Infinitesimal rotation matrix close to the identity

Question

Suppose $R_{ij}$ is a matrix that corresponds to an "infinitesimal" rotation. Then my notes mention that such matrix can be rewritten as: $$R_{ij} = \delta _{ij} + \epsilon w_{ij}$$

where $w_{ij} = - w_{ji}$ is antisymmetric and $\epsilon$ is a small quantity. Now I know that a rotation matrix must be orthogonal,but I am unsure on why is $w_{ij}$ must be antisymmetric. Also why would such a matrix represent an infinitesimal rotation?

Answer 1

$\require{enclose} \def\e{\varepsilon}$ The defining equation for a rotation matrix is the orthogonality condition $$I=R^TR$$ Add a tiny perturbation to the identity matrix, then enforce the orthogonality condition. $$\eqalign{ I &= (I+\e W)^T(I+\e W) \\ I &= I + \e W + \e W^T + (\enclose{horizontalstrike}{\e^2W^TW}) \\ 0 &= \e W + \e W^T \\ W &= -W^T \\ }$$ Conclusion:   A skew-symmetric perturbation of the identity matrix yields a rotation matrix.

Answer 2

Let's look at a path in the space of rotations: $$ c(t) = \pmatrix{\cos t & -\sin t & 0 \\ \sin t & \cos t & 0 \\ 0 & 0 & 1}. $$ Note that for any rotation $R$, we have $R^TR = I$. We'll come back to that.

Anyhow, let's look at $c'(0)$, the tangent to $c$ as it passes through the identity matrix. It's

$$ c'(0) = \left. \pmatrix{-\sin t & -\cos t & 0 \\ \cos t & -\sin t & 0 \\ 0 & 0 & 0}\right|_{t = 0} = \pmatrix{0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0}. $$ Now points on $c$ near $c(0)$ are (using the Taylor series) of the form $$ c(h) \approx c(0) + c'(0) h $$ The entries of this are $\delta_{ij} + h m_{ij}$, where $M = c'(0)$ is the skew-symmetric matrix I just wrote down.

In short:rotations about $z$ near the origin look like $I + \epsilon M$, and $M$ is skew-symmetric.

What if $c$ is some more generaal path through the identity, but still in the rotation group?

Then $$ c(t)^T c(t) = I $$ for all $t$, and differentiating both sides gives $$ c'(t)^T c(t) + c(t)^T c'(t) = 0 $$ Evaluating this at $t = 0$ (supposing that $c$ still passes through the identity matrix at time $0$), we get $$ c'(0)^T + c'(0) = 0 $$ or $$ c'(0)^T = -c'(0), $$ so just as before, $M = c'(0)$ is skew-symmetric, and the remainder of the argument carries through.