Jacobian of transformation swapping elements

Question

Suppose I have the following transformation $$ \phi(x, y) = (y, x) $$ that swaps the order of $x$ and $y$. I am told that its jacobian determinant is $1$, i..e $|\text{det}\phi'(x)|=1$. How can I show this?

Answer 1

The function is linear from $\mathbb{R}^2$ to itself. The gradient (at any point) of a linear function is the function itself. Thus, the Jacobian is just the (absolute value of the) determinant of the matrix that represents the function,

i.e., $|\det(\nabla\phi)|=\Bigl|\det\Bigl(\begin{pmatrix}0&1\\1&0\\ \end{pmatrix}\Bigl)\Bigl|=|-1|=1$.

For general functions, one must look at the definition of the gradient of a function from $\mathbb{R}^n$ to itself and compute it. Then, take the absolute value of the determinant. One can check that for linear functions, this definition gives back the matrix that represent the function itself.

Note again that linear functions have the same gradient (hence the same Jacobian) everywhere (for every $(x,y)$), but this in general depends on the point we are considering.