I just noticed this embarrassing gap in my understanding of linear algebra. This question seems to be asking something similar, although the questioner doesn't quite say it explicitly. And this question asks a version of the question which I think is a little more naive than what I'd like to say.

In finite dimensions, the Jordan decomposition of a linear endomorphism $T$ is the unique way to express $T = T_{ss} + T_n$ where $T_{ss}$ is semisimple, $T_n$ is nilpotent, and $T_{ss}$ commutes with $T_n$.

I'm wondering whether something similar holds in infinite dimensions. For a complex Banach space $X$,

  • I think a semisimple operator $T_{ss}$ should be one which commutes with a set of projection operators $P_i$ with $P_i P_j = \delta_{ij} P_i$ with $\oplus_i P_i(X)$ dense in $X$, such that the restriction of $T_{ss}$ to each $P_i(X)$ is a scalar $\lambda_i$. Please let me know if this is the wrong definition.

  • Instead of a nilpotent operator $T_n$, we'll consider a quasinilpotent operator $T_{qn}$, i.e. an operator whose spectrum is $\{0\}$, or equivalently $\lim_{k \to \infty} \|T_{qn}^k\|^{1/k} = 0$. Again let me know if there's a better notion to use here.

So then the question is: for an arbitrary bounded operator $T$ on a complex Banach space $X$, do there exist commuting bounded operators $T_{ss}, T_{qn}$ such that $T_{ss}$ is semisimple, $T_{qn}$ is quasinilpotent, and $T=T_{ss}+T_{qn}$? If the answer is no, I'd like a counterexample, and also any results establishing a positive answer on special classes of Banach spaces.

It does worry my that a positive answer to my question would immediately imply that any counterexample to the invariant subspace problem differs by a scalar from a quasinilpotent operator. Is this true/known?

EDIT As Mariano implicitly points out, the shift operator seems to scuttle this idea since it has no obvious "diagonal part" and it is not itself quasinilpotent.

But here's a modification to the question that just might be worth making. The quasinilpotent operators are exactly the limits of nilpotent operators in the norm topology. Instead we could take limits of nilpotent operators in the strong topology. Note that the shift operator is of this type. I haven't thought through the ramifications of this defintion -- If there's anything interesting to know about this class of operators, or if they have a name, I'd like to hear it.

So: does every operator $T$ on a Banach space decompose as $T = T_{ss} + T'$ where $T_{ss}$ is diagonal, $T'$ is a limit of nilpotent operators in the strong topology, and everything commutes?

EDIT As Ivan points out in his answer, although the "backward shift" operator on $\ell_p(\mathbb{N})$, say, is "strong-quasinilpotent", the "forward shift" operator is not. In fact, on $\ell_\infty(\mathbb{N})$, even the backward shift operator is not strong-quasinilpotent. Since the whole point of widening the definition is to accommodate various types of shift operators, the idea really doesn't work. A last-ditch effort would be to restrict to Hilbert spaces and consider "weak-quasinilpotent" operators $T$, i.e. those such that $T^n \to 0$ in the weak topology. At least the usual shift operators on $\ell_2(\mathbb{N})$ and $\ell_2(\mathbb{Z})$ are weak-quasinilpotent. Or maybe there's some other approach entirely to characterizing "shift-type" operators?

In the case where there is no topology, as Ivan points out, it's possible that simply specifying the matrix similarity class of a "shift operator" on a space with a countable algebraic basis might yield a sufficiently robust notion. But I don't know whether we can get a "diagonal - shift" decomposition with this definition.

  • 1
    $\begingroup$ What do you make of the shift map? $\endgroup$ Aug 24, 2014 at 3:51
  • $\begingroup$ The shift map is quasinilpotent, so we can take $T_{ss} = 0$, right? $\endgroup$
    – tcamps
    Aug 24, 2014 at 3:52
  • $\begingroup$ The shift map is actually the main reason I think that "nilpotent" needs to be changed to "quasinilpotent for infinite dimensions. $\endgroup$
    – tcamps
    Aug 24, 2014 at 4:03
  • 2
    $\begingroup$ Oh -- the shift map is not quasinilpotent, is it... $\endgroup$
    – tcamps
    Aug 24, 2014 at 16:15
  • $\begingroup$ May be my older discussion on MO is of interest as a possible example here: mathoverflow.net/questions/156958 $\endgroup$ Dec 31, 2014 at 12:30

2 Answers 2


From now on $V$ will be our Banach (Hilbert) space.

