Jacobian - Rhea

Jacobians and their applications

by Joseph Ruan

Basic Definition

The Jacobian Matrix is just a matrix that takes the partial derivatives of each element of a transformation. In general, the Jacobian Matrix of a transformation F, looks like this:

F₁,F₂, F₃... are each of the elements of the output vector and x₁,x₂, x₃ ... are each of the elements of the input vector.

So for example, in a 2 dimensional case, let T be a transformation such that T(u,v)=<x,y> then the Jacobian matrix of this function would look like this:

$ J(u,v)=\begin{bmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{bmatrix} $

To help illustrate making Jacobian matrices, let's do some examples:

Example #1:

Let's take the Transformation:

$ T(u,v) = <u*\cos v, u*\sin v> $ .

What would be the Jacobian Matrix of this Transformation?

Solution:

$ x=u*\cos v \longrightarrow \frac{\partial x}{\partial u}= \cos v , \; \frac{\partial x}{\partial v} = -u*\sin v $

$ y=u*\sin v \longrightarrow \frac{\partial y}{\partial u}= \sin v , \; \frac{\partial y}{\partial v} = u*\cos v $

$ J(u,v)=\begin{bmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{bmatrix}= \begin{bmatrix} \cos v & -u*\sin v \\ \sin v & u*\cos v \end{bmatrix} $

This example actually showcased the transformation "T" as the change from polar coordinates into Cartesian coordinates.

Example #2:

Let's take the Transformation:

$ T(u,v) = <u, v, u+v> $ .

What would be the Jacobian Matrix of this Transformation?

Solution:

Notice, that this matrix will not be square because there is a difference in dimensions of the input and output, i.e. the transformation is not injective.

$ x=u \longrightarrow \frac{\partial x}{\partial u}= 1 , \; \frac{\partial x}{\partial v} = 0 $

$ y=v \longrightarrow \frac{\partial y}{\partial u}=0 , \; \frac{\partial y}{\partial v} = 1 $

$ z=u+v \longrightarrow \frac{\partial y}{\partial u}= 1 , \; \frac{\partial y}{\partial v} = 1 $

$ J(u,v)=\begin{bmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \\ \frac{\partial z}{\partial u} & \frac{\partial z}{\partial v} \end{bmatrix}= \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ 1 & 1\end{bmatrix} $

Example #3:

Let's take the Transformation:

$ T(u,v) = <uv> $ .

What would be the Jacobian Matrix of this Transformation?

Solution:

Notice, that this matrix will not be square because there is a difference in dimensions of the input and output, i.e. the transformation is not injective.

$ x=u \longrightarrow \frac{\partial x}{\partial u}= v , \; \frac{\partial x}{\partial v} = u $

$ J(u,v)= \begin{bmatrix}\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \end{bmatrix}=\begin{bmatrix}v & u \end{bmatrix} $

Application: Jacaobian Determinants

The determinant of Example #1 gives:

$ \left|\begin{matrix} \cos v & -u * \sin v \\ \sin v & u * \cos v \end{matrix}\right|=~~ u \cos^2 v + u \sin^2 v =~~ u $

Notice that, in an integral when changing from cartesian coordinates (dxdy) to polar coordinates $ (drd\theta) $, the equation is as such:

$ dxdy=r*drd\theta $

in this case, since $ u =r $ and $ v = \theta $, then

$ dxdy=u*dudv $

It is easy to extrapolate, then, that the transformation from one set of coordinates to another set is merely

$ dC1=det(J(T))dC2 $

where C1 is the first set of coordinates, det(J(C1)) is the determinant of the Jacobian matrix made from the Transformation T, T is the Transformation from C1 to C2 and C2 is the second set of coordinates.

It is important to notice several aspects: first, the determinant is assumed to exist and be non-zero, and therefore the Jacobian matrix must be square and invertible. This makes sense because

Sources:

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