Every matrix can be decomposed to symmetrical and skew-symmetrical part with the formula: $ A=\dfrac{1}{2}(A+A^T)+\dfrac{1}{2}(A-A^T)$.

However if it is known only symmetrical part (we assume here that the whole matrix is unknown) it's impossible without additional information to reconstruct exactly skew-symmetrical part and vice versa.

In the case of 3D rotation matrix we have additional constraints and probably such reconstruction is possible. Let's look at Rodrigues formula and two (symmetrical and skew-symetrical) parts of rotation matrix:

$R(v,\theta)= \{I+(1-cos(\theta))S^2(v)\} + sin(\theta)S(v)$

$S(v)=\begin{bmatrix} v\times{i} & v\times{j}& v\times{k} \end{bmatrix}^T$, skew-symetric matrix itself ($3$ DOF set by components of an axis $v$)

One can notice that having skew-symmetrical part of rotation matrix it is relatively easy to reconstruct symmetrical part.



and the whole expression $skew(R)$ can be decomposed to the product $kK$ in such a way that the sum of squares of matrix $K$ entries
i.e. $ \begin{bmatrix} 1 & 1 & 1 \end{bmatrix} (K\circ{K}) \begin{bmatrix} 1 & 1 & 1 \end{bmatrix}^T =2 $ ...

then $k=sin(\theta)$ and $K=S(v)$
and we can calculate $ cos(\theta)$ and $ S(v)^2$
what makes possible reconstruction of symmetrical part.
Exactly we have two solutions because $sin(\theta)=sin(\pi-\theta)$.

In the second case
when we want to reconstruct skew-symmetrical part
the solution seems to be difficult to find (at least for me) so my question is:

  • how to obtain skew-symmetrical part of rotation matrix $skew(R)$ knowing its symmetrical part $sym(R)$?
  • additionally: why do such asymmetry in difficulty of solutions exist at all ? (symmetry and skew symmetry in the first formula for the decomposition of any matrix $A$ seem not to differ too much)

  • what is the situation for higher dimensions? (when we don't have a Rodrigues formula)

  • $\begingroup$ I will address your last question "why do such asymmetry in difficulty of solutions exist at all ?". The decomposition of any matrix into symmetric and skew-symmetric parts is something related to Lie algebras, while your question is related to Lie groups. The decomposition indeed generalizes to other Lie algebras, and is called I think the Cartan decomposition. $\endgroup$ – Malkoun Jul 9 '16 at 8:14
  • $\begingroup$ Interesting extension of topic... $\endgroup$ – Widawensen Jul 9 '16 at 8:19
  • $\begingroup$ I've written that skew-symmetric part of 3D rotation matrix has 3DOF from components of an axis, but these components are additionally constrained to make a unit vector so taking this into account it has 2 DOF from the axis, of course, expressed only by 3 entries and their 3 antisymmetrical ones. Summarily the whole part has 3 DOF taking into account also an angle $\theta$. Symmetrical part has 3 DOF too. $\endgroup$ – Widawensen Jul 9 '16 at 9:01
  • $\begingroup$ Ok, Rodrigues formula is basically the matrix exponential of $\theta S(v)$. In this case, you obtain the skew-symmetric part of $R$ essentially by taking the matrix logarithm of $R$. This actually gives infinitely many values, one of which being $\theta S(v)$, and the others I believe differing from $\theta S(v)$ by multiples of $2\pi S(v)$. What I wrote generalizes to higher-dimensional rotation matrices, i.e. to elements of $SO(n)$. I hope this helps. $\endgroup$ – Malkoun Jul 9 '16 at 9:02
  • $\begingroup$ ok, I realize that my previous comment does not directly address your questions, per se. If you want, I can give direct answers to your questions in an actual answer. $\endgroup$ – Malkoun Jul 9 '16 at 9:07

"how to obtain skew-symmetrical part of rotation matrix skew(R), knowing its symmetrical part sym(R)?" I will assume you are talking about rotations in $\mathbb{R}^3$. First note that a rotation about an axis of angle $\theta$ and a rotation about the same axis of angle $-\theta$ have the same symmetric parts. But apart from this ambiguity, it is possible to reconstruct the skew-symmetric part of a rotation matrix (up to a sign) by only knowing the symmetric part. Thus one can reconstruct the whole rotation matrix (apart from this ambiguity above) from the symmetric part.

