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Consider the following problem:

Let $v_1,v_2,v_3,v_4,v_5\in V$, where $V$ is a vector space over ${\mathbb R}$ and $v_i\neq 0$ for $i = 1,2,3,4,5$. If the following statement is true:

$\sum_{i=1}^{5}a_iv_i\neq0$ whenever $a_i\neq 0 $ for all $i$.

What is the possible least dimension of $V$?

All I have tried so far is considering the contrapositive of the statement. But I have no idea how to go on. Furthermore, can this problem be generalized to the $v_i(i=1,2,\cdots,n)$ case?

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    Are you sure the condition is not $\sum a_i v_i \neq 0$ whenever at least on of the $a_i$ is not $0$? In this case the answer is trivial, because you have exhibited 5 linearly independent vectors, so the dimension must be at least 5.2011-10-18
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    @PZZ: Yes, the condition is the original one: "$a_i\neq 0$" for *all* $i$. And this is the difficult part for me to draw a conclusion.2011-10-18
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    Hint: Consider $v_1 \ne v_2 = v_3 = \dots = v_n$.2011-10-18
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    @Sam: Minor wording change suggested, we want $v_1$ not to be a multiple of $v_2$. Why not post something like this as an answer to the general problem?2011-10-18
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    @AndréNicolas: I didn't want to take away the chance of discovering an example with minimal dimension on his own. I'm sure Jack could have figured it out, if he only started to think about such examples...2011-10-18

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