The article https://www.sciencedirect.com/science/article/pii/002437959190321M defines "Cauchy-Toeplitz matrices" those matrices that are simultaneously Cauchy matrices and Toeplitz ones. Then the author writes that:
It is easy to check that an arbitrary Cauchy-Toeplitz matrix of order n is of the form \begin{equation} T_n = \left( \frac{1}{g +(i-j)h} \right)_{i,j=1,\ldots,n},\label{eq}\end{equation} where g and h are some numbers. We assume that $h\ne0$ and the quotient $g/h$ is not integer.
I agree that a matrix of the above form is both a Cauchy matrix and a Toeplitz matrix. However, I have not succeeded in proving that a matrix that is both a Cauchy matrix and a Toeplitz matrix must necessarily have the above form.
Can anyone help me?
The Wikipedia article Cauchy-matrix states
If the matrix is Toeplitz, then $\,a_{i,j} = b_{i-j}\,$ for some sequence $\,\{b_i\}.\,$ In particular, this implies that
$$ a_{i,i} = \frac1{x_i-y_i} = b_0,\quad g := \frac1{b_0}, \quad a_{i+1,i} = \frac1{x_{i+1}-y_i} = b_1. $$
Use these two equations to get
$$ \frac1{b_1} - g = x_{i+1}-x_i, \quad x_i-y_i = g, \quad h := x_1-g. $$
The first differences of the sequence $\,\{x_i\}\,$ is constant and thus is an arithmetic progression. Furthermore, $$ y_i = x_i - g, \quad a_{i,j} = \frac1{(g+h\,i)-(g+h\,j-g)} = \frac1{g +h(i-j)}. $$
Conversely, suppose given a non-constant arithmetic progression $\,\{x_i\}\,$ and a non-zero constant $\,c.\,$ Define $\, y_i := x_i - c\,$ and $\,b_i := 1/(c + (x_1-x_0)i).\,$ The Toeplitz matrix generated from the sequence $\,\{b_i\}\,$ is also a Cauchy matrix.