I'm sorry this isn't a yes/no/am-I-right question but I seriously cannot see why the Jacobian/total derivative matrix is what it is?
I am also using it as LaTeX practice (for maths) hence the barely relevant blocks of math.
(note the terms in the title and this message is what I hope to get clarified, just incase they are wrong)
$f:\mathbb{R}^n\rightarrow\mathbb{R}^m$
and $f(x)=y$
I will use the notation of $x^i$ being the ith member of a vector x, that is:
$x=(x^1,x^2,x^3, … ,x^k)$ for a vector of order/length k.
Consider also the linear map: (or should I say transform?)
$A:\mathbb{R}^n\rightarrow\mathbb{R}^m$
Consider now $x,u \in \mathbb{R}^n$ (I would use delta x for u, but then my notation could backfire)
The best linear approximation we can have for the ith coordinate of the image is:
$$f(x+u)^i = f(x)^i + \sum^n_{j=1}(\frac{\partial f(x)^i}{\partial x^j}u^j)$$
Which is basically the definition of a partial derivative. Nothing new here.
Clearly now:
$$f(x) + \begin{bmatrix}
\frac{\partial f(x)^1}{\partial x^1} &
\frac{\partial f(x)^1}{\partial x^2} &
…&
\frac{\partial f(x)^1}{\partial x^n} \\
\frac{\partial f(x)^2}{\partial x^1} &
\frac{\partial f(x)^2}{\partial x^2} &
…&
\frac{\partial f(x)^2}{\partial x^n} \\
…&…&…&…\\
\frac{\partial f(x)^m}{\partial x^1} &
\frac{\partial f(x)^m}{\partial x^2} &
…&
\frac{\partial f(x)^m}{\partial x^n}
\end{bmatrix}*\begin{bmatrix}u^1\\u^2\\…\\u^n\end{bmatrix}
\approx f(x+u)$$
Let:
$$\begin{bmatrix}
\frac{\partial f(x)^1}{\partial x^1} &
\frac{\partial f(x)^1}{\partial x^2} &
…&
\frac{\partial f(x)^1}{\partial x^n} \\
\frac{\partial f(x)^2}{\partial x^1} &
\frac{\partial f(x)^2}{\partial x^2} &
…&
\frac{\partial f(x)^2}{\partial x^n} \\
…&…&…&…\\
\frac{\partial f(x)^m}{\partial x^1} &
\frac{\partial f(x)^m}{\partial x^2} &
…&
\frac{\partial f(x)^m}{\partial x^n}
\end{bmatrix}*\begin{bmatrix}u^1\\u^2\\…\\u^n\end{bmatrix}
=\begin{bmatrix}v^1\\v^2\\…\\v^m\end{bmatrix}$$
Then:
$f(x+u)\approx y+v$
That is a pretty compelling case for not transposing it!
If you transpose it you can then use row-vectors and pre-multiply, rather than post, but column-vectors and linear algebra are something that is so well established, I wont restate.
Now the total derivative matrix is listed in my notes (quite correctly IMO) as what I've just put above.
Writing it like I have above looks nice, because now let that matrix be A:
$f(x+u) \approx f(x)+Au$ isn't that lovely!
The "Jacobian of a transformation" however is written as "the determinant of the transpose of this matrix" and I must ask WHY?
Transposing a matrix does not affect it's determinant! I cannot find any reason to transpose it UNLESS there is some notation I don't know about where:
$$\frac{\partial (x,y)}{\partial(u,v)}$$ = some 2×2 matrix that is the transpose of what I claim it ought to be.
But my book says the above denotes a determinant <—NOT a matrix
So why might I want to transpose it?
Unless there are several things with "Jacobian" in the name maybe, could someone correct me!
Edit:
The books consider $A^T$ as the Jacobian Matrix (?) and "The Jacobian of the transformation" is defined using the determinant of that matrix, which is written as just:
$$\frac{\partial (y^1,y^2,…,y^n)}{\partial(x^1,x^2,..,x^n)}$$
Obviously in this case n=m
I WAS A HUGE DIPSTICK
I'm so sorry guys, this has wasted everyone's time, 'cept maybe John's…..
In the 2×2 case:
$$r_1=x_1-x_0 = (\frac{\partial x}{\partial u},\frac{\partial y}{\partial u})\Delta u$$
$$r_2=x_2-x_0 = (\frac{\partial x}{\partial v},\frac{\partial y}{\partial v})\Delta v$$
You can then find the area as $\lVert r_1 \times r_2\rVert$
And
$$r_1\times r_2 = det\begin{bmatrix}
\frac{\partial x}{\partial u} &
\frac{\partial y}{\partial u}
\\
\frac{\partial x}{\partial v} &
\frac{\partial y}{\partial v}\end{bmatrix} . \Delta u\Delta v\boldsymbol{k}$$
It is talking about the transpose of that!
I am so sorry everyone!
Best Answer
I have always used what you first have (without the transpose) as the Jacobian. It is in every book I have seen. It is also the definition on wikipedia.
https://en.wikipedia.org/wiki/Jacobian_matrix_and_determinant
So basically, you are right, where ever you are reading it is wrong.