If I have a set of linear equations
$$ \mathbf{U}x = r \\ \text{where}\, \mathbf{U} \, \text{is a positive upper triangular matrix and} \, r \, \text{ is a vector with positive elements in descending order} $$
Is it possible to write $x$ as product of a positive diagonal matrix and a vector of adjacent difference ? i.e. $x = diag(y) adj(r)$ ? where $y$ is a vector of positive elements and $adj$ is the adjacent difference operator.
NB: adjacent difference operator simply returns the a vector with difference between adjacent vector elements and last entry padded with 0. For example if $r = [7,3,1]$, then $adj(r) = [3-7,1-3,0]$.
Since the mapping $r\mapsto adj(r)$ is clearly not injective, there cannot exist a single $y$ such that $$Ux=r\implies x=diag(y)adj(r)$$
What's more, whenever the bottom right element of $U$, $u_{nn}$ and the final element of $r$, $r_{n}$ are both zero, there exists no such $y$ for which $diag(y)adj(r)=x$, because the final element of $diag(y)adj(r)$ is $0$, while the final element of $x$ is equal to $x_n = \frac{r_n}{u_{nn}}$.