Jacobian of a linear transformation I am currently solving a problem where I have to use the transformation:  
$$\begin{align*}y_1 &=\bar{x}\\
y_2 &= x_2-\bar{x}\\
y_3 &= x_3-\bar{x}\\
&\vdots   \\

y_n &= x_n-\bar{x}\\
\end{align*}$$   
My question is: how do I find the Jacobian of the above transformation. 
thank you.
 A: This is essentially what Sasha said:
We interpret $\bar x$ as ${1\over n}(x_1+\ldots + x_n)$. Then the Jacobian $J:=\bigl[{\partial y_i\over\partial x_k}\bigr]_{i,k}$ looks as follows:
$$\left[\matrix{{1\over n}&{1\over n}&\ldots&{1\over n}\cr
-{1\over n}&1-{1\over n}&\ldots&-{1\over n}\cr
\vdots\cr
-{1\over n}&-{1\over n}&\ldots&1-{1\over n}\cr}\right]\ .$$
If you add the first row to all the other rows, which does not change the determinant, you are left with the matrix
$$\left[\matrix{{1\over n}&{1\over n}&\ldots&{1\over n}\cr
0&1&\ldots&0\cr
\vdots\cr
0&0&\ldots&1\cr}\right]\ ,$$
whose determinant is obviously ${1\over n}$.
A: For the purpose of determinant computation one can drop subtraction of $\partial_{x_i} \bar{x}$ for $i>1$. Then you are to compute determinant of an identity matrix with first row replaced with constant row of $\frac{1}{n}$. It's determinant is trivially $\frac{1}{n}$ then.
The reason why $\bar{x}$ can be dropped from $y_i$, $i>2$ is that doing so you effectively would subtract from each row below the first the value of the first row, which does not change the determinant.
