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?

  • $\begingroup$ There's a specific sort of determinant (I forget its name) for working out the transformations between different coordinate systems (e.g. Cartesian and Polar as you mention). $\endgroup$ – Noldorin Jul 22 '10 at 8:04
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    $\begingroup$ The Jacobian Determinant. $\endgroup$ – user126 Jul 22 '10 at 8:08

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.

  • $\begingroup$ Could this approximation be used to linearly approximate a polar transformation? $\endgroup$ – Justin L. Jul 22 '10 at 19:43
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    $\begingroup$ Yes, if by polar transformation you mean $(x,y) \rightarrow (\sqrt{x^2+y^2},arctan(y/x)$. Basically you can do it for all differentiable functions - in fact the definition of differentiability means that you can well approximate it by a linear transformation. $\endgroup$ – Soarer Aug 8 '10 at 14:55
  • $\begingroup$ Can tensors with shape dimension higher than 2 or infinite matrices/tensors represent nonlinear transform then? $\endgroup$ – KYHSGeekCode Aug 17 '18 at 16:27

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.

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    $\begingroup$ I can't for the life of me figure out how to make those things above look like proper 2x2 matrices, any help would be greatly appreciated! $\endgroup$ – Hooked Aug 5 '10 at 17:56
  • $\begingroup$ I tried to convert to the 2D matrix. You need to use ` \\\\ ` (four of them instead of two). Also, I added a couple of \det which I presume were missing. $\endgroup$ – Aryabhata Sep 9 '10 at 22:47
  • $\begingroup$ I went ahead and changed matrix to vmatrix where appropriate. $\endgroup$ – J. M. is a poor mathematician Sep 9 '10 at 23:01

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)

  • $\begingroup$ How could this be used to represent the example Polar transformation? $\endgroup$ – Justin L. Jul 22 '10 at 9:01
  • $\begingroup$ It couldn't.... $\endgroup$ – user126 Jul 22 '10 at 9:07
  • $\begingroup$ Sorry; to clarify, I was asking if any of these "tricks" mentioned in this answer could be used to approximate or trick a matrix transformation that could approximately represent a polar transformation, to see if these "tricks" really did answer my question. $\endgroup$ – Justin L. Jul 22 '10 at 19:44

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.


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 $y=mx$)


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.

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    $\begingroup$ There are plenty of ways to legally acquire such a text, my (US) local library has a copy and if it didn't interlibrary loan usually provides. Pointing the OP to a classic reference can only enhance the answer, though I admit it is a bit sparse. $\endgroup$ – Hooked Aug 5 '10 at 17:37
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    $\begingroup$ The answer would be an order of magnitude better if he were to paraphrase whatever was on pages 12-14 of Artin's book. $\endgroup$ – Hooked Aug 6 '10 at 15:55
  • $\begingroup$ @BlueRaja, @Hooked: cleaned up some confusion regarding different revisions of this post. $\endgroup$ – Larry Wang Aug 8 '10 at 14:44
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    $\begingroup$ Please don't put anything in your answer that promotes illegal activity. If you would like to add a summary of the relevant information like Hooked suggests, post it as a comment and flag this comment so one of the moderators can add it to your answer. 97832123 created a meta question about this answer. $\endgroup$ – Larry Wang Aug 8 '10 at 20:20

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