Matrices and differentiation commute

Question

Suppose for simplicity we have a plane curve $\gamma(t)=(f(t),g(t))$. I'm just curious exactly what is the property responsible for the fact that, if $R_{90}$ is the two dimensional rotation matrix by $90$ degrees counterclockwise, then $$\frac{d}{dt}\left(R_{90}\left(\gamma(t)\right)\right)=R_{90}\left(\frac{d}{dt}(\gamma(t))\right).$$

I'm thinking this only applies to matrices $R$ that are linear isomorphisms (and for when the multiplication makes sense). What is the general property at work here?

Answer 1

You can think of this as the chain rule. Qualitatively, the derivative of a linear transformation $T$ at an arbitrary point is $T$ itself. In symbols, $DT(x) = T$ for all $x$ in the domain of $T$.

Answer 2

Note that $$R_{90}=\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix},$$ so $$R_{90}(\gamma(t))=(-g(t),f(t)).$$

Then as we take derivatives: $$\frac{d}{dt}R_{90}(\gamma(t))=(-g'(t),f'(t))=R_{90}(f'(t),g'(t))=R_{90}\left(\frac{d}{dt}\gamma(t)\right).$$

Of course, Andrew's answer illustrates a more general case.

Answer 3

The main reason is that multiplication by a matrix is a linear operation. By definition of the differentiability, $$ \gamma(t) \underset{t\to t_0}{=} \gamma(t_0) + (t-t_0)\gamma'(t_0) + o(t-t_0). $$ If $M$ is any matrix, the application $u \mapsto M(u)$ is linear and continuous, and thus $$ M(\gamma(t)) \underset{t\to t_0}{=} M(\gamma(t_0)) + (t-t_0)M(\gamma'(t_0)) + o(t-t_0). $$ (The continuity is needed to ensure that $M(o(t-t_0)) = o(t-t_0)$.)

Hence, $M\circ\gamma$ is differentiable with $(M\circ\gamma)'=M\circ\gamma'$.