diagonal vs regular

[bo198214]

"Similarity transform(ation)" in the sense of linear algebra.

If X = A * B * A^-1 then X is said to be "similar" to A; this means, it has for instance the same eigenvalues.

Gosh, that needed really explanation, I thought you were referring to similarity transforms in the geometric sense, i.e. scaling.

In the case of infinite dimension we may have, that the inverse is not unique, also we call it the "reciprocal" instead. Say Z defined to be a reciprocal to A, so that
A*Z = I
for the case of infinite size, then we may have different Z with the same reciprocity-relation.
A*Z1 = A*Z2 = A*Z3 =...= I

Also, - what I have learned here - we may have different A for a given B, such that not only

X1 = A1 * B * A1^-1

but also
X2 = A2 * B * A2^-1
...
with X1<>X2<>... and all Xk being diagonal

Then apparently it follows also, that we have multiple diagonalizations resulting in different X1,X2,X3,...
X1 = A1 * B * Z1_1 = A1 * B * Z1_2 = ...
X2 = A2 * B * Z2_1 = A2 * B * Z2_2 = ...
...

*nods*, possibly.

So the conjecture is that the eigenvalues of the truncated Carleman/Bell matrix of \( b^x \) converge to the set of powers of \( \ln(a) \) where \( a \) is the lower (the attracting) fixed point of \( b \)?

Yes, for the cases of b in the range of convergence. (maybe some excpetions: b=1 or b=exp(1) or the like)

What is the "range of convergence"?

ehmm... 1/e^e < b < e^(1/e)

I thought b=exp(1) was anyway outside the range of convergence, thatswhy I wanted to be sure what you mean by it. Also your next statement is mysteries assuming that range of convergence.

For the case of b outside this range I found that always a part of the eigenvalues(truncated matrices) converge to that logarithms, but another part vary wildly;

How can a part of the eigenvalues converge to that logarithms, if there are no real fixed points for \( b>e^{1/e} \) and hence no logarithms of that fixed points?
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