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Hi, I have given a matrix of the following form:

$M = \begin{pmatrix} a_0 & -1 & \cdots & \cdots & -1 \newline -1 & a_1 & -1 & \cdots & -1 \newline \vdots & & \ddots & & \vdots \newline \vdots & & & \ddots & \vdots \newline -1 & \cdots & -1 & \cdots & a_n \end{pmatrix}$ with $a_i \in \mathbb{Z}$, $a_i > 0$.

Is the a an easy way to write down the Smith Normalform of this matrix?

greatz Johannes

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the two by two case has a really nice SNF; the three-by-three version is already beginning to look not that nice. Perhaps the fact that $M+E$ (where $E=11^T$) is diagonal can be exploited... –  Suvrit Sep 30 '12 at 14:41
My guess is that the answer depends subtly on the gcd of the ai (or worse, subsets of the ai) try various cases where the ai are all distinct primes. –  Andy B Sep 30 '12 at 19:20
Also, when all the $a_i = n$ this becomes the computation of the critical group of the the complete graph. The SNF in this case is $(1,n,\dots,n,0)$. –  Andy B Sep 30 '12 at 19:22
I think the 3x3 should work nicely, because there is a formula for the determinant of a matrix of this kind. greatz –  Johannes Oct 4 '12 at 17:01
@Andy: the middle $n$'s should be $n+1$'s. It is easy direct calculation(see my answer below). But, I'd like to know the connection to the computation of the critical group of the the complete graph. –  i707107 Mar 5 '13 at 0:12

2 Answers 2

This is a partial answer for the special case when all $a_i$ are distinct. I will work on complex(or algebraic closure of $\mathbb{Q}$) for convenience. Writing $M-\lambda I = D_{\lambda}+E$ where $D_{\lambda}$ is the diagonal matrix with diagonal entries $a_0-\lambda+1, \cdots , a_n-\lambda+1$. With this expression, it is easy to calculate the determinant, which will give the characteristic polynomial of $M$.

If $f(\lambda)=(a_0-\lambda+1)\cdots (a_n-\lambda+1)$, then we have $$ \textrm{det}(M-\lambda I) = f(\lambda)+f'(\lambda).$$ Considering the identity $(f(t)e^t)'=(f(t)+f'(t))e^t$, we can find the roots of characteristic polynomial by looking at the critical points of $f(t)e^t$. Also, by Mean Value theorem, we know that the critical points of $f(t)e^t$ are all distinct.

Therefore, if we let $\lambda_0, \cdots, \lambda_n$ be the critical points of $f(t)e^t$, then we have the following Smith Normal form of the matrix $M-\lambda I$ over the polynomial ring $\mathbb{C}[\lambda]$, (or $\overline{\mathbb{Q}}[\lambda]$): $$ \textrm{Diag}(1,\cdots, 1, (\lambda-\lambda_0) \cdots (\lambda-\lambda_n)).$$ Hence, we obtain the Smith Normal form of $M$ in this case: $$\textrm{Diag}(1,\cdots, 1,\lambda_0 \cdots \lambda_n).$$ There is a natural way of bringing this down to $\mathbb{Z}$, then we have to deal with the irreducible factors of $f(t)+f'(t)$. This is indeed $$\textrm{Diag}(1,\cdots, 1, f(0)+f'(0)).$$ Added) This method also works for the case below:

The cardinality of $\{ j: a_i = a_j \}$ is at most $2$, for any $i=0,\cdots, n$.

Added2) This gives the SNF of $M$ over $\mathbb{Q}$. Over $\mathbb{Z}$ will be certainly more difficult.

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While I believe the full answer to your question is 'no,' I was pleasantly surprised that I can predict the first two diagonal entries of the SNF.

Permute the rows so that there is a 1 in entry $(1,1)$. Then after a first round of row and column operations, we produce a matrix:

$$ \left[ \begin{array}{c|c} 1 & \mathbf{0}^T \\\ \hline \mathbf{0} & M' \end{array}\right] $$

First observation: $1$ is the first diagonal entry.

Second observation: If every $a_i$ is equivalent to $-1$ modulo $k$ for some $k$, then $k$ divides the next diagonal entry of the SNF... and if $k$ is the largest such number, then it is the next. This is because $M'$ contains only entries such as $0$, $\pm(1 + a_i)$ or $1 - a_ia_j$.

Hope this helps!

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I would write that as $k=gcd(a_0+1,a_1+1,...,a_n+1)$. In the three-by-three case we know that the determinant is $a_0a_1a_2 - (a_0+a_1+a_2)-2$, so we can get the last entry. I suspect things get significantly more difficult already for $n=3$. –  Will Sawin Sep 30 '12 at 15:37

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