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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.

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  • $\begingroup$ It really depends on the field of scalars the Vector space is build on. If $2$ has a multiplicative inverse, then yes. $\endgroup$ – b00n heT Oct 11 '17 at 22:06
  • $\begingroup$ @b00nheT We haven't gotten to multiplicative inverse yet $\endgroup$ – L to the V Oct 11 '17 at 22:08
  • $\begingroup$ Since you said “over an arbitrary field,” I am going to assume multiplicative inverses are implied to be assumed knowledge. In this case, since it is a field, the answer to your question is “yes”. $\endgroup$ – Clayton Oct 11 '17 at 22:20
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If you are dealing with real vector spaces, for instance, then yes, it is true. If $2u\in W$, then$$u=\frac12(2u)\in W.$$

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  • $\begingroup$ I'm not really sure what you said proves $\endgroup$ – L to the V Oct 11 '17 at 22:16
  • $\begingroup$ If $2u \in W$, and $W$ is closed under scaler multiplication, then multiplying it by scaled $1/2$ means its still in $W$. But since the field is $\mathbb{R}$, we get back $u$ which is in $V$ to begin with. $\endgroup$ – Tyler Hilton Oct 11 '17 at 22:22
  • $\begingroup$ @LtotheV I am. Where do you think there is a mistake? $\endgroup$ – José Carlos Santos Oct 11 '17 at 22:27
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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 $W$.

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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 $2$.

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