I'm currently trying to get familiar with the Jordan normal form for matrices; and after some example I ask for the possible Jordan-form for the Carleman matrix for the function $f(x) = \sin(x)$ when thought as of infinite size.
The form, as used in wolfram-alpha is always $A= M \cdot J \cdot M^{-1} $ where $J$ has the near-diagonal Jordan form as an upper triangular matrix.
Well, it's not difficult to extrapolate this from small to big, even to unobservable sizes and possibly to infinite size - for instance for the Pascal-matrix, say with size $n \times n$ with $n=6$ $$ P_6= \small \begin{bmatrix} 1 & 1 & 1 & 1 & 1 & 1 \\ . & 1 & 2 & 3 & 4 & 5 \\ . & . & 1 & 3 & 6 & 10 \\ . & . & . & 1 & 4 & 10 \\ . & . & . & . & 1 & 5 \\ . & . & . & . & . & 1 \end{bmatrix} $$ we get (besides $M$ and $M^{-1}$ which are factorially rescaled versions of the matrices of Stirling numbers of first and second kind) the shape of the Jordan-matrix: $$ J_6 = \small \begin{bmatrix} 1 & 1 & . & . & . & . \\ . & 1 & 1 & . & . & . \\ . & . & 1 & 1 & . & . \\ . & . & . & 1 & 1 & . \\ . & . & . & . & 1 & 1 \\ . & . & . & . & . & 1 \end{bmatrix} $$ which extends consistently to larger sizes, and could so be conjectured to be valid also for the case where infinite size is assumed. (The Pascalmatrix $P$ is here the Carleman-matrix for the function $g(x) = 1+x $ in the sense that with a "Vandermonde-vector" of the form $V_n(x)=[1,x,x^2,x^3,...,x^{n-1}]$ we have for every size $n$ $$ V_n(x) \cdot P_n = V_n(1+x) $$
The same question using the truncated Carleman-matrices $M_n$ for $f(x)=\sin(x)$ which look like
$$ M_n= \Tiny {\begin{bmatrix}
1 & . & . & . & . & . & . & . \\
0 & 1 & . & . & . & . & . & . \\
0 & 0 & 1 & . & . & . & . & . \\
0 & -1/6 & 0 & 1 & . & . & . & . \\
0 & 0 & -1/3 & 0 & 1 & . & . & . \\
0 & 1/120 & 0 & -1/2 & 0 & 1 & . & . \\
0 & 0 & 2/45 & 0 & -2/3 & 0 & 1 & . \\
0 & -1/5040 & 0 & 13/120 & 0 & -5/6 & 0 & 1
\end{bmatrix}}$$
such that for $\lim_{n\to \infty} $ we have $$ \lim_{n\to \infty} V_n(x) \cdot M_n = V_n(\sin(x)) $$
is difficult: it gives for every $n$ a Jordan normal form with three blocks, one of them always of size $1$ and the two others of increasing sizes with increasing $n$, here the example owith $n=12$
$$ J_{n} = \Tiny{ \begin{array} {rrrrrr|rrrrr|r}
1 & . & . & . & . & . & . & . & . & . & . & . \\
1 & 1 & . & . & . & . & . & . & . & . & . & . \\
. & 1 & 1 & . & . & . & . & . & . & . & . & . \\
. & . & 1 & 1 & . & . & . & . & . & . & . & . \\
. & . & . & 1 & 1 & . & . & . & . & . & . & . \\
. & . & . & . & 1 & 1 & . & . & . & . & . & . \\
\hline \\
. & . & . & . & . & . & 1 & . & . & . & . & . \\
. & . & . & . & . & . & 1 & 1 & . & . & . & . \\
. & . & . & . & . & . & . & 1 & 1 & . & . & . \\
. & . & . & . & . & . & . & . & 1 & 1 & . & . \\
. & . & . & . & . & . & . & . & . & 1 & 1 & . \\
\hline \\
. & . & . & . & . & . & . & . & . & . & . & 1
\end{array} }
$$
and trivially it does not make sense to try to speak of two infinite-size Jordan blocks and to try to do a formula with it.
For my application I'm still confident, that -so far- the matrix-logarithm provides a reliable concept for powers of that Carleman-matrix even when thought as extended to the infinite size, but after the first successful contacts with the Jordan-form for simpler matrices (as for instance the matrix of Stirling numbers 1st and 2nd kind) this is kind of disappointing at the moment...
The view into the caracteristic polynomial seems to be of no help; I get for every size $n$ the formula $(x-1)^n$ which is also the same for the Pascal- and for the Stirling-matrices.
I've two questions:
Q1: How to explain the different Jordan forms although the characteristic polynomials are the same?
(That is something with the arithmetic and geometric multiplicity of eigenvalues, but I've not yet understood that concept and how to work with it in this cases)
Q2: Could a generalized Jordan normal form for the infinite size be given for the $f(x)=\sin(x)$ - Carlemanmatrix, too? And how should that look like?