I've had this exercise and my approach was completely different from the official solution and I am not exactly following the official solutions proof. I understand how they get there, but not the conclusion. How does this show the transformation is also uniform random variable?
The exercise: Let z be a uniform D-Dimensional RV $u ∼ p_u(u) = Unif(0, 1)$. We consider a transformation $$ f(u) = x $$ Where $$ \mathbf{x} = \begin{bmatrix} x_{1:d} \\ x_{d+1:D}\\ \end{bmatrix} = \begin{bmatrix} u_{1:d} \\ u_{d+1:D} + A(u_{1:d})\\ \end{bmatrix} $$
The first part considers $A$ to be a linear transformation. So, $f^{-1}(x) = u$ $$\begin{align} & u_{1:d} = x_{1:d} \\ & u_{d+1:D} = x_{d+1:D} - Ax_{1:d} \end{align} $$
We go on to find the Jacobian of the inverse function above
$$ \begin{bmatrix}I_d & 0\\-A & I{D-d}\end{bmatrix} $$
We conclude that the determinant of the Jacobian = 1 for an arbitrary sample $\tilde{x}$. And $$ p_x({\tilde{x}}) = p_z({f^{-1}\tilde{x}})\cdot |det\frac{\partial f^{-1}(\tilde{x})}{\partial x} | = q(f^{-1}(\tilde{x})) = 1. $$
It says that we use the fact that $\tilde{z} = f^{-1}(x) \in [0,1]^D$. Therefore, $q(\tilde{z}) = 1$.
The next part proving that for arbitrarily complex functions $A$, $x$ is still distributed uniformly because the Jacobian determinant doesn't change.
You help in understanding this is much appreciated. If someone can format the matrix and vectors correctly I am grateful as well.