Deleting duplicate columns from an infinite matrix: can we always finish? Consider a countably infinite matrix $M$ whose entries are drawn from a finite set $S$. Let $C$ be a set of representatives of the set of equivalence classes under entrywise equality of columns of $M$. Let $M'$ be the matrix produced by deleting column $i$ and row $i$ from $M$ whenever $\vec{m}_i \notin C$. Call $M'$ a "reduct" of $M$. Call a matrix $A$ "reduced" if $A$ is a reduct of itself. Let $M_1,M_2,...$ be a sequence of matrices where $M_1 = M$ and $M_{n+1}$ is a reduct of $M_n$.
Question: Does the sequence eventually reach a reduced matrix?
The idea is that we want to delete all of the duplicate columns from $M$, and their corresponding rows. But a reduct $M'$ of $M$ might not be free of duplicates (reduced), if two columns in $M'$ differed in $M$ only in rows that we deleted. If the matrix is finite we always get a reduced matrix eventually. But if it's infinite we might not as far as I can tell, and I'm wondering if anybody can produce an example where this happens.
 A: Let $M_{ij} = 0$ for $j<i+2$. Other entries should be $1$ on rows which are square numbers and $2$ otherwise. In other words the top right region is filled with rows of $1$ and $2$, but offset so the first two columns are both zero.
$$\left( \begin{array}{ccc}
0 & 0 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 &... \\
0 & 0 & 0 & 2 & 2 & 2 & 2 & 2 & 2 & 2 & 2 &\\
0 & 0 & 0 & 0 & 2 & 2 & 2 & 2 & 2 & 2 & 2 &\\
0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 & 1 & 1 &\\
0 & 0 & 0 & 0 & 0 & 0 & 2 & 2 & 2 & 2 & 2 &\\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 2 & 2 & 2 & 2 &\\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 2 & 2 & 2 &\\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 2 & 2 &\\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 &\\
...&&&&&&&&&&&...
\end{array} \right)$$
Only two columns are the same, $1$ and $2$, and whichever we choose removing the corresponding row and column leaves us with a new matrix where once again only two columns are the same. Each time we remove a row a new pair of columns will match. This is most obvious if we keep choosing to remove the first row and column. The alternating 2 1 pattern prevents the matrix becoming a Reduct of itself.
