0
$\begingroup$

Let $V$ be the vector space of infinite dimension on the field $\mathbb{Z}_2$. Let's say $$ V=\langle v_1\rangle+\langle v_2\rangle+\dots+ \langle v_n\rangle+\dots$$ where each $v_i$ has order $2$ and form a basis of the vector space V.

Let $\alpha$ be the linear transformation of $V$ such that $$v_i\mapsto v_1+v_2+\dots+v_i, \text{ for each } i\geq 1$$

Is there a proper $\alpha$-invariant infinite subspace of $V$?

$\endgroup$
  • $\begingroup$ The kernel of $\alpha$ would appear to be infinite-dimensional, no? (It contains $v_i$ with $i$ even...) $\endgroup$ – Nick Gill Feb 19 '15 at 10:08
  • $\begingroup$ I think $\alpha$ is bijective so the kernel is $0$. In fact, $v_2$ for instance is mapped in $v_1+v_2$. $\endgroup$ – W4cc0 Feb 19 '15 at 10:13
  • $\begingroup$ Oh, sorry, I misread the definition, you are quite right. $\endgroup$ – Nick Gill Feb 19 '15 at 10:13
1
$\begingroup$

No there is no proper $\alpha$-invariant infinite dimensional subspace.

For $k \ge 0$, let $V_k$ be the subspace of $V$ spanned by $v_1,\ldots,v_k$. Then $V_k$ is $\alpha$-invariant, and we claim that these are the only proper $\alpha$-invariant subspaces.

Let $W$ be an $\alpha$-invariant nonzero subspace of $V$, let $w \in W$,and let $k$ be the highest $v_k$ occurring in the sum of the basis elements for $w$. We claim that $V_k \le W$. Since $w + \alpha(w) \in W$, and has highest term $v_{k-1}$, this is a straightforward induction of $k$.

So either there is a maximal $v_k$ occurring as highest term in the elements of $W$, in which case $W=V_k$, or there is no such highest term, in which case $W$ contains $V_k$ for all $k$, so $W=V$.

$\endgroup$
  • $\begingroup$ I think I love you, I'm sorry. Thanks Derek! $\endgroup$ – W4cc0 Feb 19 '15 at 10:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.