Can non-linear transformations be represented as Transformation Matrices?

Question

I just came back from an intense linear algebra lecture which showed that linear transformations could be represented by transformation matrices; with more generalization, it was later shown that affine transformations (linear + translation) could be represented by matrix multiplication as well.

This got me to thinking about all those other transformations I've picked up over the past years I've been studying mathematics. For example, polar transformations -- transforming $x$ and $y$ to two new variables $r$ and $\theta$.

If you mapped $r$ to the $r$ axis and $\theta$ to the $y$ axis, you'd basically have a coordinate transformation. A rather warped one, at that.

Is there a way to represent this using a transformation matrix? I've tried fiddling around with the numbers but everything I've tried to work with has fallen apart quite embarrassingly.

More importantly, is there a way to, given a specific non-linear transformation, construct a transformation matrix from it?

Answer 1

As Harry says, you can't (the example of affine transformations can be tweaked to work because they're just linear ones with the origin translated). However, approximating a nonlinear function by a linear one is something we do all the time in calculus through the derivative, and is what we often have to do to make a mathematical model of some real-world phenomenon tractable.

Answer 2

As others have already mentioned, the Jacobian determinant transforms one coordinate system to another by relating infinitesimal areas (or volumes) from one system to another. Consider going from Cartesian to Polar coordinates:

\begin{align} J &= \det\frac{\partial(x,y)}{\partial(r,\theta)} =\begin{vmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \\\\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} \\\\ \end{vmatrix} \\&=\begin{vmatrix} \cos\theta & -r\sin\theta \\\\ \sin\theta & r\cos\theta \\\\ \end{vmatrix} =r\cos^2\theta + r\sin^2\theta = r \end{align}

This is useful because:

$$\mathrm{d}A = J\;\mathrm{d}r\,\mathrm{d}\theta = r\,\mathrm{d}r\,\mathrm{d}\theta$$

$$\iint_\mathbf{R} f(r,\theta)\,\mathrm{d}A = \int_a^b \int_0^{r(\theta)} f(r,\theta) r\,\mathrm{d}r\,\mathrm{d}\theta$$

Which tells you that if you have a function $f(r, \theta)$ you can compute the integral as long as you remember to add a factor of $r$. The common transformations have all been worked out and can be found here on Wikipedia.

Answer 3

You can't represent a non linear transformation with a matrix, however there are some tricks (for want of a better word) available if you use homogenous co-ordinates. For example, $3\text{D}$ translation is a non-linear transformation in a $3\times3$ $3\text{D}$ transformation matrix, but is a linear transformation in $3\text{D}$ homogenous co-ordinates using a $4\times4$ transformation matrix. The same is true of other things like perspective projections. This is why $4\times4$ matrices are used in $3\text{D}$ graphics as the homogenous co-ordinate system simplifies things a lot.

To clarify - using homogenous co-ordinates increases the range of transformations representable using matrices from plain linear transformations to affine transformations and some projections, but it doesn't make all non-linear transformations representable using matrices. The non-linear transformation provided as an example is still beyond representation as an affine transformation (Thanks to @Harry for prompting this clarification in the comments)

Answer 4

You can represent some non-linear transforms (like translation) of an $n$-dimensional vector with an $(n+1)$-dimensional matrix. However, converting the vector to its $(n+1)$-dimensional homogeneous version and back is not a linear transformation and also not representable as a matrix.

More is explained here.

Answer 5

The point of transformation matrices is that the images of the $n$ basis vectors is sufficient to determine the action of the entire transformation -- this is true for linear transformations, but not an arbitrary transformation.

However, nonlinear transformations (the smooth ones, anyway), can be locally approximated as linear transformations. With a bit of calculus, you get the "Jacobian matrix", which acts on the tangent vector space at every point on a manifold. This is a generalisation of transformation matrices in the sense that linear transformation's Jacobian is equal to its matrix representation, i.e. in the same sense that the derivative generalises the slope (which completely determines a linear function $y=mx$)

Answer 6

No. Everything is determined by a choice of basis. For a more in-depth answer, I would need to explain the first two weeks of linear algebra and draw some commutative diagrams.

If you'd like a better explanation, see pages 12-14 of Emil Artin's monograph Geometric Algebra.