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I am working on a undergrad research project with some other guys. Now the conjecture (unrelated to this question) we are trying to prove boils down to a final subproblem:

Let $A (n \times n)$ be a lower triangular Toeplitz matrix (that is, a Toeplitz matrix without the upper triangular part). The Smith invariant factors of $A([k-1,\dots,n],[1,\dots,k-1])$ are also Smith invariant factors of $A([k,\dots,n],[1,\dots,k])$ including multiplicities, where $A([i_1,\dots,i_2],[j_1,\dots,j_2])$ denotes the submatrix of A formed by $i_1$ to $i_2$th rows and $j_1$ to $j_2$th columns and $k\leq [n/2]$; or equivalently, the gcd of all m by m minors of $A([k-1,\dots,n],[1,\dots,k-1])$ equals the gcd of all m by m minors of $A([k,\dots,n],[1,\dots,k])$. $(m\leq k-1)$

Since Toeplitz matrix is rather special and this property (though only confirmed by maple) seems quite nice, I am wondering if anyone else has seen this or something similar before?

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  • $\begingroup$ Or any ideas are welcome $\endgroup$
    – Epsilon
    Commented Jul 20, 2010 at 22:11
  • $\begingroup$ I took the liberty of "sprucing up" some of the formatting in your post; if this is undesirable, please feel free to change it back to the original. $\endgroup$
    – Yemon Choi
    Commented Jul 20, 2010 at 22:36

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Some friends (including the original poster!) and I wrote up the proof of the result in this paper. The proof involves symmetric function theory.

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