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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?

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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.2011-10-02
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    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.2011-10-02
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    $T$ has a non-empty spectrum if and only if $V$ has a $T$-invariant finite dimensional subspace. But why this need be true?2011-10-02
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    Yeah, I don't know. Presumably some argument involving countability of basis works, but one that fails for an uncountable basis.2011-10-02
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    Also, why is the forward implication of what you said true?2011-10-02
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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?2011-10-02
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    And are you restricting to *bounded* operators, or to any linear operator?2011-10-02
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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.2011-10-02
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    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.2011-10-02
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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.2011-10-02
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    @Arturo: Also I am looking at any linear operator and not necessarily bounded ones.2011-10-02
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    @Giuseppe: I did not define anything in the question I asked, so I am not sure what exactly you mean by strange.2011-10-02
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    @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.2011-10-02
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    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.2011-10-02
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    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.2011-10-02
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    @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.2011-10-02
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    @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.2011-10-02
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    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.2011-10-02
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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.2011-10-02
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    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 $\mathbb{C}$.2011-10-03
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    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 {$(T -\lambda Id)^{-1}$ : $\lambda \in \mathbb{C}$}.2011-10-03
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    Okay, just a note on convention: when I say $\mathbb{C}(x)$, I mean the fraction field of $\mathbb{C}[x]$.2011-10-03
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    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).2011-10-03
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    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.2011-10-03
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    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.2011-10-03

2 Answers 2

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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.

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    Indeed, this is a better way of putting it. Thanks!2011-10-03
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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))

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    Dear Rankeya, please see my "answer".2011-10-03
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    Ingenious indeed!2011-10-03
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    I think this proof also clearly shows that if one has a vector space $V$ over any algebraically closed field $k$, as long as the dimension of $V$ over $k$ is strictly less than the cardinality of $k$, then any linear operator on $V$ has a non-empty spectrum. I wonder what some of the applications of this result could be.2011-10-04
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    I should have mentioned "both proofs" instead of just my proof in the comment above.2011-10-04