Proof: Case 1: $\operatorname{sym}(R)-I \neq 0$. By Rodrigues formula, $\operatorname{sym}(R)-I = (1-\operatorname{cos}(\theta))S^2(v)$. For a square matrix $A$, you can define its norm by $||A||^2 = \frac{1}{2} Tr(A^TA)$. Then by normalizing $\operatorname{sym}(R)-I$, one gets rid of $1-\operatorname{cos}(\theta)$, and gets $S^2(v)$. The kernel of $S^2(v)$ is the axis of rotation. It remains to recover $\operatorname{sin}(\theta)$, up to a sign. Well, in $\operatorname{sym}(R)-I$, the factor multiplying $S^2(v)$ is $1-\operatorname{cos}(\theta)$, so $\operatorname{cos}(\theta)$ is known, and from it we can get $\operatorname{sin}(\theta)$ up to a sign.

Case 2: $\operatorname{sym}(R)-I=0$ By Rodrigues formula, this corresponds to $\operatorname{cos}(\theta) = 1$, i.e. to $R = I$. This finishes the proof.

  • $\begingroup$ Oops, I will fix my answer. $\endgroup$ – Malkoun Jul 9 '16 at 9:22
  • $\begingroup$ Hmm, what do you mean by normalizing $sym(R)−I$ ? I feel that it is the step in good direction but I don't know what it means exactly in this context.. when $sym(R)−I$ is "normalized"? $\endgroup$ – Widawensen Jul 9 '16 at 9:58
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    $\begingroup$ Well, if you have a square matrix $A$, you can say define its norm by $||A||^2 = 1/2 Tr(A^T A)$. To be honest, there are various norms one can define on the space of square matrices, and this is only one possible choice. $\endgroup$ – Malkoun Jul 9 '16 at 10:02
  • $\begingroup$ The point is that, independently of the choice of $v$ (with $||v||=1$), $||S^2(v)||$ is just some (positive) constant. $\endgroup$ – Malkoun Jul 9 '16 at 10:05
  • $\begingroup$ yes but, just think of it like this. In some orthonormal basis $e_1$, $e_2$, $e_3$, $S^2(v)$ simply maps $e_1$ to $0$, and $e_2$, $e_3$ to $-e_2$ and $-e_3$ respectively (basically you take $e_1$ to be $v$, and then complete the orthonormal basis). $\endgroup$ – Malkoun Jul 9 '16 at 10:21

Let $A\in O(n)$. By an orthonormal change of basis, we can write $A$ in the form $A=diag(R_{\theta_1},\cdots,R_{\theta_p},I_q,-I_r)$ with $R_{\theta_i}=\begin{pmatrix}\cos(\theta_i)&-\sin(\theta_i)\\\sin(\theta_i)&\cos(\theta_i)\end{pmatrix}$, $\theta_i\in(0,2\pi)\setminus\{\pi\}$ and $2p+q+r=n$. Then the considered decomposition is $A=S+K$ with $S=diag(\cos(\theta_1)I_2,\cdots,\cos(\theta_p)I_2,I_q,-I_r)$.

Conversely, assume that we know $S$, the symmetric part of $A$; then by an orthonormal change of basis, we can write $S$ in the form $S=diag(\cos(\theta_1)I_2,\cdots,\cos(\theta_p)I_2,I_q,-I_r)$.

