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How would you prove that if V is a vector space over $\mathbb{C}$ of countably infinite dimension, and $T$ is a linear operator on V, then Spectrum($T$) is non-empty?

Rankeya
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  • So, I figured that it suffices to prove the above statement for an $\mathbb{C}$-linear automorphism T, and I also know the proof for the case when V is finite dimensional (in that case the spectrum is just the set of roots of the characteristic polynomial), but I have no idea how to proceed with this. – Rankeya Oct 02 '11 at 02:18
  • Also, (correct me if I am wrong on this one) Lang in his Algebra and Steven Roman in Advanced Linear Algebra seem to talk about the Spectrum of linear operators only in the finite dimensional case. – Rankeya Oct 02 '11 at 02:28
  • $T$ has a non-empty spectrum if and only if $V$ has a $T$-invariant finite dimensional subspace. But why this need be true? – Manos Oct 02 '11 at 02:44
  • Yeah, I don't know. Presumably some argument involving countability of basis works, but one that fails for an uncountable basis. – Rankeya Oct 02 '11 at 02:55
  • Also, why is the forward implication of what you said true? – Rankeya Oct 02 '11 at 02:58
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    @Rankeya: What is your definition of spectrum for infinite dimensional vector spaces? For finite dimensional spaces, it is precisely the set of eigenvectors. Do you mean the set of values $\lambda$ for which $T-\lambda I$ is not invertible? – Arturo Magidin Oct 02 '11 at 06:20
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    And are you restricting to bounded operators, or to any linear operator? – Arturo Magidin Oct 02 '11 at 06:26
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    @Manos: Depending on exactly what Rankeya means by "spectrum", your assertion could be wrong. If the Spectrum is the set of all $\lambda$ for which $T-\lambda I$ is noninvertible, then the right-shift operator on the direct sum $\oplus_{i=1}^{\infty}\mathbb{C}$ has no finite dimensional invariant subspace, but $0$ is in the spectrum since $T$ is not invertible. – Arturo Magidin Oct 02 '11 at 06:54
  • I agree with Arturo that there is something strange in this definition. Indeed, a vector space of countable dimension cannot be a Banach space for topological reasons (Baire category theorem). As far as I know, completeness of the underlying space is a key ingredient in infinite-dimensional spectral theory. – Giuseppe Negro Oct 02 '11 at 10:57
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    @Arturo: Okay, so my definition of a Spectrum of $T$ is just the set of $\lambda \in \mathbb{C}$ such that $T - \lambda Id$ is not invertible. I am not yet willing to jump to the conclusion that some additional hypotheses are necessary. – Rankeya Oct 02 '11 at 15:57
  • @Arturo: Also I am looking at any linear operator and not necessarily bounded ones. – Rankeya Oct 02 '11 at 16:06
  • @Giuseppe: I did not define anything in the question I asked, so I am not sure what exactly you mean by strange. – Rankeya Oct 02 '11 at 16:08
  • @Rankeya: It's strange because "spectrum" is usually reserved for the case of Banach spaces (and bounded operators); infinite countable-dimensional vector spaces cannot be Banach spaces, so the definition does not usually apply to it. – Arturo Magidin Oct 02 '11 at 18:33
  • Right, sorry. I mean: there is something weird in this problem. Indeed I've always seen the word "spectrum" either in finite-dimensional or in Banach case. – Giuseppe Negro Oct 02 '11 at 18:45
  • Anyway, if you admit for operators that are not defined on the whole $V$ then the claim is false. For example, let $D(T)$ be the space of complex polynomials that vanish at the origin and $T \colon D(T)\subset \mathbb{C}[x] \to \mathbb{C}[x]$ be $(Tf)(x)=f'(x)$. This operator has empty spectrum, meaning that $T-\lambda I$ is bijective for every $\lambda \in \mathbb{C}$. Hope this helps. – Giuseppe Negro Oct 02 '11 at 18:51
  • @Arturo and Giuseppe: So, do you both think that there is a reasonable level of doubt as to the correctness of this question for me to go and ask the professor about this? Also, the question says that $T$ is an endomorphism of V, so it is defined on all of V. So, Giuseppe, your excellent example is still not a counter-example. This is really strange, because although the statement does not seem to be correct, there is no easily constructed counterexample. – Rankeya Oct 02 '11 at 19:12
  • @Rankeya: It's not a question about the "correctness of the question". It's a question as to the meaning of the terms in the question. There's no problem with using defined terms, so long as they are either defined explicitly, or in common use. "Spectrum of an operator" is not usually used for infinite-countably dimensional vector spaces; doesn't mean it cannot be used. – Arturo Magidin Oct 02 '11 at 20:30
  • I see what you are saying. Sorry, I misunderstood what you were saying before. Well, I have been thinking about this for a while, and I am not close to a solution. Hopefully, someone knows how to prove this statement. – Rankeya Oct 02 '11 at 22:01
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    Is there some context? Where does this exercise come from? Did you receive (explicit or implicit) hints? I've tried something but with no avail: both the finite-dimensional approach (based on determinants) and the Banach approach (based on theorems on holomorphic functions) don't seem to adapt to this particular case. – Giuseppe Negro Oct 02 '11 at 23:21
  • Okay I solved it. Here the solution: Suppose there is a linear operator $T$ on $V$, such that Spectrum($T$) is empty. This means that for all $\lambda \in \mathbb{C}$, $T - \lambda Id$ is an isomorphism. This implies that the map $\mathbb{C}(x) \longrightarrow End(V)$ that maps $X \rightarrow T$ is an injection. Let $\mathbb{C}(T)$ denote the image of this map. It is easily seen that $\mathbb{C}(x)$ has an uncountable basis as a $\mathbb{C}$ vector space as the set of {$1/x-\lambda$ : $\lambda \in \mathbb{C}$} is linearly independent over $\mathbb{C}$. – Rankeya Oct 03 '11 at 03:05
  • Thus $\mathbb{C}(T)$ has uncountable basis as a vector space over $\mathbb{C}$. Now the set {$(T -\lambda Id)^{-1}$ : $\lambda \in \mathbb{C}$} is linearly independent over $\mathbb{C}$. Since, V has a countable basis, let v be a non-zero element of one such basis. Then the set {$(T - \lambda Id)^{-1}$(v) : $\lambda \in \mathbb{C}$} is linearly dependent over $\mathbb{C}$ because V has a countable basis. But, this is contradicts linear independence of {$(T -\lambda Id)^{-1}$ : $\lambda \in \mathbb{C}$}. – Rankeya Oct 03 '11 at 03:22
  • Okay, just a note on convention: when I say $\mathbb{C}(x)$, I mean the fraction field of $\mathbb{C}[x]$. – Rankeya Oct 03 '11 at 03:30
  • The proof as given above is quite terse, but it works in my opinion. Note, that because any element $p(x)/q(x) \in \mathbb{C}(x)$ can be written as a quotient of linear factors, and since by our assumption any $T - \lambda Id$ is an isomorphism, the map from $\mathbb{C}(x) \longrightarrow End(V)$ is really injective. (This observation is crucial to our proof). – Rankeya Oct 03 '11 at 04:11
  • Do you mind writing the solution as an answer? It will be more readable and, if it is correct, you'll be able later to accept it. This way other users will know immediately that the question has been settled. Thank you. – Giuseppe Negro Oct 03 '11 at 08:28
  • Okay, I posted an answer. I still have not said everything explicitly, as I think it is a bit tedious to write out all the details. I am not sure if Stack Overflow wants me to write detailed proofs. I am still new to this thing. – Rankeya Oct 03 '11 at 08:55

