# Linearly Independent Vectors and Colinearity (Proof Verification)

I'm practicing my proof-writing and was hoping you could let me know if this proof looks good. I would like to know if the proof is incorrect, if there are parts that are overly wordy/complicated, or if I'm missing some element of a proof that is helpful to see, if not strictly necessary.

The Prompt (from here):

Two nonzero vectors are linearly independent if and only if they are not colinear (or proportional, i.e. for two vectors (u,v), there exists λ ∈ R such that u = λ.v)

My Proof:

Sufficiency: Suppose two nonzero vectors v1 and v2 are linearly independent, but let them be colinear so that v1 = λv2. Then:

t1v1 + t2v2 = 0

λt1v2 + t2v2 = 0

v2(λt1 + t2) = 0

This implies that λt1 = -t2 even for nonzero values of t1 and t2, which contradicts the fact that v1 and v2 are linearly independent, so v1 and v2 must not be colinear.

Necessity: We prove the necessary condition by the contrapositive. Suppose v1 and v2 are not linearly independent, so that there is some nonzero t ∈ R2 such that t1v1 + t2v2=0. Then t1v1 = -t2v2 so v1 = -(t2/t1)v2 so v1 and v2 are proportional (by the scalar -t2/t1) and colinear. Since v1 and v2 not being linearly independent implies them being colinear, if they are not colinear then they are linearly independent.

Your proof is basically fine apart from the following points.

You shouldn't have $$=0$$' in each line in your chain of equations in the proof of sufficiency, and there is the possibility of one of the vectors being zero to be (briefly) considered. What I would suggest is something like this.

Suppose that $$v_1$$ and $$v_2$$ are linearly independent but that they are collinear. If $$v_1$$ or $$v_2$$ are equal to zero then $$v_1$$ and $$v_2$$ cannot be linearly independent, so suppose that $$v_1,v_2\neq 0$$. Since $$v_1$$ and $$v_2$$ are collinear we have $$v_1=\lambda v_2$$ for some $$\lambda\neq 0$$. Now for any $$t_1,t_2\in\mathbb{R}$$ we have

$$\begin{array}{ll} & t_1v_1+ t_2v_2\\ = & \lambda t_1v_2 + t_2v_2\\ = & v_2(\lambda t_1 + t_2)\end{array}$$

But if $$t_1\neq 0$$ and $$t_2=-\lambda t_1$$ we get $$v_2(\lambda t_1 + t_2)=0$$, which contradicts the fact that $$v_1$$ and $$v_2$$ are linearly independent. Therefore $$v_1$$ and $$v_2$$ are not collinear.

Of course it's important to note that at least one of $$t_1,t_2$$ is non-zero (as you had in your own proof) to make sure you have a contradiction.

In your proof of necessity you have $$t$$ instead of $$t_1$$ and $$t_2$$. I would suggest something like ... so there are $$t_1,t_2\in\mathbb{R}$$, not both zero, such that $$t_1v_1+t_2v_2=0$$. Suppose that $$t_1\neq 0$$ (the case where $$t_2\neq 0$$ is similar). Then $$t_1v_1=-t_2v_2$$...'

Finally, the word is usually spelled collinear. Apart from that it's a fine proof.