Case 1. The $(\cos(\theta_j)_j$ are distinct. Let $P_j$ be the eigenplane of $S$ associated to the double eigenvalue $\cos(\theta_j)$; note that $P_j$ is the invariant plane of $K$ s.t. the eigenvalues of $K_{|P_j}$ are $\pm i\sin(\theta_j)$ and that the $(\pm i\sin(\theta_j))_j$ are distinct; moreover, let $E$ be a subspace s.t. $S_{|E}=\pm I$, then $K_{|E}=0$. Thus $K$ is in the form $diag(U_1,\cdots,U_p,0_q,0_r)$ and, clearly, there are only two choices for each $U_j$: $\pm\begin{pmatrix}0&-\sin(\theta_j)\\\sin(\theta_j)&0\end{pmatrix}$. Finally, there are $2^p$ choices for the skew symmetric matrix $K$.

Case 2. The $(\cos(\theta_j)_j$ are not distinct. For example, assume that $n=4$ and $A$ is similar to $diag(R_{\theta},R_{\theta})$. Thus we know $S=\cos(\theta)I_4$. Note that if $P\in O(4)$, then $P^{-1}(\cos(\theta)I_4+K)P=\cos(\theta)I_4+P^{-1}KP$; consequently, the set of solutions in the skew matrix $K$ is composed with the $P^{-1}K_0P$ where $P\in O(4)$, $K_0=diag(\begin{pmatrix}0&-\sin(\theta)\\\sin(\theta)&0\end{pmatrix},\begin{pmatrix}0&-\sin(\theta)\\\sin(\theta)&0\end{pmatrix}),$. There is an infinity of such matrices $K$.

EDIT. Answer to @ Widawensen . Assume that $A$ is a real normal matrix ($AA^T=A^TA$). Then $SK=KS$ and there is a unitary matrix $P$ s.t. $S=P^{-1}diag(a_1 I_2,\cdots, a_p I_2,c_1,\cdots,c_q)P,K=P^{-1}diag(ib_1,-ib_1,\cdots,ib_p,-ib_p,0_q)P$ where $2p+q=n$, $(a_j,c_j)_j\in\mathbb{R}$ and $(b_j)_j\in\mathbb{R}^*$. More precisely, there is $P\in O(n)$ s.t. $S=P^{-1}diag(a_1 I_2,\cdots, a_p I_2,c_1,\cdots,c_q)P,K=P^{-1}diag(U_1,\cdots,U_p,0_q)P$ where $U_j=\begin{pmatrix}0&-b_j\\b_j&0\end{pmatrix}$.

Conversely, assume that we know $S$, that is $diag(a_1 I_2,\cdots, a_p I_2,c_1,\cdots,c_q)$. If $a_1,\cdots,a_p,c_1,\cdots,c_q$ are distinct, then $K$ is in the form $diag(U_1,\cdots,U_p,0_q)$; yet, unlike the case where $A\in O(n)$, we know nothing about the $(b_j)_j$ values.

  • $\begingroup$ Very fine solution and extremely interesting relation between real and imaginary parts of eigenvalues - so for 3D can we calculate real parts from S and imaginary from K and then simply to sum up them obtaining whole eigenvalues ? Is it property only for rotations or also for some other matrix decomposed to S and K? $\endgroup$ – Widawensen Jul 11 '16 at 7:08
  • $\begingroup$ I upvoted here your answer as much as I could, your answer is very enlightening and gives a real flavor of rotations .. $\endgroup$ – Widawensen Jul 11 '16 at 15:24
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    $\begingroup$ Off-topic: why do you hate matrix cookbook? You are excellent in matrices... $\endgroup$ – Widawensen Jul 11 '16 at 15:41
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    $\begingroup$ Good question... There is another similar (more advanced) cookbook here: research.microsoft.com/en-us/um/people/minka/papers/matrix/… You can read the first 3 lines: " Warning: This paper contains a large number of matrix identities which cannot be absorbed by mere reading. The reader is encouraged to take time and check each equation by hand and work out the examples. This is advanced material; see Searle (1982) for basic results." $\endgroup$ – loup blanc Jul 13 '16 at 8:55
  • $\begingroup$ This is exactly the problem with these publications; in general, an interested student copies the formula (when it is in the book) without understanding; when it is not, he writes a question on this forum ! $\endgroup$ – loup blanc Jul 13 '16 at 8:55

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