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Hi I am a 2nd year undergraduate student studying Mathematics. I have been given the following question on an example sheet and I've been trying to figure it out for a while now and I can't get my head around it.

Consider the triple $(V, \oplus, \odot)$. With $V=\Re$, $u \oplus v = u + 2v$ and $\lambda \odot u=\lambda u$ for any $u, v \in V$ and any $\lambda \in \Re$.

The question is, does $V$ form a vector space over $\Re$? Giving a reason.

I think that you have to show that $V$ satisfies the 8 properties of vector spaces. However the definition of $\oplus$ is confusing me.

When I try to prove the Associativity of addition property - $\forall u, v, w \in V (u \oplus v) \oplus w = u \oplus (v \oplus w)$

I am not sure how to deal with the introduction of $w$ (with $w \in V$).

Does $(u \oplus v) \oplus w = (u + 2v) + 2w$ or does $= (u+2v) + w$? Either way, I can't see how this property can be satisfied

Any help will be greatly appreciated. Many thanks

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    $(u \oplus v) \oplus w = (u \oplus v) + 2w = (u + 2v) + 2w$. And $u \oplus (v \oplus w) = u + 2(v \oplus w) = u + 2(v + 2w) = u + 2v + 4w$. Conclusion...? – 2017-02-08
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    Notice that your question lacks some details. **In the definition of $\oplus$, there is an other operation $+$, which is nowhere defined.** Since you haven't written what $V$ is, it can mean anything. The same applies to $\odot$. Let $V=\{ 0 \}$ where $0$ is the real zero, and $+$, $\cdot$ the usual arithmetic operations. Then the triple $(V,\oplus, \odot)$ is in fact a vector space! – 2017-02-08

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Hint: The first one is correct. Now what happens when you change the association by moving the grouping symbols to the pair in the other $\oplus$? For associativity to hold, the results must be equal for all choices of $u,v,w$. Is that the case?

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    Thanks for the hint. So in doing as you suggest $u \oplus (v \oplus w) = u + 2(v \oplus w) = u + 2(v + 2w) = u + 2v + 4w$. So in conclusion $(u \oplus v) \oplus w \neq u \oplus (v \oplus w)$? – 2017-02-08
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When I have issues like this, I often like to completely change the variables I'm using. $u$ and $v$ are the dummy variables you use in the pointwise definition of $\oplus$? Then use dummy variables $a,b,c$ in the formulation of associativity.

When you're trying to compute $(a \oplus b) \oplus c$, the fact you use a completely different set of variables removes one source of difficulty in matching up the arguments of the operation to the dummy variables in the pointwise formula.

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    Thank you for the help. So by doing what you say do you get $u \oplus (v \oplus w) = u + 2(v \oplus w) = u + 2(v + 2w) = u + 2v + 4w$ . So in conclusion $(u \oplus v) \oplus w \neq u \oplus (v \oplus w)$? – 2017-02-08
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You can't see how this property can be satisfied because with this definition, no, it can't be satisfied. You have just proved that this thing fails to be a vector space. That is how you do it.

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    Okay thanks, but this question along with another which is a $2*2$ matrix with $u \oplus v = {(0,0),(0,0)}$ and $\lambda \odot v = v$ for any .... are worth 20 marks? where would I get the 20 marks from? I'm assuming that the second one is a vector space over $\Re$? – 2017-02-08
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    Part (a) You show clearly how the above fails to be a vector space (write out the two additions clearly and show the results are different, then state it is not a vector space because it fails the associativity of addition property)) and that is your first ten points. Part (B) I think fails one of the distributive properties. Is $(\lambda_1 + \lambda_2)v = \lambda_1v + \lambda_2v$? Show clearly how and why it fails and that is your second ten points. – 2017-02-08