A block Hankel generalized confluent Vandermonde matrix

Abstract: Vandermonde matrices are well known. They have a number of interesting properties and play a role in (Lagrange) interpolation problems, partial fraction expansions, and finding solutions to linear ordinary differential equations, to mention just a few applications. Usually, one takes these matrices square, $q\times q$ say, in which case the $i$-th column is given by $u(z_i)$, where we write $u(z)=(1,z,...,z^{q-1})^\top$. If all the $z_i$ ($i=1,...,q$) are different, the Vandermonde matrix is non-singular, otherwise not. The latter case obviously takes place when all $z_i$ are the same, $z$ say, in which case one could speak of a confluent Vandermonde matrix. Non-singularity is obtained if one considers the matrix $V(z)$ whose $i$-th column ($i=1,...,q$) is given by the $(i-1)$-th derivative $u^{(i-1)}(z)^\top$. We will consider generalizations of the confluent Vandermonde matrix $V(z)$ by considering matrices obtained by using as building blocks the matrices $M(z)=u(z)w(z)$, with $u(z)$ as above and $w(z)=(1,z,...,z^{r-1})$, together with its derivatives $M^{(k)}(z)$. Specifically, we will look at matrices whose $ij$-th block is given by $M^{(i+j)}(z)$, where the indices $i,j$ by convention have initial value zero. These in general non-square matrices exhibit a block-Hankel structure. We will answer a number of elementary questions for this matrix. What is the rank? What is the null-space? Can the latter be parametrized in a simple way? Does it depend on $z$? What are left or right inverses? It turns out that answers can be obtained by factorizing the matrix into a product of other matrix polynomials having a simple structure. The answers depend on the size of the matrix $M(z)$ and the number of derivatives $M^{(k)}(z)$ that is involved. The results are obtained by mostly elementary methods, no specific knowledge of the theory of matrix polynomials is needed.