# Can I assign the term “is eigenvector” and “is eigenmatrix” of matrix **P** in my specific (infinite-size) case?

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

• 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 $\ell^2$. – lcv Dec 6 '19 at 12:54
• @lcv - done. (link is at the introductional remark) – Gottfried Helms Dec 6 '19 at 13:56
• I would say that "eigenvector" and "eigenvalue" are totally natural terms in this context. "Eigenmatrix" I have never heard, sounds highly non-standard. "Diagonalization" typically means, also in infinite dimensional context, that one has found a basis of 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. – Kostya_I Jan 9 at 19:37
• I've never heard of the term ••eigenmatrix••, except in the context that something is acting on a space of matrices (so an eigenvector). I would not use that term, but eigenvector, eigenvalue, are OK in this general context. – David Handelman Jan 9 at 21:00
• I guess there's no ambiguity that eigenvalue $t$ (for a linear endomorphism $A$ of a $K$-linear space $V$) means $A-t$ is non-injective (and eigenvector is defined in the obvious way). I've encountered "algebraic spectral value" to mean that $A-t$ is non-bijective, and topological spectral value to mean, when $A$ is continuous on some topological vector space, that $A-t$ is not (bijective with continuous inverse). Of course most results only carry over finite dimension only. Viewing $V$ as module over the PID $K[t]$ yields some useful information anyway in infinite dimension. – YCor Jun 8 at 14:27

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.