If $2u$ is an element of the subspace $W$, is $u$ also an element of $W$? Suppose $V$ is a vector space over an arbitrary field $\mathbf{k}$, $W$ is a subspace of $V$, and $u \in V$.
The question is: if $2u \in   W$, then is $u \in W$?.
Intuitively I think this is true, because if $W$ is a subspace then it's closed under scalar multiplication, so if $u\in W$, then $cu\in W$ for all $c \in \mathbf{R}$. But I'm not sure how to show this as a proof, or if my statement is enough.
 A: If you are dealing with real vector spaces, for instance, then yes, it is true. If $2u\in W$, then$$u=\frac12(2u)\in W.$$
A: You said that you don't know what it means to be invertible. If you have a field $\mathbf{k}$, an element $x\in \mathbf{k}$ is said to be invertible if there exists some element $y\in \mathbf{k}$ such that $xy=1$. Take for example any nonzero element in $\mathbf{R}$ or $\mathbf{C}$.
Let $\mathbf{k}$ be our field, and let $\lambda\in \mathbf{k}$ such that $\lambda$ is invertible. Then for a $\mathbf{k}-$Vector space $V$, with a subspace $W$ such that $v\in W$, we have 
$$\lambda v\in W\iff \lambda^{-1}(\lambda v)\in W\iff (\lambda^{-1}\lambda)v\in W\iff v\in W.$$
This follows from closure of subspaces under scalar multiplication by $\mathbf{k}$. Of course, in the case where $\lambda$ is not invertible, this might not be true. Take $\mathbf{Z}/p\mathbf{Z}$, for $p$ some prime. Then $pv\in W$ for any choice of $v$, because $pv=0$, but $v$ might not be an element of $W$.
A: If $2\in {\bf k}$ is invertible, which is equivalent to $2\neq 0$ since ${\bf k}$ is a field, then $$2u\in W \implies u = 2^{-1}(2u)\in W.$$
The problem is when the field ${\bf k}$ is of characteristic $2$.
