TITLE: How can skew-symmetric matrices be thought of as infinitesimal rotations?
QUESTION [5 upvotes]: I've recently stumbled upon the fact that skew-symmetric matrices represent somehow infinitesimal rotations. Having never encountered them, I looked them up and learnt they have to do with Lie algebras and groups, but this is beyond what I've studied so far.
Is it possible to have a more intuitive understanding of this? 
Also, from Wikipedia:

skew-symmetric matrices are derivatives, while an actual infinitesimal rotation matrix has the form $I+Ad\theta$ where $d\theta$ is vanishingly small and $A ∈ so(3)$.

Having read this is about derivatives and has applications in physics, that "lonely" $d\theta$ is actually a bit suspicious. What about it?

REPLY [4 votes]: I would like to say how bad the formula "infinitesimal rotation matrix" $=I+Ad\theta$ is.
Firstly the correct equality is $R(\theta)=R(0)+\theta A+O(\theta^2)$ where $R(0)=I$ and $A\in SK$ (a skew symmetric matrix).
Secondly, one must not believe that this type of formula is valid only when the angle varies; we can make it valid when the angle AND the axis vary and more generally,  for $O(n,\mathbb{R})$, the group of orthogonal real $n\times n$ matrices.
Thirdly, it's simply the Taylor formula in $O(t^2)$. We will see below that this type of formula can be easily extended in $O(t^3)$.
Proposition. Let $t\in I\rightarrow R(t)\in O(n,\mathbb{R})$   be a $C^3$ function where $I\subset \mathbb{R}$ is a neighborhood of $0$. Then there are two skew symmetric matrices $K,L$ s.t.
$R(t)=(I+tK+t^2/2(K^2+L))R(0)+0(t^3).$ 
Proof. The beginning is very well-known. $RR^T=I$ implies that $R'R^T+RR'^T=0$, that is $R'R^T=K\in SK$ and finally $R'=KR$.
Second step. $R''R^T+2R'R'^T+RR''^T=0$, that implies $R"R^T+RR''^T-2K^2=0$. 
Then $R''R^T-K^2=L\in SK$ and consequently $R''=(K^2+L)R$.
We finish by applying the Taylor formula
$R(t)=R(0)+tR'(0)+t^2/2R''(0)+O(t^3)$.
Remark. $O(n,\mathbb{R})$ is an algebraic real group that admits $SK$ as tangent vector space in $I$; therefore, $O(n,\mathbb{R})$ has dimension $n(n-1)/2$. Moreover, if $K\in SK$, then $t\rightarrow e^{tK}R$ is the geodesic of $O(n,\mathbb{R})$ through $R$ in the direction $KR$.