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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. –  user12014 Oct 18 '11 at 1:32
@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. –  Jack Oct 18 '11 at 1:36
Hint: Consider $v_1 \ne v_2 = v_3 = \dots = v_n$. –  Sam Oct 18 '11 at 1:44
@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? –  André Nicolas Oct 18 '11 at 2:00
@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... –  Sam Oct 18 '11 at 12:42

2 Answers 2

up vote 0 down vote accepted

The answer should be $5$. Let $A$ be the matrix with column vectors $v_1,v_2,v_3,v_4,v_5$ and assume that $\dim(V)=5$. Up to isomorphism, we can let $V=R^m$. Then $A$ is a linear transform $A:{\mathbb R}^5\to{\mathbb R}^m$. We have $$ \dim(\ker(A))=0 $$ since $Aa=0$ where $a\in{\mathbb R}^5$ has unique solution $a=0$, which is an equivalent condition to that in OP. By the rank-nullity theorem, $$ \dim(range(A))=5-0=5. $$ It follows that $m\geq 5$.

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1 dimensional is out: If $v_1=c_2v_2=c_3v_3=c_4v_4=c_5v_5$ then $5v_1-c_2v_2-\cdots-c_5v_5=0$ (all coefficients non-zero).

On the other hand, $v_1=(1,0)$ and $v_2=\cdots=v_5=(0,1)$ then $a_1v_1+\dots+a_5v_5 \not=0$ if just $a_1 \not=0$. This gives a 2 dimensional example.

So the minimum is 2.

I must say, what an odd question.

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@Jack: always try examples, for all problems. –  Mariano Suárez-Alvarez Oct 18 '11 at 3:28
It took me a few minutes to see what was going on. My first thought was to try from the top down (why not dim 5,4,etc.)...FAIL. Misreading the question actually helped solve it more quickly. I hadn't noticed the $v_i \not=0$ hypothesis. Without that the dimension is 1: $v_1=1$, all else zero. With the nonzero hypothesis, you just need a slight adjustment. –  Bill Cook Oct 18 '11 at 3:33

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