Generalization of the Jordan form for infinite matrices

Under what conditions is it the case that for a matrix $M$ whose rows and columns are indexed by a countably infinite set $S$ one has a Hamel basis consisting of generalized eigenvectors (i.e. $v \in \ker(M - \lambda I)^n$) of $M$? Must $M$ be a compact operator (I have a norm)?

The matrix I am working with has non-negative entries, row sums not exceeding $1$ (substochastic), is irreducible and aperiodic. However, I suspect this question may be of general interest to others, so any solution not employing these properties would be all the more useful.

EDIT

Here is some more information: the matrix $M$ which I am working with is $R$-positive. This means that none of the sequences $\{ M^n_{ij}\}_{n \in \mathbb{N}}$, $i,j \in S$, converge to $0$, where $$R^{-1} := \lim_{n \to \infty} (M_{ij}^n)^{1/n}.$$ In such a case, it is known that $R^{-1}$ is the spectral radius of $M$, and moreover that $R^{-1}$ is an eigenvalue for $M$ for which there are unique left and right eigenvectors $\alpha,\beta$ which are strictly positive and satisfy $$\sum_{k \in S} \alpha(k) \beta(k) < \infty.$$ In particular the set of eigenvalues for $M$ cannot be empty.

Thanks!

• I think you can look up Jacobson. – Bombyx mori May 28 '14 at 2:23
• @Bombyxmori can you give a reference? I haven't found anything which seems connected. Thanks! – Rookatu May 29 '14 at 21:50
• I think the Pascal-matrix can be Jordan-decomposed (having the matrices of Stirling-numbers as generalized eigenvectors) and the identity matrix with additional first subdiagonal containing the unit as Jordan form. This is consistent for any size nxn, and I've always assumed that this is this also valid for the infinite case. Similarly I assume the same generalization from finite to infinite size is true for the Jordandecomposition of the (Carleman-)matrix containing the Stirling numbers, which map $x \to \exp(x)-1$ and the inverse. But I've no such theoretical background as the other answers. – Gottfried Helms Dec 31 '14 at 9:30

Let $H$ be a countably-dimensional Hilbert space with basis $\{e_n\}$. Let $T$ be the weighted shift given by $$Te_n=\frac1n\,e_{n+1}.$$ This operator is compact (actually, Hilbert-Schmidt).

If $Tv=\lambda v$, with $v=\sum_n\alpha_n e_n$, then $$\sum_{n=1}^\infty\alpha_n\lambda e_n=\sum_{n=1}^\infty\alpha_n Te_n=\sum_{n=1}^\infty\alpha_n\,\frac1n\,e_{n+1}=\sum_{n=2}^\infty\frac{\alpha_{n-1}}{n-1}\,e_n.$$ If $\lambda=0$, we deduce that $\alpha_n=0$ for all $n$, so $v=0$. If $\lambda\ne0$, then $\alpha_1=0$, and $\alpha_{n+1}=\alpha_n/n$, implying again that $\alpha_n=0$ for all $n$. So $v=0$. This shows that $T-\lambda I$ has trivial kernel for all $\lambda$.

So $T$ has no nonzero generalized eigenvectors, and it cannot have a Jordan form, at least in the obvious sense.

• Thank you for the thoughtful answer. However, if I am not mistaken, the matrix of this operator would not be irreducible, correct? My question is really about what conditions allow for the desired representation, as well as whether the properties I listed are sufficient. Any thoughts? Thanks! – Rookatu May 28 '14 at 3:51
• What do you mean by irreducible? The way I use the word is that the commutant of $T$ is trivial, which it certainly is. – Martin Argerami May 28 '14 at 11:22
• My meaning is that the matrix I'm dealing with, which is substochastic, has the property that for any $i,j$ in the indexing set $S$, there exists an $n \in \mathbb{N}$ with the property that $M^n_{ij} > 0$. It seems to me that you can "transition" forward, from $e_n$ to $e_{n+1}$ but never backward and hence that the corresponding matrix would not be irreducible, though I must say that functional analysis is not my strong suit and I may have misinterpreted. Thanks! – Rookatu May 28 '14 at 11:37
• I see. I cannot really imagine how it would be possible to check that condition but in the simplest of cases. – Martin Argerami May 28 '14 at 23:09
• I don't really know. But let me tell you this: you are thinking in terms of matrices, and not of operators. But eigenvalues and eigenvectors are properties of the operator, not the matrix. While being R-positive is a property of the matrix and not of the operator. – Martin Argerami Jun 5 '14 at 1:54

