Commutation of differentiation with any linear map Consider $A$ and $B$ are two linear operators in finite-dimensional vector space. So to commute, they should share something special. For matrix $A$ to commute with all other is quite restrictive: it would only true for multiply of identity.
However differential is also linear operator and it commutes with any matrix $A$:
$$
f(x) = A\phi(x), [D_{(x_0)}A\circ\phi(x)](h) = A[D_{(x_0)}\phi(x)](h).
$$
This is quite a puzzle for me from this perspective.
 A: First of all, the space of differential forms $\Omega^*(M)$ on a manifold $M$ is not a finite dimensional vector space: it is a finite rank $\mathcal{C}^{\infty}(M)$-module, and an infinite dimensional vector space over $\Bbb R$.
Moreover, in this specific context, the exterior differential $d\colon \Omega^*(M)\to \Omega^*(M)$ does not commute with all linear operators.
For instance, if $X$ is a vector field, the interior multiplication $\iota_X$, given by
$$
\iota_X\omega = \omega(X,\cdot,\ldots,\cdot)
$$
does not commute with $d$ for a generic $X$.
In your example, the matrix you consider has constant coefficient: it is not surprising that $d$ commutes with it since $d$ is a differential operator.
We could expect a sort of Leibniz rule, like
$$
d(A\phi) = dA . \phi + A d\phi,
$$
and $A$ being with constant coefficients, we would have $dA=0$
(caution, what I said here is not totally rigorous).
A: The differentiation map doesn’t commute with every other linear map. Let $V$ denote the vector space of differentiable functions from $R \to R$. Define $D : V \to V$ by $$Df = f’$$ and define $ T : V \to V$ by $$Tf = xf(x)$$
So, for example, $T(x^2) = xx^2 =x^3$. It’s a bit of an abuse of notation.
As you can see that $$(DT)(x^2)= 3x^2$$ Whereas $$(TD)(x^2) = 2x^2$$
A: It's because the derivative of a linear map $T$ is again just the linear map $T$. This is because the derivative is the best linear approximation, but if you map is already linear, then there is no need to approximate it. For example, in one dimension, this is why $T:\mathbb{R}\to \mathbb{R}$ by $T(x)=\lambda x$ has derivative $\lambda$.
