Linearly Independent Vectors and Colinearity (Proof Verification)

Question

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.

Answer 1

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.