A block Hankel generalized confluent Vandermonde matrix

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• 作者：André ; Klein^a ; ^{a.a.b.klein@uva.nl" class="auth_mail} ; Peter Spreij^b ; ^{spreij@uva.nl" class="auth_mail}
• 关键词：15A09 ; 15A23 ; 15A24 ; 15B99
• 刊名：Linear Algebra and its Applications
• 出版年：15 August 2014
• 年：2014
• 卷：455
• 期：Complete
• 页码：32-72
• 全文大小：470 K

文摘

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×q$q\times q$ say, in which case the i  -th column is given by u(z_i)$u(z_i)$, where we write u(z)=(1,z,…,z^q−1)^鈯?/sup>$u(z)={(1,z,\dots ,z^{q-1})}^{鈯?/mo>}$. If all the z_i$z_i$ (i=1,…,q$i=1,\dots ,q$) are different, the Vandermonde matrix is non-singular, otherwise not. The latter case obviously takes place when all z_i$z_i$ 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)$V(z)$ whose i  -th column (i=1,…,q$i=1,\dots ,q$) is given by the (i−1)$(i-1)$-th derivative u⁽ⁱ⁻¹⁾(z)^鈯?/sup>$u^{(i-1)}{(z)}^{鈯?/mo>}$.

We will consider generalizations of the confluent Vandermonde matrix V(z)$V(z)$ by considering matrices obtained by using as building blocks the matrices M(z)=u(z)w(z)$M(z)=u(z)w(z)$, with u(z)$u(z)$ as above and 22d3501d0e89a0f8ef68f2dda" title="Click to view the MathML source">w(z)=(1,z,…,z^r−1)$w(z)=(1,z,\dots ,z^{r-1})$, together with its derivatives M^(k)(z)$M^{(k)}(z)$. Specifically, we will look at matrices whose ij  -th block is given by M^(i+j)(z)$M^{(i+j)}(z)$, where the indices i,j$i,j$ 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)$M(z)$ and the number of derivatives M^(k)(z)$M^{(k)}(z)$ that is involved. The results are obtained by mostly elementary methods, no specific knowledge of the theory of matrix polynomials is needed.