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I'm new to linear algebra, and I've stumbled upon a hypothesis that I'm not sure if it's right or not.

Assume $V$ is a group of vectors, where each of its vectors belong to $\mathbb R^4$. This means that each of the vectors components belongs to $\mathbb R$.

Based on this, can I assume that $V$ itself is a subgroup of $\mathbb R^4$ and therefore of $\mathbb R$? (I'm not necessarily asking about a subspace, just about a subgroup)


Thank you!

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When you write "assume $V$ is a group of vectors", do you really mean group or do you just mean set? – Brad Nov 13 '12 at 14:18
Apparently I mean a group indeed, sorry for being unclear, I'm trying to translate the concepts into English. – vondip Nov 13 '12 at 14:27
up vote 2 down vote accepted

Well you can assume all sorts of things, but in this case assuming is not nearly as good as concluding something.

If you really mean it when you write "group of vectors," then you are probably assuming that adding two of them together results in a vector that is still in the group, and that the negative version of each vector is in the group, so that the addition operation is the vector operation we all know.

In that case, you would be right that the set $V$ is a subgroup of $R^4$. That's exactly the definition of a subgroup of a group: a subset that is closed under the operation of the containing group, and contains inverses to its own elements.

It is not necessarily a subspace, because there is the possibility it does not contain all scalar multiples of its elements.

Writing "$V$ is a subgroup of $R^4$ and therefore of $R$" does not make much sense, though, since $R^4$ is not a subset of $R$.

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I know that I'm slipping into a new topic, but perhaps you could explain the connection between R4 and R? – vondip Nov 13 '12 at 14:31
@vondip Sure, no problem! $R^4$ is the set of ordered 4-tuples of real numbers. That means each thing in it looks like $(a,b,c,d)$ where $a,b,c,d$ are real numbers. We use the addition and multiplication in $R$ to make $R^4$ into a vector space. For addition: $(a,b,c,d)+(e,f,g,h)=(a+e,b+f,c+g,d+h)$, and $k(a,b,c,d)=(ka,kb,kc,kd)$. The addition and multiplication inside the parenthesis are $R$'s operations. – rschwieb Nov 13 '12 at 14:34
I see, so would it be more correct to say that R4 is a vectorial space over the field of R ? – vondip Nov 13 '12 at 14:47
Yes, that's right... although in English the term is "vector space." – rschwieb Nov 13 '12 at 14:54
Thanks! this makes so much more sense now. – vondip Nov 13 '12 at 15:01

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