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Let $W$ be the subset of $\mathbb{R}^5$ consisting of all vectors an odd number of the entries in which are equal to $0$. Is $W$ a subspace of $\mathbb{R}^5$?

I'm not sure how to do this. Any solutions or hints are greatly appreciated. I know that in order for anything to be a subspace of something the zero vector must be in it. How would I go about this? What exactly do we mean by subset here? Is it any $5$-tuple or could it be $1,2,3,4,5$-tuples?

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No. For instance, let $v=(0,1,1,1,1)$ and $w=(1,0,1,1,1)$. Then $v,w\in W$ but $v+w=(1,1,2,2,2)\notin W$.

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  • $\begingroup$ So this is a counterexample that shows that $W$ isn't closed under addition so it can't be a subspace? $\endgroup$ – Mike Aug 28 '15 at 3:20
  • $\begingroup$ Yes, that is correct. $\endgroup$ – Jason Aug 28 '15 at 3:21
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Since $W$ is a subspace of $\mathbb{R}^{5}$ if and only if (i) $0_{\mathbb{R}^{5}} \in W$, (ii) $x,y \in W$ implies $x+y \in W$, and (iii) $c \in \mathbb{R}$ and $x \in W$ imply $cx \in W$, and since $(1,2,0,3,4) \in W$ and $(4,0,3,2,1) \in W$ but their sum is a vector having no zero components, so $W$ is not a subspace of $\mathbb{R}^{5}$.

I misunderstood your question. :)

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  • $\begingroup$ oh ok great, thank you. Should I also do the same thing when I consider a $5$-tuple with $1$ zero and $5$ zeros since it said an odd number of entries are zero? $\endgroup$ – Mike Aug 28 '15 at 3:10
  • $\begingroup$ This does not describe all of $W$. $\endgroup$ – Jason Aug 28 '15 at 3:11

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