*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

**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**

*no***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"?

**One argument which is possibly against the use of the concept of**

*Update, added**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*

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