I have an infinite matrix $A$ that is in row echelon form, except zeros are to the right of each pivot, as shown below. The $*$ symbol denotes any integer, though most of them are zero.
$A= \left( \begin{array}{ccccccc} 1&0&0&0&0&0&\dots \\ *&*&1&0&0&0&\dots \\ *&*&*&1&0&0&\dots \\ *&*&*&*&*&1&\dots \\ \vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\ddots \\ \end{array} \right)$
I want to calculate the column Hermite normal form of $A$. Without doing any calculations, it seems like an infinite number of column operations would yield
$H= \left( \begin{array}{ccccccc} 1&0&0&0&0&0&\dots \\ 0&0&1&0&0&0&\dots \\ 0&0&0&1&0&0&\dots \\ 0&0&0&0&0&1&\dots \\ \vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\ddots \\ \end{array} \right)$
Is that correct? Also, once I have the column Hermite normal form $H$, how can I use it to solve $A \mathbf{x}=\mathbf{b}$ with integer solutions for $\mathbf{x}$?