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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?

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    $\begingroup$ Q1 as stated is a little confusing: Why do different JNF (with the same characteristic polynomial) need an "explanation"? $\endgroup$ – Christian Remling Jun 7 '14 at 18:59
  • $\begingroup$ @ChristianRemling : When I asked this I was relatively new to the method of Jordan decomposition and the question "how to explain" should have better been asked as "how to understand". After much reading and exercising I think I've understood this now. $\endgroup$ – Gottfried Helms Oct 4 '14 at 7:57
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[update] I've solved this now using the shown idea for the ordering of the J-matrix. For the first three eigenvectors (associated to the first three eigenvalues) I've then the freedom of norming, all other eigenvectors can be successively be computed. Also I changed the orientation of the matrix to meet my own conventions (transposed display)


The idea is simply to permute rows and columns in the finite case, to arrive at a continuously extensible pattern and then try to show, that this pattern makes sense for the infinite case.

It is possible, simply to shift the non-diagonal entries to the next subdiagonal, such to get $$ J^*_8=\small \begin{bmatrix} 1 & . & . & . & . & . & . & . \\ . & 1 & . & . & . & . & . & . \\ 1 & . & 1 & . & . & . & . & . \\ . & 1 & . & 1 & . & . & . & . \\ . & . & 1 & . & 1 & . & . & . \\ . & . & . & 1 & . & 1 & . & . \\ . & . & . & . & 1 & . & 1 & . \\ . & . & . & . & . & 1 & . & 1 \end{bmatrix} $$ just to interweave the two jordan block of size $\gt 1$.
The complete solution would then look like $$ \Large J_\infty = \small \begin{array} {r|rrrrrrrrr} 1 & . & . & . & . & . & . & . & . & \cdots\\ \hline . & 1 & . & . & . & . & . & . & . & \cdots\\ . & . & 1 & . & . & . & . & . & . & \cdots\\ . & 1 & . & 1 & . & . & . & . & . & \cdots\\ . & . & 1 & . & 1 & . & . & . & . & \cdots\\ . & . & . & 1 & . & 1 & . & . & . & \cdots\\ . & . & . & . & 1 & . & 1 & . & . & \cdots\\ . & . & . & . & . & 1 & . & 1 & . & \cdots\\ . & . & . & . & . & . & 1 & . & 1 & \cdots\\ \vdots &\vdots &\vdots &\vdots & \vdots& \vdots&&&&\ddots \end{array}$$

This can now smoothly be extended to the case of arbitrary size, and I assume also to the case of infinite size (which is what I'm of course originally interested in).

The matrix S of the Jordandecomposition $M = S \cdot J \cdot S^{-1}$ where the leading entries of the first three eigenvectors are normed to $1$ looks like
$\qquad \qquad$ image

If the rows are scaled by factorials we get integer entries only and have the form

$\qquad \qquad$ image2

Final remark: It was extremely interesting for me, that the matrix S of eigenvectors for lower triangular Carlemanmatrices with unit-diagonal (because associated to schlicht functions) in general provide the generating functions for the iterates of their associated functions; and the fractional iterates are then binomially weighted sums of integer-iterates, thus "implementing" the Newton-formula for finite differences applied to fractional iterates (see for instance L. Comtet, Advanced combinatorics) . In this specific case we get, if we formulate the generating functions as dot-product of a vector with consecutive powers of x with the matrix S in the form $V(x) \cdot S = Y(x) $ then the entries $y_k$ of $Y(x)$ contain the formal power series for $$ \small \begin{array} {} y_0(x) &=& 1 \\ y_1(x) &=& x & y_2(x) &=& x^2\\ y_3(x) &=& \sin(x) - x & y_4(x) &=& \sin(x)^2 - x^2\\ y_5(x) &=& \sin(\sin(x)) -2 \sin(x) + x & y_6(x) &=& \sin(\sin(x))^2 -2 \sin(x)^2 + x^2 \\ \vdots && \vdots &\vdots&& \ddots \end{array}$$

The formal powerseries for the integer and fractional iterates of iterationheight h are then determined by the binomial coefficients which occur in the column 2 of the h'th power of $J$. Explicitely can this for fractional powers be computed using the matrix logarithm and -exponential. But the result in the relevant 2nd column is finally simply the set of the generalized binomial-coefficients $\binom{h}{k}$ where $k$ is $(r-1)/2$ of the odd $r$ only and $r$ is the row-index (beginning at zero). We have thus $$ \sin°^{h}(x) = \sum_{k=0}^N \binom{h}{k} y_{1+2k}(x) $$ where $N=h$ if $h$ is integer and $N=\infty$ if $h$ is fractional or negative.

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  • $\begingroup$ $J_8^*$ has a double eigenvalue $1$, the algebraic for multiplicity $J_8$ was $8$. $\endgroup$ – Christian Remling Jun 7 '14 at 19:03

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