Let $\textbf A\in \mathbb Z^{n \times (n-1)}$ be a full-rank integer matrix and $\textbf w \in \mathbb Z^n$ be a vector such that $\textbf w^T \textbf A = \textbf 0$.
Now, consider the Smith normal form $\textbf A_{\sf smith} \in \mathbb Z^{n \times (n-1)}$ of $\textbf A$ and let $\alpha_1, \cdots, \alpha_{n - 1} \in \mathbb Z$ be the non-zero diagonal entries of $\textbf A_{\sf smith}$, where $\alpha_i | \alpha_{i + 1}$ for $\forall 1 \le i < n - 1$.
Question: Is there a way to upper bound the size of $\alpha_{n - 1}$ by using only $\textbf w$??
E.g., $\alpha_{n - 1} \le \|\textbf w\|_2 / d(\textbf w)$, where $d(\textbf w)$ is the greatest common divisor of all the entries in $\textbf w$.
The motivation is that, I have a matrix $\textbf A$ and a vector $\textbf w$ with the above property, and I only need to have the upper bound on the value of $\alpha_{n - 1}$ in order for the algorithm to work. So, in particular, it would be much more efficient if I didn't have to compute the Smith normal form.