Derivative of a linear operator

In a quip, A' = A
More accurately,
(Ax)' = A
That is, the derivative of a -linear transform-
is a -constant-,
just as (cx)' = c (in 1-dimension)
The quip reflects that we often write a matrix for the map it
represents:
the derivative of the (function) is the (constant)

Concretely, the Jacobian of a linear transform
is the linear transform itself (w/r/t any basis).

Very concretely, let
A = [1 2
     3 4]
So Ax = 1x1 + 2x2
        3x1 + 4x2
...and J(A) =
1\partial x1 + 2\partial x2
3\partial x1 + 4\partial x2
= A(\partial x)
...if that clarifies anything
[actually in J(A) you should only write the matrix,
 not the equations, as J(A) is the Jacobian matrix/operator...
 should actually write
 J(A)(\partial x) = A(\partial x)
 ]

This "constant" reflects that linear transforms are (infintesimally)
homogeneous (and indeed this -defines- them): they behave the same at
every point.
...and at every point the infintesimal behavior is exactly the global
behavior.

Equivalently, they sorta commute with translation.
They are not -isotropic- though;
that's essentially just dilation/scalars
(you might argue that rotations in R^2 are isotropic,
 seeing as they are central in SO(2) (as it's abelian),
 but this is sorta vague: they're not central in O(2),
 but they're still "the same" in every direction:
 they're central under -some transitive subgroup-)
