# Vector spaces and infinite cyclic linear transformations II

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 $V$?

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

## 1 Answer

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 $W=V$.

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