# Smith normal form of a multivariate matrix associated with partitions

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## Abstract

Consideration of a question of E. R. Berlekamp led Carlitz, Roselle, and Scoville to give a combinatorial interpretation of the entries of certain matrices of determinant 1 in terms of lattice paths. Here we generalize this result by refining the matrix entries to be multivariate polynomials, and by determining not only the determinant but also the Smith normal form of these matrices. A priori the Smith form need not exist but its existence follows from the explicit computation. It will be more convenient for us to state our results in terms of partitions rather than lattice paths.

## Keywords

Lattice paths Partitions Determinants Smith normal form \(q\)-Catalan numbers.## AMS Classification

05A15 05A17 05A30Berlekamp [1, 2] raised a question concerning the entries of certain matrices of determinant 1. (Originally, Berlekamp was interested only in the entries modulo 2.) Carlitz, Roselle, and Scoville [3] gave a combinatorial interpretation of the entries (over the integers, not just modulo 2) in terms of lattice paths. Here we will generalize the result of Carlitz, Roselle, and Scoville in two ways: (a) we will refine the matrix entries so that they are multivariate polynomials, and (b) we compute not just the determinant of these matrices, but more strongly their Smith normal form (SNF). A priori our matrices need not have a Smith normal form, since they are not defined over a principal ideal domain, but the existence of SNF will follow from its explicit computation. A special case is a determinant of \(q\)-Catalan numbers. It will be more convenient for us to state our results in terms of partitions rather than lattice paths.

*extended partition*\(\lambda ^*\). Let \((r,s)\) denote the square in the \(r\)th row and \(s\)th column of \(\lambda ^*\). If \((r,s)\in \lambda ^*\), then let \(\lambda (r,s)\) be the partition whose diagram consists of all squares \((u,v)\) of \(\lambda \) satisfying \(u\ge r\) and \(v\ge s\). Thus, \(\lambda (1,1)=\lambda \), while \(\lambda (r,s)=\emptyset \) (the empty partition) if \((r,s)\in \lambda ^*{\setminus }\lambda \). Associate with the square \((i,j)\) of \(\lambda \) an indeterminate \(x_{ij}\). Now for each square \((r,s)\) of \(\lambda ^*\), associate a polynomial \(P_{rs}\) in the variables \(x_{ij}\), defined as follows:

*rank*of \(\lambda (i,j)\) (the number of squares on the main diagonal, or equivalently, the largest \(k\) for which \(\lambda (i,j)_k\ge k\)), then \(m=\rho _{ij}+1\). Let \(M(i,j)\) denote the matrix obtained by inserting in each square \((r,s)\) of \(S(i,j)\) the polynomial \(P_{rs}\). For instance, for the partition \(\lambda =(3,2)\) of Fig. 1, the matrix \(M(1,1)\) is given by

*Smith normal form*(SNF) over \(R\) if there exist matrices \(P\in {{\mathrm{GL}}}(m,R)\) (the set of \(m\times m\) matrices over \(R\) which have an inverse whose entries also lie in \(R\), so that \(\det P\) is a unit in \(R\)), \(Q\in {{\mathrm{GL}}}(n,R)\), such that \(PMQ\) has the form (w.l.o.g., here \(m\le n\), the other case is dual)

### **Theorem 1**

We can now state the recurrence relation for \(P_{rs}\).

### **Theorem 2**

### *Proof of Theorem 2*

When \(j=1\), the Inclusion–Exclusion process works exactly as before, except that the term \(A_{11}\) is never canceled from \(\tau _0 P_{11} = P_{11}\). Hence the theorem is also true for \(j=1\). \(\square \)

With this result at hand, we can now embark on the proof of Theorem 1. This is done by induction on \(\rho \), the result being trivial for \(\rho =0\) (so \(\lambda =\emptyset \)). Assume the assertion holds for partitions of rank less than \(\rho \), and let rank\((\lambda )=\rho \). For each \(1\le i\le \rho \), multiply row \(i+1\) of \(M(1,1)\) by \((-1)^i\tau _i\) and add it to the first row. By Theorem 2 we get a matrix \(M'\) whose first row is \([A_{11},0,0,\dots ,0]\). Now by symmetry we can perform the analogous operations on the *columns* of \(M'\). We then get a matrix in the block diagonal form \(\left[ \begin{array}{c@{\quad }c} A_{11} &{} 0\\ 0 &{} M(2,2) \end{array} \right] \). The row and column operations that we have performed are equivalent to computing \(P'MQ'\) for upper and lower unitriangular matrices \(P',Q'\in {{\mathrm{SL}}}(\rho +1,R)\), respectively. The proof now follows by induction. \(\square \)

*Note* The determinant above can also easily be evaluated by the Lindström–Wilf–Gessel–Viennot method of nonintersecting lattice paths, but it seems impossible to extend this method to a computation of SNF.

We now come to the second approach toward the SNF which does not use Theorem 2. Indeed, we will prove the more general version below where the weight matrix is not necessarily square; while this is not expounded here, the previous proof may also be extended easily to any rectangular subarray (regarded as a matrix) of \(\lambda ^*\) whose lower right-hand corner lies in \(\lambda ^*{\setminus }\lambda \). The inductive proof below will not involve Inclusion–Exclusion arguments; again, suitable transformation matrices are computed explicitly stepwise along the way.

### **Theorem 3**

### *Proof*

We note that the claim clearly holds for \(1\times e\) rectangles as then \(A_{1,e}=1=P_{1,e}\). We use induction on the size of \(\lambda \). For \(\lambda =\emptyset \), \(\lambda ^*=(1)\) and \(F\) can only be a \(1\times 1\) rectangle. We now assume that the result holds for all rectangles as above in partitions of \(n\). We take a partition \(\lambda '\) of \(n+1\) and a rectangle \(F'\) with its corner on the rim \(\lambda '^*{\setminus }\lambda '\), where we may assume that \(F'\) has at least two rows, we will produce the required transformation matrices inductively along the way.

First we assume that we can remove a square \(s=(a,b)\) from \(\lambda '\) and obtain a partition \(\lambda =\lambda '{\setminus }{s}\) such that \(F'\subseteq \lambda ^* \). By induction, we thus have the result for \(\lambda \) and \(F=F'\subset \lambda ^*\), say this is a rectangle with corner \((d,e)\), \(d\le e\). Then, \(a<d\) and \(b\ge e\), or \(a\ge d\) and \(b<e\). We discuss the first case, the other case is analogous. Set \(z=x_{ab}\).

## Notes

### Acknowledgments

R. P. Stanley’s contribution is based upon work supported by the National Science Foundation under Grant No. DMS-1068625.

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