Abstract algebraic link between two problems involving polynomials and (generalized) Vandermonde matrices?

Question

Let two integers $L\leq N$, a sequence of $L+1$ distinct real numbers $\alpha_0,\dots,\alpha_L$, and integers $k_0,\dots,k_L\in\mathbb N$ such that $N=L+\sum_{i=0}^Lk_i$.

I noticed that the two following results, both in the field of polynomial interpolation, involve Vandermonde matrices, or their generalization known as confluent Vandermonde matrices (better explained here if you have access):

1. Generalized Hermite Interpolation solves the following linear system over polynomials $P\in\mathbb R_N[X]$ : $$\forall i=0\dots L,\quad \forall j=0\dots k_i,\quad \frac1{j!}P^{(j)}(\alpha_i)=a_i^{(j)}$$ for given target values $\{a_i^{(j)}\}_{i=0\dots L,\; j=0\dots k_i}$. When expressed in the canonical basis $(1,X,\dots,X^N)$, this system is represented by the confluent Vandermonde matrix $V(\alpha_0,\dots,\alpha_L,k_0,\dots,k_L)$.

2. When studying the theory of splines, I came across the following Proposition: the family of polynomials $$\Big\{\binom{N}{j} (X+\alpha_i)^{N-j}\Big\}_{i=0\dots L,\; j=0\dots k_i}$$ is always free (and thus, a basis of $\mathbb R_N[X]$). Indeed, when expressed in the basis $(X^N,\dots,\binom{N}{k}X^{N-k},\dots,1)$, this family is again represented by the confluent Vandermonde matrix $V(\alpha_0,\dots,\alpha_L,k_0,\dots,k_L).$

My question is : is there a deeper, algebraic (coordinate free) relation between the two results ? For example : could we prove the second proposition (show that the given family of polynomials is always free) simply by casting it into some form of generalized Hermite interpolation problem ? Or is the similarity between the two results "just by chance", simply because Vandermonde matrices are everywhere ?

NOTE: just to make explicit the easier instantiation of these problems: When $k_i=0$ for all $i$, we recover classic (Legendre) polynomial interpolation ($\forall i,P(\alpha_i)=a_i$), classic Vandermonde matrices, and the family involved in the second proposition is simply $\{(X+\alpha_i)^N\}_{i=0\dots N}$.

Answer 1

Well, after some more thinking I'm going to answer my own question. It was pretty much just a matter of linking all elements together.

Here are my notations, in $\mathbb R_N[X]$.

• Note $\partial$ the derivation operator : $(\partial P)(X)=P'(X)$.

• Note $\tau_z$ the translation operator : $(\tau_z P)(X) = P(X+z)$. Note that $\tau_z$ is a polynomial in $\partial$, owing to Taylor's formula ($\tau_z = \sum_{n=0}^N \frac{z^n}{n!} \partial^n$).

• Note $\delta_z$ the evaluation functional at $z$ : $\langle\delta_z,P\rangle=P(z)$.

• For $f\in{\cal L}(\mathbb R_N[X])$, note $f^*\in{\cal L}(\mathbb R_N[X]^*)$ its algebraic transpose.

• Note multiplicatively all linear applications and compositions as long as they remain within $\mathbb R_N[X]$ (resp. $\mathbb R_N[X]^*$). For example, for $P\in\mathbb R_N[X]$, $L\in\mathbb R_N[X]^*$, and $f,g\in{\cal L}(\mathbb R_N[X])$, I can write $\langle L,fgP\rangle=\langle g^*f^* L,P\rangle$.

We now note that the family of linear forms

$$ \Big( \delta_0,\; \partial^*\delta_0,\; \dots,\; (\partial^*)^{N-1}\delta_0,\; (\partial^*)^N\delta_0 \Big) $$ is a basis of the dual space $\mathbb R_N[X]^*$. (Indeed, it is the canonical dual basis, up to scaling by $n!$.)

Now let $Q$ be any fixed polynomial of degree exactly $N$. (My initial question is recovered when setting $Q=X^N$.) By a simple consideration on degrees, the family $$ \Big( Q,\; \partial Q,\; \dots,\; \partial^{N-1} Q,\; \partial^N Q \Big) $$ is a basis of $\mathbb R_N[X]$.

The two previous lines tell us that operator $\partial^*$ is cyclic starting from vector $\delta_0$, and operator $\partial$ is cyclic starting from vector $Q$. Hence, there exists a unique isomorphism $g:\mathbb R_N[X]\to \mathbb R_N[X]^*$ such that $$ g\circ\partial = \partial^*\circ g \quad{\rm and}\quad g(Q)=\delta_0 $$

Isomorphism $g$ is precisely the one I was looking for. First, since $\tau_z$ is a polynomial in $\partial$ (with Taylor's formula), it falls easily that $g\circ\tau_z = \tau_z^*\circ g$. Then, for every integer $k$ and real number $z$, we can write $$ g\big(Q^{(k)}(X+z)\big) = g(\tau_z\partial^k Q) = (\tau_z\partial^k)^*\circ g(Q) = (\partial^*)^k\tau_z^*\delta_0 = \delta_z^{(k)} $$ noting the linear form $\langle\delta_z^{(k)},P\rangle:=P^{(k)}(z)$. Thus, isomorphism $g$ establishes a correspondence between the two following families : $$ \left\{ Q^{(j)}(X+z_i)\right\}_{i=0\dots L, j=0\dots k_i} \xrightarrow{g}\left\{ \delta_{z_i}^{(j)} \right\}_{i=0\dots L, j=0\dots k_i} $$ which is exactly the link I was looking for.