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.
 A: 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.
