Carleman matrix

Remarks

Example: for $f(x) = ax$ with $a \in K$, one has $f(x)^j = a^j x^j$, so $M[f]$ is diagonal with entries $1, a, a^2, \ldots$. For $f(x) = x$ it is the identity matrix.

Non-example: the diagonal matrix with entries $1, 2, 3, \ldots$ is not a Carleman matrix. Its row $1$ would force $f(x) = 2x$, and then row $2$ would have to list the coefficients of $4x^2$, not of $3x^2$.

If $g(0) = 0$, then $M[f \circ g] = M[f] \, M[g]$, a product of infinite matrices in which each entry is a finite sum. Check: $f(g(x))^j = \sum_k M[f]_{jk} \, g(x)^k = \sum_k M[f]_{jk} \sum_l M[g]_{kl} x^l$, and $M[g]_{kl} = 0$ for $k > l$ because $g(x)^k$ starts at $x^k$. In particular, if $f(0) = 0$, then $M[f^n] = M[f]^n$ for the iterates of $f$, and $M[f]$ is upper triangular with diagonal entries $f_1^j$. The transpose of $M[f]$ is the Bell matrix of $f$, which reverses the order of products.