# Linear independence question (do 2 vectors who are not multiples of one another and a third which is not in their span form R^3?

The True/False question is:

Suppose that $v_1, v_2, v_3$ are in $\Bbb R^5$, $v_2$ is not a multiple of $v_1$, and $v_3$ is not a linear combination of $v_1$ and $v_2$. Then $\{v_1, v_2, v_3\}$ is linearly independent.

In the book it says it is a False statement, after looking in the web I found a source saying it was True. (neither of which give an explanation.

Personally I thought it was True, take $v_1$, and $v_2$, together they could span $R^2$, and if indeed $V_3$ is not in this span, all three would span $R^3$.

Is this correct or, if not why is the statement false?

Thank you

• It doesn't make sense to say that vectors in $\mathbb R^5$ span $\mathbb R^2$ or $\mathbb R^3$. – Santiago Canez Dec 24 '14 at 15:06
• Why? I thought span was the subset of R^n generated by a set of vectors. If you are in R^3, by taking 2 vectors which are not multiples of each other, could you not span R^2 with them. That is to say, a plane inside R^3? – user115919 Dec 24 '14 at 15:55
• @user11519 You can span a set which is isomorphic to $\Bbb R^2$ (a plane), but how can you span the set $\Bbb R^2 = \{(a_1,a_2)\mid a_i \in \Bbb R\}$ with any number of vectors $(b_1, b_2, b_3) \in \Bbb R^3$ (notice the difference in sizes)? – user137731 Dec 24 '14 at 17:08
• I see, thanks a lot! – user115919 Dec 24 '14 at 18:57

Admittedly, this is a bit of a tricky one. The statement is false because we could have $v_1 = \vec 0$, as in the counterexample $$v_1 = \vec 0\\ v_2 = (1,0,0,0,0)\\ v_3 = (0,1,0,0,0)$$
If we add to the statement "$v_1 \neq 0$", we would prove that it's true as follows:
suppose $c_i \in \Bbb R$ are such that $$c_1 v_1 + c_2 v_2 + c_3 v_3 = \vec 0$$ If $c_3 \neq 0$, then we can write $$v_3 = -\frac{c_1}{c_3}v_1 -\frac{c_2}{c_3}v_2$$ which is to say that $v_3$ is a linear combination of $v_1$ and $v_2$, contradicting our premise. So, $c_3 = 0$.
Now, suppose $c_2 \neq 0$. Then $$v_2 = -\frac{c_1}{c_2} v_1$$ which is to say that $v_2$ is a multiple of $v_1$, contradicting our premise. So, $c_2 = 0$
From here, we have $c_1 v_1 = \vec 0$. If $v_1 \neq \vec 0$, then we can conclude $c_1 = 0$.
Thus, $$c_1 v_1 + c_2 v_2 + c_3 v_3 = \vec 0 \implies c_1 = c_2 = c_3 = 0$$