linear algebra - How can skew-symmetric matrices be thought of as infinitesimal rotations?

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?

Answer

If you have a sufficiently smooth function $F$ from a real variable $t$ to real $n\times n$ matrices, then you can differentiate each element of the matrix with respect to $t$, and therefore give meaning to $F'(t)$.

Now if we know that $F(t)$ is always a rotation matrix and that $F(t_0)=I$, then it turns out that $F'(t_0)$ will always be skew-symmetric. And conversely, every skew-symmetric matrix will be the derivative of some $F$ that satisfies these conditions.

In this way we can consider the skew-symmetic $F'(t_0)$ to encode which way and how fast the new coordinates given by $F(t)$ rotate at time $t=t_0$.

Without the assumption that $F(t_0)=I$ we can still say that $F(t)^{-1}F'(t)$ and $F'(t)F(t)^{-1}$ are always skew-symmetric; these encode the instantaneous rotation at any time in two different ways.