# Thread: Jacobian of a function

1. ## Jacobian of a function

Show that the following transformation doesn't change volumes.
$\begin{cases} x_1=u_1 \\ x_2=u_1+u_2 \\ ... \\ x_n=u_1+...+u_n \end{cases}$.
My attempt : I believe I must show that the Jacobian of the transformation is worth $1$. I'm having a hard time finding the Jacobian though. I've no clue.
$f(x_i)=\sum _{j=1}^{i} u_i$.

$\begin{bmatrix} \frac{\partial f(x_1)}{\partial u_1}, \frac{\partial f(x_2)}{\partial u_1}, ... , \frac{\partial f(x_n)}{\partial u_1} \\ \frac{\partial f(x_1)}{\partial u_2}, \frac{\partial f(x_2)}{\partial u_2}, ... , \frac{\partial f(x_n)}{\partial u_2} \\ ......................... \\ \frac{\partial f(x_1)}{\partial u_n}, \frac{\partial f(x_2)}{\partial u_n}, ... , \frac{\partial f(x_n)}{\partial u_n} \end{bmatrix}$
I know it's wrong, but it's my attempt.

2. Hi,

But that's the way !

However, you can note that the Jacobian matrix is just :

$\begin{pmatrix}
1 & 1 & 1 & \ldots & 1 \\
0 & 1 & 1 & \ldots & 1 \\
0 & 0 & \ddots & \ddots & \vdots \\
\vdots & \vdots & \vdots & \ddots & 1\\
0 & 0 & 0 & \ldots & 1
\end{pmatrix}$

Since it's an upper triangular matrix, its determinant is just the product of the terms that are in the diagonal, namely 1.

3. I think I made a mistake with the Jacobian matrix, it should read the transpose of the matrix I put. The determinant won't change though.