# Proving a subspace of $\mathbb{R}^{2}$ is still a vector space with redefined addition / scalar multiplication operations?

Question: Let $S =$ { $(a,b) \in \mathbb{R}^{2}$ | $(a,b)$ has the form $(1, x)$ }.

Redefine addition and scalar multiplication operations as follows:

$(1, y) + (1, y') = (1, y + y')$

$k(1, y) = (1, ky)$

Can I now say that because $\mathbb{R}^{2}$ is a vector space and $S \subset \mathbb{R}^{2}$, that if additive & scalar multiplicative closures are shown to hold, then $S$ is a vector space?

Or do I first have to show that $\mathbb{R}^{2}$ is still a vector space with the newly defined operations?

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Although I suppose a proof of additive and scalar multiplicative closures for $S$ with the newly defined operations would be the exact same as a proof for $\mathbb{R}^{2}, so perhaps the second question is redundant... – robjb Jun 7 '11 at 5:30 Thanks Jonas, edited to remove the numbering. – robjb Jun 7 '11 at 5:30 ## 1 Answer The operations you've defined do make$S$into a vector space, but it is not a subspace of$\mathbb{R}^2$. A subspace of a vector space$V$is a subset$W\subseteq V$that's a vector space under the same operations as$V$. So, you need to show all of the axioms for a vector space hold for$S$. Also, I don't know what you mean by "$\mathbb{R}^2$is still a vector space with the newly defined operations"; how do you intend to extend the definitions$(1, y) + (1, y^\prime) = (1, y + y^\prime)$and$k(1, y) = (1, ky)$to all of$\mathbb{R}^2$, in a way that makes$\mathbb{R}^2$a vector space? - +1, I see what you mean. I can't very well define those operations over all of$\mathbb{R}^{2}\$... you've helped me see where I'm going wrong in my thought process. – robjb Jun 7 '11 at 5:33
@Rob: I'm glad to have helped! It's certainly easy to forget to check those kinds of things. – Zev Chonoles Jun 7 '11 at 5:46