This lower triangular matrix ( with entries $\pm 1$) \begin{eqnarray*} \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 \\ -1 & 1 & 0 & 0 & 0 & 0 \\ 1 & -1 & 1 & 0 & 0 & 0 \\-1 & 1 & -1 & 1 & 0 & 0 \\1 & -1 & 1 & -1 & 1 &0 \\ -1 & 1 & -1 & 1 & -1 & 1 \end{pmatrix} \end{eqnarray*}
has some interesting properties. The main diagonal is all $1$ and the other diagonals are signed but constant. I wish to know if these kind of matrices are studied in the literature. Is there a name for such matrices? What is shown here is a $6\times 6$ version, but this generalizes to any square dimension.
For starters, this is the inverse of a constant band matrix. That is.,
\begin{eqnarray*} \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 \\1 & 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 & 0 & 0 \\0 & 0 & 1 & 1 & 0 & 0 \\0 & 0 & 0 & 1 & 1 & 0 \\0 & 0 & 0 & 0 & 1 & 1\end{pmatrix}^{-1} &=& \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 \\ -1 & 1 & 0 & 0 & 0 & 0 \\ 1 & -1 & 1 & 0 & 0 & 0 \\ -1 & 1 & -1 & 1 & 0 & 0 \\ 1 & -1 & 1 & -1 & 1 &0 \\ -1 & 1 & -1 & 1 & -1 & 1 \end{pmatrix} \end{eqnarray*}
I am aware that band matrices are well studied in the literature, but curious about work done on their inverses and their properties/structure etc., Walsh, Hadamard, Sierpinski, any such connections?
Specific questions that I am looking for are:
$1.$ Is there a name for such matrices?
$2.$ Is there any prior work done on such matrices, especially geometric or graph theoretic connections?
$3.$ Any known matrix related to these family of matrices?
Some special constant matrices have their own names, such as Hilbert matrix or Kac matrix, but names are far more often given not to such constant matrices, but to classes of matrices, such as Toplitz matrices, Vandermonde matrices, Cauchy matrices, Hankel matrices, Hessenberg matrices, Hadamard matrices, Hermitian matrices, circulant matrices, stochastic matrices, $M$-matrices, triangular matrices or diagonal matrices.
The matrix in your question is a constant matrix, not a class of matrices. It is just the inverse of a unipotent Jordan block. Maybe it has a name, but I don't think it is special enough to deserve one.