2 Answers2

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This is a comment to Rankeya's answer posted as a community wiki answer.

Dear Rankeya, your answer looks correct to me, but I think the writing could be improved. Here is how I would have written it. I don't claim that my own writing is good. I'm just trying to help. Dear users, you're all welcome to suggest a better wording.

Suppose by contradiction there is a linear operator $T$ on $V$ such that

(1) $T-\lambda$ is an isomorphism for all $\lambda\in\mathbb C$.

Claim. There is a (unique) $\mathbb C$-algebra morphism $\phi$ from $\mathbb C(X)$ to $\text{End}(V)$ which maps $X$ to $T$. (Here $X$ is an indeterminate.)

Proof of the Claim. Let $p\in\mathbb C[X]$ be nonzero. It suffices to check that $p(T)$ is invertible. But this follows from (1). QED

The formula $fv:=\phi(f)v$, for $f\in\mathbb C[X]$ and $v\in V$, turns $V$ into a nonzero $\mathbb C(X)$-vector space. To get the sought-for contradiction, it suffices to prove that the dimension of $\mathbb C(X)$ over $\mathbb C$ is uncountable. But otherwise the set of all $(X-\lambda)^{-1}$, where $\lambda$ runs over $\mathbb C$, would be $\mathbb C$-linearly dependent, contradicting the uniqueness of the partial fraction decomposition.

Pierre-Yves Gaillard
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Here is a solution: Suppose there is a linear operator $T$ on V, such that Spectrum($T$) is empty. This means that for all $\lambda \in \mathbb{C}$, $T− \lambda Id$ is an isomorphism. This implies that the map $\mathbb{C}(x) \longrightarrow End(V)$ that maps $X \rightarrow T$ is an injective ring map, and also a $\mathbb{C}$ -linear map. Let $\mathbb{C}(T)$ denote the image of this map. It is easily seen that $\mathbb{C}(x)$ has an uncountable basis as a $\mathbb{C}$ vector space as the set of {$1/(x−\lambda) : \lambda \in \mathbb{C}$} is linearly independent over $\mathbb{C}$. Thus $\mathbb{C}(T)$ has uncountable basis as a vector space over $\mathbb{C}$. Now the set {$(T− \lambda Id)^{-1} : \lambda \in \mathbb{C}$} is linearly independent over $\mathbb{C}$ (since it is the image of the linearly independent set {$1/(x−\lambda) : \lambda \in \mathbb{C}$} under the injective map). Since, V has a countable basis, let $v$ be a non-zero element of one such basis. Then the set {$(T− \lambda Id)^{-1}(v) : \lambda \in \mathbb{C}$} is linearly dependent over $\mathbb{C}$ because V has a countable basis. But, this contradicts linear independence of {$(T−\lambda Id)^{-1} : \lambda \in \mathbb{C}$}. ((To see that the last step is really true, note that any non-zero element of $\mathbb{C}(T)$ is an isomorphism, since $\mathbb{C}(T)$ is a field. Thus, any non-zero element of $\mathbb{C}(T)$ must map a non-zero element of the vector space V, to a non-zero element of V))

Rankeya
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