1
$\begingroup$

I'm looking for any known results on a shift operator commutated by a Vandermonde matrix. That is, let

$$T=\begin{bmatrix}0 & 1 & 0 & 0 & \cdots \\ 0 & 0 & 1 & 0 & \cdots \\ 0 & 0 & 0 & 1 & \cdots \\ \vdots &\vdots & \vdots &\vdots &\ddots \end{bmatrix} $$ be the (infinite-dimensional) shift operator, and let

$$C=\begin{bmatrix}1 & c_0 & c_0^2 & \cdots \\ 1 & c_1 & c_1^2 & \cdots \\ 1 & c_2 & c_2^2 & \cdots \\ \vdots &\vdots &\vdots &\ddots \end{bmatrix} $$ be a Vandermonde matrix. Can anything be said about the product $$C^{-1}TC$$ (assuming, of course that $C$ is invertible)? (So, "of course", shift operators are one of the principle topics of study in ergodic theory and entire books have been written about them. $T$ is the basic "Bernoulli shift"; many things can be said about it. That is known. The similarity matrix is what confounds the present problem.)

The motivation is that it shows up in a simpler problem, which is to find a shift function $f(x)$ satisfying $f(c_k)=c_{k+1}$ and expressing it as an analytic series $f(x)=\sum_n a_n x^n$. This promptly leads to a matrix equation with the Vandermonde matrix:

$$\begin{bmatrix}1 & c_0 & c_0^2 & \cdots \\ 1 & c_1 & c_1^2 & \cdots \\ 1 & c_2 & c_2^2 & \cdots \\ \vdots &\vdots &\vdots &\ddots \end{bmatrix} \begin{bmatrix} a_0 \\ a_1 \\ a_2 \\ \vdots \end{bmatrix}= \begin{bmatrix} c_1 \\ c_2 \\ c_3 \\ \vdots \end{bmatrix} $$

Note that the column on the far right is a shift of the second column on the left.

The required solution just isolates one column: $$ \begin{bmatrix} a_0 \\ a_1 \\ a_2 \\ \vdots \end{bmatrix}= C^{-1}TC \begin{bmatrix} 0 \\ 1 \\ 0 \\ \vdots \end{bmatrix} $$ Because $C^{-1}TC$ looks so terribly ... generic... it seems like the right thing to ask questions about.

(My specific problem does have additional structure, but I don't really want to burden the question with the additional details. But, if it helps: my $c_k$ are the roots of certain (rather simple) polynomials, and so my particular Vandermonde matrix has a lot of regularity to it. The $c_k^n$ for $n$ greater than the order of the polynomial can be reduced. Polynomials being what they are, each row ends up being a chaotic iterate of that polynomial for that row. Each of polynomial has only one real root, and these appear in descending order. My particular polynomials are pseudo-cyclotomic; they're cyclotomic-like but a bit contorted (the complex roots almost lie in a circle, but not quite...) Because of this, I vaguely suspect there might be existing results on finite fields, where e.g. the polynomials really are cyclotomic ... although in my case I am working with the reals.)

$\endgroup$
2
  • $\begingroup$ Is your shift operator nilpotent or cyclic ? I mean, is there a zero or a one at last row first column ? $\endgroup$ Oct 23, 2020 at 2:55
  • 1
    $\begingroup$ I'ts infinite-dimensional. It's the Bernoulli shift. $\endgroup$
    – Linas
    Oct 23, 2020 at 2:57

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.