Dimension of a vector space. Let $v_1,v_2,v_3,v_4$ and $v_5$ be the non-zero vectors of a vector space $V$ such that 
$a_1v_1+a_2v_2+a_3v_3+a_4v_4+a_5v_5\neq0 \hspace{1cm} (\forall a_i\neq0,\, 1\leq i\leq5)$
Then what is the smallest possible dimension of $V$?
 A: Assuming $a_i\ne 0, i=1,\cdots,5$ then, if we assume that $\vec{v_i}\ne \vec{0}, i=1,\cdots,5,$ then the smallest dimension is $2:$ consider $\vec{v_1}=\cdots =\vec{v_4}\ne \vec{v_5}.$
Note that the dimension can't be one. In such a case $\vec{v_i}=\alpha_i\vec{v_1}, i=2,3,4,5.$ Now, $$-(\alpha_2+\alpha_3+\alpha_4+\alpha_5)\vec{v_1}+\vec{v_2}+\vec{v_3}+\vec{v_4}+\vec{v_5}=\vec{0}.$$ If $\alpha_2+\alpha_3+\alpha_4+\alpha_5\ne 0$ we have done. Note that $\alpha_2+\alpha_3+\alpha_4+\alpha_5=0$ means $\vec{v_2}+\vec{v_3}+\vec{v_4}+\vec{v_5}=\vec{0}.$
In other case, write $\vec{v_i}=\beta_i\vec{v_5}, i=1,2,3,4.$ Now, $$\vec{v_1}+\vec{v_2}+\vec{v_3}+\vec{v_4}-(\beta_1+\beta_2+\beta_3+\beta_4)\vec{v_5}=\vec{0}.$$ If $\beta_1+\beta_2+\beta_3+\beta_4\ne 0$ we have done. Note that $\beta_1+\beta_2+\beta_3+\beta_4=0$ means $\vec{v_1}+\vec{v_2}+\vec{v_3}+\vec{v_4}=\vec{0}.$
So we have to show that 
$$\alpha_2+\alpha_3+\alpha_4+\alpha_5=0=\beta_1+\beta_2+\beta_3+\beta_4$$ is not possible. In such a case $$\vec{v_1}+2\vec{v_2}+2\vec{v_3}+2\vec{v_4}+\vec{v_5}=(\vec{v_1}+\vec{v_2}+\vec{v_3}+\vec{v_4})+(\vec{v_2}+\vec{v_3}+\vec{v_4}+\vec{v_5})=\vec{0}+\vec{0}=\vec{0},$$ which is not possible by assumption.
