Question with regards to logic of proof:

Question

Proof: Let $V$ be a vector space over $\mathbb{R}$. Let $a,b \in V$ and $a,b \neq 0$ . If $a, b$ are linearly dependent, then there exists w $\in \mathbb{R}$ such that $a=wb.$

I know that if $a,b$ are linearly dependent then the equation $$K_1a+K_2b=0$$means that $K_1=K_2=0\ $ is not the only solution, so $K_1$ or $K_2$ is non-zero, and if I set $$w = -K_2/K_1$$ (WLOG assuming $K_1$ is non-zero) I have showed the existence of the scalar. But does the following show the uniqueness of the scalar up to coeffiecients?

$$K_1a+K_2b=0\implies K_1a=-K_2b,\ \text{ and so }\ a= \frac{-K_2}{K_1}b$$

Edit:

Thank you guys for your replies. I was thinking about this in terms of logic. To show uniqueness one can show through contradiction or to show the existence of an element that satisfies the condition and then proving that for any other element x that satisfies the condition for x implies x is the existing element. What I have shown does not show uniqueness of the solutions, because $K_1=K_2=0$ is already a solution. What I have shown is that if I strictly work under the assumption that $K_1 \neq 0$ then the solution is unique.

Answer 1

Yes, for two non-zero vectors, being dependent means that each is a multiple of the other, and uniqueness of multiplication means that there is only one multiple that works. You do want to be a little careful in not assuming that your $K_1 \neq 0$ in the operation $-\frac{K_2}{K_1}$ (but you can easily verify that separately).

In other words, if $a = w_1 b$ and $a = w_2b$ for $w_1 \neq w_2$ then

$$b = \frac{w_1 - w_2}{w_1 - w_2} b = \frac{w_1b - w_2b}{w_1-w_2} = \frac{a - a}{w_1 - w_2} = 0$$

Answer 2

It looks alright except that you may be dividing by zero. Note that if $K_1$ is zero then $K_2b = 0$ too, which implies $K_2 = 0$, in contradiction. From here your proof is fine.

Answer 3

No, because the other solution is always $$K_1 = K_2 = 0.$$

(Yes, there must be at least one other, which you selected.)

By the way, at the end you expressed unambiguosly the vector $a$, NOT the pair of scalars $K_1, K_2.$