# Why aren't different coordinate systems of interest in linear algebra?

In linear algebra we used the standard basis and talked about vector spaces. But we never mentioned coordinate systems.

In physics we made it clear which coordinate system we were using (cartesian, cylindrical, spherical).

Why aren't different coordinate systems of interest in linear algebra?

• The whole point of linear algebra is to consider transformations between different affine coordinate systems. Cylindrical/spherical etc. coordinate systems are not linear or affine. The study of non-linear coordinate systems and the transformations between them is part of the field of differential geometry. – Chill2Macht May 27 '17 at 9:24
• @Chill2Macht: that deserves to be an answer. – Dan Uznanski May 27 '17 at 9:29
• Because the (Fréchet) derivative of a linear map is the linear map itself: no differential calculus is needed. – egreg May 27 '17 at 9:48
• Also note that linear algebra in most parts works without basis or coordinates – JJR May 27 '17 at 9:50

(Note: this answer is not particularly well-written and is just meant to expand somewhat on my terse comment above, as was requested by others.)

The whole point of linear algebra is to consider transformations between different affine coordinate systems.

Cylindrical/spherical etc. coordinate systems are not linear or affine.

Specifically, a coordinate system is determined by a set of coordinate functions. Linear algebra can be considered to include the study of linear coordinate functions (see linear functional and dual space), while the study of more general coordinate systems (i.e. ones whose coordinate functions are not necessarily linear) requires techniques of differential geometry.

For example, given two vectors in $\mathbb{R}^2$, $x, y$, we can consider their polar coordinates $r, \theta$ to be two functions $\mathbb{R}^2 \to \mathbb{R}$. When I say that polar coordinates aren't "linear" or "affine", I just mean that we do not necessarily have that $r(x + y) = r(x) + r(y)$ or that $\theta(x+y) = \theta(x) + \theta(y)$.

The study of non-linear coordinate systems and the transformations between them is part of the field of differential geometry.

Describing how to change and move between arbitrary coordinate systems, not just ones whose coordinate functions are linear, are historically what motivated Riemann to develop the concept of "chart" and "atlas", and thus ultimately of manifold.

The key idea to remember here is that any Euclidean space is a manifold, and that considering more general types of coordinate systems besides linear ones requires more and different techniques besides just those offered by linear algebra. It turns out that a lot of these techniques are unchanged when considering objects which look "locally" like Euclidean space, called manifolds. So a lot of the motivations for the concepts of differential geometry can be found and studied in surprising depth just by considering non-linear coordinate systems for Euclidean spaces.

It's mostly because in the other coordinate systems, a lot of things are more difficult.

In cylindrical and spherical coordinates, addition of vectors is so incredibly annoying that the standard method is to convert to cartesian coordinates, do the addition, and convert back.

Linear transformations, built as they often are from addition of vectors, are no better.

The dot and cross products can be applied to get relative angle information, which looks like it could make spherical and cylindrical coordinates useful, but the relative angles in cylindrical coordinates are entirely worthless because that only measures it in the plane, and in spherical coordinates the calculation of relative angle requires considerable use of trigonometric functions - the aviation formulary suggests five forward and one inverse - to get an answer. Similar problems arise when trying to use cross product for its obvious purpose of finding a normal.

I guess a simple way to put it is this: in linear algebra, lines remain lines, and planes remain planes. Circles and spheres don't.