lambda v$. But $S$ is skew-symmetric then $0=v^tSv=\lambda v^tv=\lambda$. Thus, $Sv=0$, which is a contradiction.

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  • $\begingroup$ I've just noticed there's an error: $RS$ is skew symmetric if and only if $$ -SR = (SR)^T = R^TS^T = - R^TS, $$ if and only if $S = RSR$. $\endgroup$ – Branimir Ćaćić May 31 '14 at 2:22
  • $\begingroup$ @BranimirĆaćić Thanks for catching this mistake. I hope it is ok now. $\endgroup$ – Daniel May 31 '14 at 7:33

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