Why does the $k$-th invariant factor of some matrix depend only on the first $k$ columns

Question

Let $A$ be an $n\times n$ integral matrix and $\xi$ an integral vector of dimension $n$. Let $$W_i=[\xi,A\xi,\ldots,A^{i-1}\xi]$$ for each positive $i$. Let $d_{i,j}$ (where $1\le j\le i$) denote the $j$-th invariant factor of $W_i$. It seems that $d_{i,j}$ only depends on $j$, that is $d_{j,j}=d_{j+1,j}=\cdots=d_{n,j}=d_{n+1,j}=\cdots$. By Cayley-Hamilton Theorem, we know that $W_i$ is equivalent (in $\mathbb{Z}$) to $[W_n,0_{n\times (i-n)}]$ for $i>n$. Thus we only need to consider the case that $i\le n$.

For example, let \begin{equation*} A=\left(\begin{matrix} 2&2&3&3\\ 1&3&0&0\\ 3&3&0&3\\ 0&0&3&2\\ \end{matrix} \right),\quad \xi=\left(\begin{matrix} 1\\ 2\\ 3\\ 0\\ \end{matrix} \right) \end{equation*} Then \begin{equation*} W_1=\left(\begin{matrix} 1\\ 2\\ 3\\ 0\\ \end{matrix} \right), W_2=\left(\begin{matrix} 1&15\\ 2&7\\ 3&9\\ 0&9\\ \end{matrix} \right),W_3=\left(\begin{matrix} 1&15&98\\ 2&7&36\\ 3&9&93\\ 0&9&45\\ \end{matrix} \right), W_4=\left(\begin{matrix} 1&15&98&682\\ 2&7&36&206\\ 3&9&93&537\\ 0&9&45&369\\ \end{matrix} \right) \end{equation*} The Smith normal forms of these matrices are \begin{equation*} \left(\begin{matrix} 1\\ 0\\ 0\\ 0\\ \end{matrix} \right), \left(\begin{matrix} 1&0\\ 0&1\\ 0&0\\ 0&0\\ \end{matrix} \right),\left(\begin{matrix} 1&0&0\\ 0&1&0\\ 0&0&3\\ 0&0&0\\ \end{matrix} \right), \left(\begin{matrix} 1&0&0&0\\ 0&1&0&0\\ 0&0&3&0\\ 0&0&0&9090\\ \end{matrix} \right) \end{equation*} For a fixed $j$, why these $j$-th invariant factors of $W_i$'s are all the same?

Answer 1

You can find the Smith normal form of $W_i$ using Gaussian Elimination. In this way, if you do this, let us say, to $W_4$, for instance, you start eliminating from the first columns, so you are doing a Gaussian elimination to $W_j$, $j<4$.

Please see Algebraically-nice general solution for last step of Gaussian elimination to Smith Normal Form? and Computing the Smith Normal Form on Math.StackExchange, and find more searching for "\(A\) Smith Normal Form" on SearchOnMath.