Can non-linear makeovers be stood for as Transformation Matrices?

I simply returned from an extreme linear algebra lecture which revealed that straight makeovers can be stood for by transformation matrices; with even more generalization, it was later on revealed that affine makeovers (straight + translation) can be stood for by matrix reproduction too.

This obtained me to thinking of all those various other makeovers I've grabbed over the previous years I've been researching maths. As an example, polar makeovers-- changing $x$ and also $y$ to 2 new variables $r$ and also $\theta$.

If you mapped $r$ to the $r$ axis and also $\theta$ to the $y$ axis, you 'd primarily have a coordinate transformation. An instead deformed one, at that.

Exists a means to represent this making use of a transformation matrix? I've attempted messing about with the numbers yet every little thing I've attempted to collaborate with has actually crumbled fairly embarrassingly.

Extra notably, exists a means to, offered a details non-linear transformation, construct a transformation matrix from it?

2019-05-04 17:26:58
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Answers: 5

As others have actually currently stated, the Jacobian component changes one coordinate system to an additional by connecting infinitesimal locations (or quantities) from one system to an additional. Take into consideration going from Cartesian to Polar works with:

\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 serves due to the fact that:

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

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

Which informs you that if you have a function $f(r, \theta)$ you can calculate the indispensable as long as you bear in mind to add a variable of $r$. The usual makeovers have actually all been exercised and also can be located here on Wikipedia.

2019-05-09 02:08:36

You can stand for some non - straight changes (like translation) of an $n$ - dimensional vector with an $(n+1)$ - dimensional matrix. Nonetheless, transforming the vector to its $(n+1)$ - dimensional uniform variation and also back is not a straight transformation as well as additionally not representable as a matrix.

Extra is clarified here.

2019-05-09 02:06:48

As Harry claims, you can not (the instance of affine makeovers can be fine-tuned to function due to the fact that they're simply straight ones with the beginning converted). Nonetheless, estimating a nonlinear function by a straight one is something we do regularly in calculus via the by-product, and also is what we usually need to do to make a mathematical version of some actual - globe sensation tractable.

2019-05-08 07:54:41

No. Every little thing is established by a selection of basis. For an extra in - deepness solution, I would certainly require to clarify the first 2 weeks of linear algebra and also attract some commutative layouts.

If you would certainly such as a far better description, see web pages 12 - 14 of Emil Artin's essay Geometric Algebra .

2019-05-08 06:05:13

You can not stand for a non straight transformation with a matrix, nonetheless there are some methods (for desire of a far better word) readily available if you make use of uniform co-ordinates. As an example, $3\text{D}$ translation is a non-linear transformation in a $3\times3$ $3\text{D}$ transformation matrix, yet is a straight transformation in $3\text{D}$ uniform co-ordinates making use of a $4\times4$ transformation matrix. The very same holds true of various other points like viewpoint estimates. This is why $4\times4$ matrices are made use of in $3\text{D}$ graphics as the uniform co-ordinate system streamlines points a whole lot.

To make clear - making use of uniform co-ordinates raises the series of makeovers representable making use of matrices from simple straight makeovers to affine makeovers and also some estimates, yet it does not make all non-linear makeovers representable making use of matrices. The non-linear transformation given as an instance is still past depiction as an affine transformation (Thanks to @Harry for motivating this explanation in the remarks)

2019-05-08 02:23:00