I am not sure if this answer is useful in your case.

Theorem: Exists a basis formed by generalized eigenvectors of $T:V\rightarrow V$, even if $V$ is an infinite dimensional vector space over a field $F$, if we assume the existence of a polynomial $p(x)\in F[x]$ with all roots in $F$ such that $p(T)=0$.

Of course, we need to prove first for nilpotent operators. Then we must use the Primary decomposition theorem to extend for arbitrary operators satisfying this hypothesis. Notice that this hypothesis is always true in finite dimension by Cayley-Hamilton's theorem, if $F=\mathbb{C}$. The proof is almost the same of the finite dimensional case.

The proof for nilpotent operators is an induction on the nilpotency index $($the smallest $k$ such that $T^k=0)$ which does not depend on the dimension of $V$.

Let us prove for nilpotent operators $($i.e. when $p(x)=x^k)$.

Proof: Let $T:V\rightarrow V$ be a nilpotent operator. Let $k$ be the nilpotency index. If $k=1$ the theorem is trivial. Suppose $k>1$.

Since $k>1$ then $T\neq0$ and $\Im(T)\neq 0$. Define $T':\Im(T)\rightarrow\Im(T)$, such that $T'(x)=T(x)$.

The nilpotency index of $T'$ is smaller than the index of $T$.Thus by induction hypothesis exists a basis $\alpha$ of $\Im(T)$ such that

1. $\alpha=\cup_{i\in I}\alpha_i$
2. $\alpha_i=\{v_1^i,\ldots,v_{s_i}^i\}$, $s_i<k$
3. $T(v_l^i)=v_{l-1}^{i}$ for $1<l\leq s_i$ and $T(v_1^i)=0$

Next for each $v_{s_i}\in\alpha_i\subset\Im(T)$, choose $v_{s_{i+1}}^i$ such that $T(v_{s_{i+1}}^i)=v_{s_{i}}^i$.

Consider the following preimage of $\alpha$: $\cup_{i\in I}\beta_i$, where $\beta_i=\{v_2^i,\ldots,v_{s_i}^i,v_{s_{i+1}}^i\}$ Now, let $W$ be a subspace of $V$ generated by $\cup_{i\in I}\beta_i$. Notice that $\ker(T)\oplus W=V$ and $\cup_{i\in I}\beta_i$ is a basis of $W$.

Now, $\{v_1^i,i\in I\}$ is a basis of $\ker(T)\cap\Im(T)$. (It is straightfoward)

Let $R$ be a subspace of $\ker(T)$ such that $R\oplus (\ker(T)\cap\Im(T))=\ker(T)$. Let $\{r_j, j\in J\}$ be a basis of $R$.

Finally the required basis of $V=\ker(T)\oplus W=R\oplus (\ker(T)\cap\Im(T))\oplus W$ is $$\{r_j, j\in J\}\cup \{v_1^i,i\in I\}\cup (\cup_{i\in I}\{v_2^i,\ldots,v_{s_i}^i,v_{s_{i+1}}^i\})=$$ $$=\{r_j, j\in J\}\cup (\cup_{i\in I}\{v_1^i, v_2^i,\ldots,v_{s_i}^i,v_{s_{i+1}}^i\}).$$ $\square$