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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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@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 2 '11 at 6:20
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And are you restricting to bounded operators, or to any linear operator? –  Arturo Magidin Oct 2 '11 at 6:26
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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 2 '11 at 6:54
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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 2 '11 at 15:57
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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 2 '11 at 23:21

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! –  Rankeya Oct 3 '11 at 17:31

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". –  Pierre-Yves Gaillard Oct 3 '11 at 15:18
    
Ingenious indeed! –  Giuseppe Negro Oct 3 '11 at 22:53
    
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. –  Rankeya Oct 4 '11 at 2:28
    
I should have mentioned "both proofs" instead of just my proof in the comment above. –  Rankeya Oct 4 '11 at 4:53

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