# One more question about eigenvalues in infinite dimension

Consider the following statement:

If $U$ is a subspace of $V$ that is invariant under every operator on $V$ then $U=\{0\}$ or $U=V$.

I can prove it easily assuming that $dim(V) = n$ is finite: If $U$ is neither zero nor the whole space then there is a non-zero vector $v\in V$ not in $U$. Extend it to a basis of $v$ and pick the operator that swaps any vector in the basis with $v$. This yields a contradiction.

But can the statement be proved for $U,V$ infinite dimensional? Of course the argument I used in the finite case also works for the infinite case... if there is a basis for $V$ that extends $v$. Is there?

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Have you proved a result like "every (possibly infinite dimensional) vector space has a basis"? Alternatively, are you allowed to use the axiom of choice? – walcher Aug 15 '13 at 9:02
the axiom of choice is not required for this result. – Ittay Weiss Aug 15 '13 at 9:22
@Ittay: Yes it is. – Asaf Karagila Aug 15 '13 at 9:24
@AsafKaragila OP's question requires AC? – Ittay Weiss Aug 15 '13 at 9:25
@Ittay: It is consistent to have vector space that all their endomorphisms are scalar multiplication (dubbed Lauchli spaces). Every subspace is invariant under scalar multiplication. Therefore if such Lauchli space exists, every subspace is invariant under all the endomorphisms, but it doesn't have to be trivial. – Asaf Karagila Aug 15 '13 at 9:26

The argument requires the axiom of choice. This is an axiom from set theory (which has a lot of equivalents from all over mathematics, e.g. Zorn's lemma). The axiom itself is less relevant, but it does allow us to prove the existence of many objects which we cannot exhibit explicitly with a formula. For example a basis for a vector space. If we have the axiom of choice then we can prove it in the following way:

Given $U\subsetneq V$ which is non-zero we can find $u\in U$ and $v\in V\setminus U$. In particular $\{u,v\}$ is a linearly independent set. Therefore we can extend it to a basis of $V$, say $B$. Now consider the operator which is defined by extending $T(u)=v, T(v)=u, T(w)=w$ for $w\in B\setminus\{u,v\}$. It is easy to see that $U$ is not invariant under $T$.

On the use of the axiom of choice. We use the fact that the axiom of choice guarantees us that every vector space has a basis, and that every linearly independent set can be extended to a basis. These two facts are in fact equivalent to the axiom of choice itself.

I'm not sure, but it seems probable that a weaker version of the axiom of choice would suffice to prove this. On the other hand, some axiom of choice is definitely needed.

In my masters thesis I extended the work of Lauchli and showed that it is consistent that the axiom of choice fails and we have a vector space over the real numbers (or any other pre-chosen field really) which is not finitely generated, but its endomorphisms are exactly scalar multiplication from the field.

Since every subspace is invariant under scalar multiplication, such vector space would have the property that every subspace is invariant under all linear operators.

I gave the outline of the construction in my answer: Axiom of choice and automorphisms of vector spaces, I later generalized this method in my thesis. (The construction doesn't really depend on the field being $\Bbb F_2$, just the result for that particular question. You can fully repeat it with other fields instead, but you may need to slightly adjust the cardinality of your atoms.)

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Perhaps I'm wrong, but I don't think my proof uses AC? – walcher Aug 15 '13 at 9:37
@walcher: You and Ittay made the same mistake. In your answer, what is $f(cu+v)$? Your operator is not linear. – Asaf Karagila Aug 15 '13 at 9:38
Yes, of course, you're right. Thanks! – walcher Aug 15 '13 at 9:39
But isn't $f(cu + v)$ equal to $0$? @walcher – newb Aug 15 '13 at 9:41
@newb Yes, but it should be equal to $cv$ – walcher Aug 15 '13 at 9:42