*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*

basisof the space in question (in the appropriate sense, e. g. Hilbert basis) in which the operator acts diagonally. Since this is far from truth in your case, I would refrain from using this term. $\endgroup$ – Kostya_I Jan 9 at 19:37