# Linear independence

I'm given the following question:
Given $v_1, v_2, v_3$ in $\mathbb{R}^5$; $v_2$ not a multiple of $v_1$; $v_3$ not a linear combination of $v_1$ and $v_2$. Must $\{v_1, v_2, v_3\}$ be linearly independent?
There is an equivalent question in the Ch. 1 Supplementary Exercises for Linear Algebra and Its Applications (Lay), the answer being true (the set must be linearly independent).
However, I think that there are cases where such a set could be linearly dependent. For example, the set $${ v_1= (0,0,0,0,0), \hspace{5pt}v_2= (1,1,1,1,2), \hspace{5pt}v_3=(1,1,1,1,3) }$$ The given conditions seem to hold, and $1\cdot v_1 + 0\cdot v_2 + 0\cdot v_3 = (0,0,0,0,0)$.
What do you think?

• I edited you title since it was too long and hard to read. Your'e welcome to revert my changes and name it differently. – Dennis Gulko Oct 17 '12 at 16:50
• You are right, this example shows that they are not l.i. – N. S. Oct 17 '12 at 16:50

Just a long "short" comment, didn't fit as a comment.

Checking linearly independence means solving

$$a_1v_1+a_2v_2+a_3v_3 =0 \,.$$

Now we first claim that $a_3=0$. Indeed, if this is not true, we get

$$v_3=-\frac{a_1}{a_3}v_1-\frac{a_2}{a_3}v_2 \,,$$

which is not possible.

if $a_3=0$ then we get

$$a_1v_1+a_2v_2 =0$$

exactly the same way you can prove $a_2=0$.

Thus you get

$$a_1v_1=0 \,.$$

So either $v_1=\bf 0$ or $a_1=0$.

So under the conditions of the problem, you have either $v_1=0$ or the vectors are linearly independent. As you observed the fist condition means the statement is wrong, but it also suggests that if one adds the extra condition $v_1\neq 0$ (of course this becomes a completely new problem), then the new problem/statement is true.

Your example is ok. You need an additional condition - that $v_1$ is not a multiple of $v_2$.