MacMahon's master theorem

Revision en1, by adamant, 2023-12-04 17:18:03

Hi everyone!

Mandatory orz to Elegia whose blog introduced me to these concepts.

Today, I would like to write about MacMahon's master theorem (MMT). It is a nice result on the intersection of combinatorics and linear algebra that provides an easy prove to some particularly obscure combinatorial identities, in particular the Dixon's identity:

$$$ \sum\limits_k (-1)^k \binom{a+b}{a+k} \binom{b+c}{b+k} \binom{c+a}{c+k} = \binom{a+b+c}{a,b,c}. $$$

To begin with, let's formulate MacMahon's master theorem.


MMT. Let $$$\mathbf A = [a_{ij}]_{n\times n}$$$, $$$\mathbf X = \operatorname{diag}(x_1,\dots,x_n)$$$ and $$$\mathbf k = (k_1,\dots,k_n) \geq 0$$$. Then,

$$$ \boxed{[\mathbf t^\mathbf k] \prod\limits_{i=1}^n \left(\sum\limits_{j=1}^n a_{ij} t_j\right)^{k_i} = [\mathbf x^\mathbf k] \det(\mathbf I-\mathbf X \mathbf A)^{-1}} $$$

With MMT, the Dixon's identity is proven as elegantly as this:

Proof

Now, let's look into MMT and try to understand what does it actually stand for? First of all, we can multiply both sides by $$$\mathbf x^\mathbf k$$$ and sum them up over all $$$\mathbf k \geq 0$$$. In this case, the RHS, by definition, will turn into $$$\det(\mathbf I - \mathbf X \mathbf A)^{-1}$$$, while the LHS will turn into

$$$ \sum\limits_{\mathbf k \geq 0} [\mathbf t^\mathbf k] \prod\limits_{i=1}^n\left(\sum\limits_{j=1}^n a_{ij} x_i t_j\right)^{k_i}. $$$

We can now do the substitution $$$\mathbf B = \mathbf X \mathbf A$$$ and use $$$b_{ij} = a_{ij} x_i$$$ to reformulate MMT in the equivalent form

$$$ \boxed{\sum\limits_{\mathbf k \geq 0} [\mathbf t^\mathbf k] \prod\limits_{i=1}^n \left(\sum\limits_{j=1}^n b_{ij} t_j\right)^{k_i} = \det(\mathbf I - \mathbf B)^{-1}} $$$

This form is more convenient for us to work with, as it doesn't depend on any variables in RHS. Now, let's take a closer look on the individual summands of the LHS. What do they enumerate?

$$$ [\mathbf t^\mathbf k]\prod\limits_{i=1}^n \left(\sum\limits_{j=1}^n b_{ij} t_j\right)^{k_i} = \sum\limits_{(j_1,\dots,j_k)} \prod\limits_{s=1}^k b_{i_s j_s}, $$$

where $$$(i_1,\dots,i_k)$$$ is a tuple such that $$$1 \leq i_1 \leq \dots \leq i_k \leq n$$$ and each $$$i$$$ occurs exactly $$$k_i$$$ times in it. Correspondingly, $$$(j_1,\dots,j_k)$$$ are all possible rearrangements of such tuple. Wait a second. We go over all rearrangements of a certain set of indices and sum up the products of $$$b_{i_s j_s}$$$ over them. Doesn't it sound familiar?

If we consider all permutations of $$$(1,\dots,k)$$$ rather than direct rearrangements of $$$(i_1,\dots,i_k)$$$, this will just be a permanent of a $$$(k_1+\dots+k_n) \times (k_1+\dots+k_n)$$$ matrix, in which the element $$$b_{ij}$$$ occurs as a block of size $$$k_i \times k_j$$$.

This concept usually occurs in literature defined even for two distinct vectors $$$\mathbf k \geq 0$$$ and $$$\mathbf l \geq 0$$$, so that $$$b_{ij}$$$ occurs in a block of size $$$k_i \times l_j$$$. The resulting matrix is typically denoted as $$$\mathbf B^{(\mathbf k, \mathbf l)}$$$. So, e.g. for $$$n = 3$$$, $$$\mathbf k = (3, 2, 1)$$$ and $$$\mathbf l = (1, 2, 3)$$$, $$$\mathbf B^{(\mathbf k, \mathbf l)}$$$ would look like

$$$ \mathbf B = \begin{pmatrix} b_{11} & b_{12} & b_{13} \\ b_{21} & b_{22} & b_{23} \\ b_{31} & b_{32} & b_{33} \end{pmatrix} \mapsto \left( \begin{array}{c|c|c} \begin{matrix} b_{11} \\ b_{11} \\ b_{11} \end{matrix} & \begin{matrix} b_{12} & b_{12} \\ b_{12} & b_{12} \\ b_{12} & b_{12} \end{matrix} & \begin{matrix} b_{13} & b_{13} & b_{13} \\ b_{13} & b_{13} & b_{13} \\ b_{13} & b_{33} & b_{33} \end{matrix} \\ \hline \begin{matrix} b_{21} \\ b_{21} \end{matrix} & \begin{matrix} b_{22} & b_{22} \\ b_{22} & b_{22}\end{matrix} & \begin{matrix} b_{23} & b_{23} & b_{23} \\ b_{23} & b_{23} & b_{23} \end{matrix} \\ \hline \begin{matrix}b_{31}\end{matrix} & \begin{matrix} b_{32} & b_{32} \end{matrix} & \begin{matrix} b_{33} & b_{33} & b_{33} \end{matrix} \end{array} \right) = \mathbf B^{(\mathbf k, \mathbf l)}. $$$

Note that to go back to rearrangements of $$$(i_1,\dots,i_k)$$$ rather than permutations of $$$(1,\dots,k)$$$ we should divide the permanent by $$$k_1 ! \dots k_n!$$$, which is commonly denoted in literature as $$$\mathbf k!$$$. In this terms, MMT rewrites concisely as

$$$ \boxed{\sum\limits_{\mathbf k \geq 0} \frac{\operatorname{per}\mathbf B^{(\mathbf k, \mathbf k)}}{\mathbf k!} = \det(\mathbf I - \mathbf B)^{-1}} $$$

Or, in terms of the original matrix $$$\mathbf A$$$, as

$$$ \boxed{\sum\limits_{\mathbf k \geq 0} \operatorname{per}\mathbf A^{(\mathbf k, \mathbf k)} \frac{\mathbf x^\mathbf k}{\mathbf k!} = \det(\mathbf I - \mathbf X\mathbf A)^{-1}} $$$
Tags combinatorics, generating functions, linear algebra, tensor

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en8 English adamant 2023-12-05 01:16:04 548
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