bo198214 Wrote:Hey Gottfried,

did you ever thought about the spectrum of the Carleman matrix of ?

... oh so many times...

Quote:For finite matrices the spectrum is just the set of eigenvalues. However for an infinite matrix or more generally for a linear operator on a Banach space the spectrum is defined as all values such that is not invertible.

We saw that the eigenvalues of the truncated Carleman matrices of somehow diverge. So it would be very interesting to know the spectrum of the infinite matrix. As this also has consequences on taking non-integer powers of those matrices.

Or is the spectrum just complete because is not invertible itself?

Well, I've had at most 100 explanative pages about infinite matrices in my hands, so I cannot cite an authoritative statement about the cardinality, as well as about the characteristics of the non-invertibility.

But what I decided to use as hypotheses:

1)

I didn't discuss about a single eigenvalue or some arbitrarily formed set yet. I used always the restriction, that a) an infinite set of eigenvalues exists, which has also a unique structure: it is defined by the consecutive powers of one base-parameter.

and that b) there are *infinitely many* such sets of eigenvalues, according to the infinitude of fixpoints: each fixpoint defines one of such sets.

While one single eigenvalue defines "

is non-invertible", the definition of such a set is more informative, since with this we can say

where dV(x) denotes, as usual, a diagonalmatrix of consecutive powers of x - and

each of these powers is an eigenvalue so that

,

,

,

,... are all non-invertible.

The complete set of eigenvalues I "know of" is thus the set of all fixpoints and all of their (nonnegative) integer powers, which is then

N x N.

And

W is an infinite set of invariant column-vectors according to the selection of lambda and its power.

There may be other eigenvalues, but I didn't think about this yet.

--------------------------------------------

2)

The idea, that the spectrum is complete

because of non-invertibility seems to be a non sequitur.

I see the noninvertibility of

A for one single reason. First:

A is decomposbale into two well known triangular matrices Binomial

P and Stirling-kind2

S2 (factorially scaled).

Both factors

P and

S2 are invertible, so

and formally hte inverse is possible

where

S1 is the matrix of Stirlingnumbers 1st kind, also factorially scaled.

The reason why

A is not invertible, is that in

the dotproduct first row by second column is infinite and exactly gives log(0).

If by some measure this multiplication can be avoided in a more complex matrix-operation, then the whole formula behaves like if

A^-1 would exist (we discussed this in the context of fixpoint-shift)

Gottfried