Let's call $S^+$ the "$+1$" shifting operator and $S_-$ the "$-1$" shifting operator. Well, in your definition $S_-$ is quasinilpotent, the other is not. In my opinion that's the main point of the question.

We have to generalize the notion of "generalized eigenspace" as follow.

Let's call $W_{\lambda} := \bigcup_{n} \ \operatorname{ker}(T-\lambda\operatorname{Id})^n$

Since clearly in the case of $T= S_-$ we get that $V = W_0$ we obtain that in the case of $T= \lambda\operatorname{Id}+S_-$ then $V = W_{\lambda}$. Now it's sounding as Jordan Normal Form.

But still we are far, what about $S^+$?

I think it's time to admit that we have to take it as a brick of our "Generalized Jordan form".

I'd move as follow. Let $\lambda \in \operatorname{Spec}(T)$, consider $(T-\lambda \operatorname{Id})^k$ and, as in Jordan normal form, find a basis of increasing kernels, obtaining that the restriction of $T$ to $W_{\lambda}$ is similar to $\sum_i (\lambda\operatorname{Id} + N_i) $ where $N_i$ are nilpotent operators or $S_-$ operators on suitable subspaces of $W_{\lambda}$.

Actually I believe that $\{ W_{\lambda} \}_{\lambda \in \operatorname{Spec}(T)}$ is a family of subspaces in direct sum. I've no idea to prove it but may be obvious from here .

Now let's take any vector $v$ outside $\oplus_{\lambda} W_{\lambda}$. If we consider $\{ T^k(v) \}_k$ we may generate a supplementary space for $\oplus_{\lambda} W_{\lambda}$ or may not but if we don't we can take a vector outside $(\oplus_{\lambda} W_{\lambda}) \bigoplus \operatorname{Span}(\{ T^k(v) \}_k) $ and go on till space is over. Clearly the restriction of $T$ to any of ${Span}(\{ T^k(v) \}_k)$ is similar to a $S^+$ operator.

So, at the end $T \sim \sum_{ik} (\lambda_k\operatorname{Id} + \ ^kN_i) + \sum_j S_j$ where $N_i$ are nilpotent operators or $S_-$ operators on suitable subspaces of $V$ and $S_j$ are $S^+$ operators on suitable subspaces of V.

What do you think? I've be thinking for better approaches but I can't find anything.

I'd like to point out that the notion of "quasinilpotent" is not intuitive as one would like. Consider $$e_i \mapsto \frac{1}{2^{i+1}}e_{i+1}.$$ This is similar to $S^+$ but it's quasinilpotent. (If we ask for continuous operators things are tamer).

I think that all I've said is correct for algebraic basis on Banach Spaces but would be wrong for Hilbert bases on Hilbert Spaces (and continuous operators) because base changes might not be continuous.

  • $\begingroup$ Thanks for pointing out what I should have noticed -- there's a "shift operator" $S^+$ which is not "strong-quasinilpotent"! Notice, in fact, that on $\ell_\infty(\mathbb{N})$, even $S_-$ fails to be "strong-quasinilpotent"! However, I don't think your approach works. The step where you say "clearly" is very unclear :) -- it sounds like this argument should equally prove that every operator is a shift operator. $\endgroup$
    – tcamps
    Jan 6, 2015 at 23:33
  • $\begingroup$ But I suppose one might hope for a diagonal + shift operator decomposition in an algebraic basis (since this means throwing topological and metric considerations to the wind, I wouldn't refer to the space as a Banach space in this case): infinite-dimensional discrete vector spaces are so little-studied that it's possible such a simple fact about them might not be widely known... $\endgroup$
    – tcamps
    Jan 6, 2015 at 23:48

If $T$ is a spectral operator then it is known that $T=S+N$ where $S$ is of scalar type, i.e. $S=\int z E(dz)$ where $E$ is the resolution of the identity for $T$, and $N$ is quasinilpotent, i.e. $\operatorname{Spec}(N)=\{0\}.$


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