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What I want to prove is that for infinite dimensional vector space, $0$ is the only eigenvalue doesn't imply $T$ is nilpotent.

But I am not sure how to find eigenvalues of infinite dimensional linear operator $T$. Since we normally find eigenvalues by find zeros of characteristic polynomials, we even cannot find the characteristic polynomial for this situation.

I am specifically interested in the differential operator on the vector space of all formal power series.

Esther Jacob
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    In general, it is very difficult to determine the eigenvalues of an operator on an infinite-dimensional space. The idea is to come up with some very special operator for which it is easy. Are there any interesting examples of operators on infinite-dimensional spaces that you have seen that you might try to use as a starting point? – Eric Wofsey Oct 18 '15 at 23:59
  • @Eric Wofsey I am gonna try the differential operator on the vector space of all format power series over $\mathbb{F}$. Does it work? – Esther Jacob Oct 19 '15 at 00:09
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    That's a good idea of something to try. Can you figure out whether it works? If it doesn't work, can you see any way you might modify it so that it does work? – Eric Wofsey Oct 19 '15 at 00:29
  • @Eric Wofsey But it may not work. Since I cannot find a basis for that vector space. – Esther Jacob Oct 19 '15 at 02:17

2 Answers2

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Let $\mathcal{H}$ be an infinite-dimensional Hilbert space and let $S$ be any operator on $\mathcal{H}$ that has no eigenvalues [for example, take $\mathcal{H} = L^2[0, 1]$ and let $S$ be the operator on $\mathcal{H}$ defined by $(Sf)(x) = xf(x)$.] Now define an operator $T$ on $\mathcal{H} \times \mathcal{H}$ by $$T(f, g) = (0, Sg).$$ Then $0$ is the only eigenvalue of $T$, but $T$ is not nilpotent.

Sheldon Axler
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  • Sorry, I haven't learnt Hilbert space yet. Could you give me another example? – Esther Jacob Oct 19 '15 at 02:18
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    If you want to avoid Hilbert spaces, in my example above take $\mathcal{H}$ to be the vector space of polynomials and $S$ to be the operator of multiplication by $x$ (this operator $S$ has no eigenvalues). – Sheldon Axler Oct 19 '15 at 05:41
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There's no general way to find eigenvalues of an operator in an infinite dimensional space. Actually, some operators don't have any eigenvalue.

You may consider the vector space $k[X]$ of polynomials over the field $k$, and the operator $D : P \mapsto P'$. This operator has only $0$ as an eigenvalue, but is not nilpotent (although it's not far from being nilpotent since for all $P$, there exists $n_P$ such that $D^{n_P}(P) = 0$ if you take $n_P = \deg P + 1$).

The operator $\sigma : P(X) \mapsto X P(X)$ is an example of operator with no eigenvalues.

Joel Cohen
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  • Joel, would you mind to have a look at my earlier question http://math.stackexchange.com/questions/459454/ I thought I have there an example of eigenvalues of an infinite-sized operator and I think (but I'm not an expert so I feel insecure) that this is just the example that you provide. (If needed we can also meet in chat.) – Gottfried Helms Oct 26 '15 at 17:02
  • The operator $\sigma$ does have an eigenvalue, namely $1$ (all elements of $k \setminus {0}$ are eigenvectors). – Boccherini May 13 '22 at 16:19
  • @Boccherini you're right, thanks! Changed it for another operator with no eigenvalue. – Joel Cohen Jun 24 '22 at 05:09