remark: I asked this in MSE, the question got views and votes but seemingly no one had an answer so far.

Background: I'm rereading a couple of my exploratory (surely not research-level) math-essays and want to fix some possible wrong or misleading expressions. I've used the following notions/expressions a couple of years already but I would like to confirm that I can really use it in revisions of my web-essays.

Consider the (upper triangular) infinite "Pascal"/"binomial"-matrix P with top-left element as $$\small \begin{bmatrix} 1 & 1 & 1 & 1 & 1 & 1 \\ . & 1 & 2 & 3 & 4 & 5 \\ . & . & 1 & 3 & 6 & 10 \\ . & . & . & 1 & 4 & 10 \\ . & . & . & . & 1 & 5 \\ . & . & . & . & . & 1 \end{bmatrix}$$ Rightmultiplying it with the columnvector $E_1 = [1,1/1!,1/2!,1/3!, \cdots]$ gives $$ P \cdot E_1 = e \cdot E_1 $$ which has the form of an eigenvector equation as known from the cases with matrices of finite size. However, using $P$ and $E_1$ truncated to finite size $P^\star$ and $E_1^\star$ this would never be correct since $P^\star$ has no diagonalization.
Back to infinite size: in general with some columnvector $E(x)=[1, x^1, x^2/2!, x^3/3!, \cdots]$ we have $$ P \cdot E(x) = e^x \cdot E(x) $$ thus for each $x$ we had that $P$ has $E(x)$ as eigenvector to eigenvalue $e^x$.

Now what I'm discussing in a couple of essays are a second type of infinite matrices, namely the concatenation of vectors $E_n=E(n)$ to a matrix $$EZ=[E_0,E_1,E_2,E_3,...]$$
and following the example I could write $$ P \cdot EZ = EZ \cdot \,^dV(e) \\\qquad \qquad \qquad \small{\text{ where $\,^dV(e)$ = diagonal}([1,e,e^2,e^3...])} $$ which has again the form of an eigenmatrix-decomposition (or "diagonalization").
I always tended to say, that

  • "EZ is an eigenmatrix of P" (or is matrix-of-eigenvectors), or that
  • "P of infinite size has a diagonalization"

and used this at several places in my manuscripts.

But because for the case of finite size P$\,^\star$ has no diagonalization (it has only a Jordan-form), I feel it might be too sloppy to formulate this as an Eigenmatrix-relation or even as "diagonalization of P" (the latter is even more problematic since the matrix EZ has no inverse/reciprocal and we cannot write $\text{P}=\text{EZ} \cdot \,^dV(e) \cdot \text{EZ}^{-1}$).

Q: How could I correctly express that relation, even in a informal context? Can I still apply the terms "matrix of eigenvectors", "...of eigenvalues" and "diagonalization"?

Update, added One argument which is possibly against the use of the concept of diagonalization here, is perhaps that of the existence of a Jordan-decomposition for the finite-size case $P^\star$ . The top-left $6 \times 6$ -truncation of that (finite-size) Jordan-decomposition $P^\star = S^\star \cdot J^\star \cdot S^{\star -1}$

shows known matrices $S^\star$ (from Stirlingnumbers $1$st kind, left hand, factorially scaled) , the simple matrix $J^\star$ (in the middle) and $S^{\star-1}$ (from Stirlingnumbers $2$st kind, right hand, factorially scaled)
(or in a non-canonical rescaled version, but Stirlingnumbers nicer recognizable):
I don't know, whether it is more appropriate to apply the generalization to the case of infinite-size for the Jordan-decomposition, but if we do this, than we had a parallel between "diagonalization" and "Jordan-decomposition" which likely points to some incompatibility here with respect to the "diagonalization"-concept for the case of infinite size.

P.s.: don't know the best tagging for MO. Please feel free and improve if you think fits

  • $\begingroup$ from the eigenvalue equation ${\rm det}\,(\lambda-P)=0$ the conclusion would be that all eigenvalues of $P$ are equal to 1 (not some power of $e$). $\endgroup$ – Carlo Beenakker Dec 6 '19 at 12:41
  • $\begingroup$ @CarloBeenakker - yes, that's a concurring observation. But because for the case of infinite matrix-size my observation is also true I need some reliable terminology for that. $\endgroup$ – Gottfried Helms Dec 6 '19 at 12:45
  • $\begingroup$ @CarloBeenakker - I don't know whether the following is a meaningful remark here, but let's see. For the case of infinite size we can also have multiple left-inverses (even if the matrix is lower triangular). An example is the factorially scaled matrix of Stirlingnumbers second kind. Its ("principal") inverse for the case of infinite size is the factorially scaled matrix of Stirlingnumbers first kind. (As for instance described in the NIST - handbook) But there are also infinitely many other left-inverses possible, according to the multiplicy of $\exp(\log(1+x)+2k\pi i)$. Maybe it's related. $\endgroup$ – Gottfried Helms Dec 6 '19 at 12:54
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    $\begingroup$ The definition of eigenvalues, eigenvector, and even 'eigenmatrix' don't rely on the space being finite dimensional. The only problem is that in the infinite dimensional case you have to make sure that your objects actually belong to the space. A meaningful space for your problem is $\ell^2$. $\endgroup$ – lcv Dec 6 '19 at 12:54
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    $\begingroup$ @lcv - done. (link is at the introductional remark) $\endgroup$ – Gottfried Helms Dec 6 '19 at 13:56

I would suggest to call $E(x)=(1, x^1, x^2/2!, x^3/3!, \cdots)$ a fixed point of the linear map $E\mapsto M_\lambda\cdot E$, with $M_\lambda=\lambda P$ and scale factor $\lambda=e^{-x}$. In this way you can avoid the words eigenvector and eigenvalue, which mean something different in this context.

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  • $\begingroup$ Ah, well, the formulation "a fixed point of the linear map" did not yet come to my mind. Thanks for that! But perhaps I should expand my question. Because in other cases of such infinite triangular matrices we have consecutive powers of a certain constant in the diagonal and might easier say, that "we have an eigensystem property": The idea from functional iteration which leads to the well known "Schroeder-function" can -via the connotation of the so-called "Carlemanmatrices"- be identified with the diagonalization in the cases of finite matrix size ... $\endgroup$ – Gottfried Helms Dec 10 '19 at 18:38
  • $\begingroup$ ... where the matrix of eigenvectors (and its inverse) in the infinite case are associated to that Schroeder-functions. I've read such expressions elsewhere, for instance in R.Aldrovandi's articles on iterated exponeniation/Tetration. (But I think also elsewhere). From this I think, -if the use of the terms from diagonalization are not appropriate for my given case- then it would be good for the reader, why in this case we can not apply that termini but must resort to something else. - hmm. Sorry if I'm just confusing things more than explicate... $\endgroup$ – Gottfried Helms Dec 10 '19 at 18:43

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