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# parameterization of equitable matrices

A $n\times n$ matrix is *equitable* if and only if it can be expressed in the form

$m_{{ij}}=\exp(\lambda_{i}-\lambda_{j})$ |

for real numbers $\lambda_{1},\lambda_{2},\ldots,\lambda_{n}$ with $\lambda_{1}=0$.

Proof: Assume that $m_{{ij}}$ are the entries of an equitable matrix.

Since all the elements of an equitable matrix are positive by definition, we can write

$m_{{ij}}=\exp\mu_{{ij}}$ |

with the quantities $\mu_{{ij}}$ being real numbers (which may be positive, negative or zero).

In terms of this representation, the defining identity for an equitable matrix becomes

$\mu_{{ik}}=\mu_{{ij}}+\mu_{{jk}}$ |

Since this comprises a system of linear equations for the quantities $\mu_{{ij}}$, we could solve it using the usual methods of matrix theory. However, for this particular system of linear equations, there is a much simpler approach.

Consider the special case of the identity when $i=j=k$:

$\mu_{{ii}}=\mu_{{ii}}+\mu_{{ii}}.$ |

This simplifies to

$\mu_{{ii}}=0.$ |

Consider the case when $i=k$ (but does not equal $j$).

$\mu_{{ij}}+\mu_{{ji}}=\mu_{{ii}}$ |

By wat we have just shown, the right hand side of this equation equals zero. Hence, we have

$\mu_{{ij}}=-\mu_{{ji}}.$ |

In other words, the matrix of $\mu$’s is antisymmetric.

We may express any entry in terms of the $n$ entries $\mu_{{i1}}$:

$\mu_{{ij}}=\mu_{{i1}}+\mu_{{1j}}=\mu_{{i1}}-\mu_{{j1}}$ |

We will conclude by noting that if, given any $n$ numbers $\lambda_{i}$ with $\lambda_{1}=0$, but the remaining $\lambda$’s arbitrary, we define

$\mu_{{ij}}=\lambda_{i}-\lambda_{j},$ |

then

$\mu_{{ij}}+\mu_{{jk}}=\lambda_{i}-\lambda_{j}+\lambda_{j}-\lambda_{k}=\lambda_% {i}-\lambda_{k}=\mu_{{ik}}$ |

Hence, we obtain a solution of the equations

$\mu_{{ik}}=\mu_{{ij}}+\mu_{{jk}}.$ |

Moreover, by what we what we have seen, if we set $\lambda_{i}=\mu_{{i1}}$, all solutions of these equations can be so described.

Q.E.D.

## Mathematics Subject Classification

15-00*no label found*

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