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Lets say we have a function $f : \mathbb{R}^3\rightarrow \mathbb{R}^3$, as defined below, with its value being denoted as $(a, b, c)$ for convenient reference.

$$f(x,y,z) = (x^2, y^2, z^2) = (a, b, c)$$

The Jacobian matrix of $f$ and subsequently the Jacobian determinant would then be:

$$ \begin{bmatrix} a'\\ b'\\ c' \end{bmatrix} = \begin{bmatrix} \frac {\partial a}{\partial x} & \frac {\partial a}{\partial y} & \frac {\partial a}{\partial z}\\ \frac {\partial b}{\partial x} &\frac {\partial b}{\partial y} & \frac {\partial b}{\partial z}\\ \frac {\partial c}{\partial x} &\frac {\partial c}{\partial y} & \frac {\partial c}{\partial z} \end{bmatrix} = \begin{bmatrix} 2x & 0 &0 \\ 0 & 2y &0 \\ 0 & 0 &2z \end{bmatrix} $$

$$ \begin{vmatrix} a'\\ b'\\ c' \end{vmatrix} = \begin{vmatrix} \frac {\partial a}{\partial x} & \frac {\partial a}{\partial y} & \frac {\partial a}{\partial z}\\ \frac {\partial b}{\partial x} &\frac {\partial b}{\partial y} & \frac {\partial b}{\partial z}\\ \frac {\partial c}{\partial x} &\frac {\partial c}{\partial y} & \frac {\partial c}{\partial z} \end{vmatrix} = \begin{vmatrix} 2x & 0 &0 \\ 0 & 2y &0 \\ 0 & 0 &2z \end{vmatrix} = 2x2y2z $$ Ok sure, this makes sense. It's kind of just like normal calculus but expanding everything out into a matrix.

Now I look at the shorthand notation for the Jacobian determinant: $$ \frac {\partial(a,b,c)}{\partial(x,y,z)} = 2x2y2z $$ Where did this even come from? Why are there partials there. How does it convey the same amount of information? How do I even read this "shorthand notation". It just seems so left field - out of nowhere.

  • How do I read it
  • How did this arguably rather cryptic notation come about?
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  • $\begingroup$ Probably the way it came about is simply by imitation of the 1D case where you have (substitution rule) $du=\dfrac{du}{dx}\,dx$. For the 3D case, things are more complicated but you end up with $da\,db\,dc=\dfrac{\partial(a,b,c)}{\partial(x,y,z)}\,dx\,dy\,dz$ as the change of variables for going from $(a,b,c)$ coordinates to $(x,y,z)$ coordinates. The change from $d$ to $\partial$ is because each $a,b,c$ can depend on $(x,y,z)$ in their own way. (Why the determinant is correct is a different matter entirely...) $\endgroup$ – Semiclassical Mar 24 '16 at 3:14
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It is a little awkward, but using the partial derivative notation with vectors basically means:

$$\begin{align}\dfrac{\partial (a, b, c)}{\partial (x, y, z)} ~=~& \det\left((\dfrac{\partial}{\partial x}, \dfrac{\partial}{\partial y}, \dfrac{\partial}{\partial z})^\top(a,b,c) \right)^\top \\[1ex] =~& \begin{vmatrix}\dfrac{\partial a}{\partial x}&\dfrac{\partial a}{\partial y}&\dfrac{\partial a}{\partial z}\\ \dfrac{\partial b}{\partial x}&\dfrac{\partial b}{\partial y}&\dfrac{\partial b}{\partial z}\\\dfrac{\partial c}{\partial x}&\dfrac{\partial c}{\partial y}&\dfrac{\partial c}{\partial z}\end{vmatrix} \end{align}$$

The notation summarises the essentials of the Jacobian determinant.   You are taking the partial derivatives of $a, b, c$ each with respect to $x,y,z$, constructing a matrix of the result and evaluating its determinant.

$$\dfrac{\partial (a, b, c)}{\partial (x, y, z)}$$

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  • $\begingroup$ Thanks for the explanation. A bit of tangent - but is there such a thing as a Jacobian for matrix-valued functions? Or are they typically just for vector-valued ones? $\endgroup$ – AlanSTACK Mar 24 '16 at 4:11

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