(I'm not sure if the tittle is a good description for this question, since I'm not even sure what the question is really asking. Hopefully you can point me in the right direction, and edit the title if necessary.)
Question:
$\begin{array}{l}{\text { Let } C \text { be an } m \times n \text { matrix with real entries. For vectors } u, v \in \mathbb{R}^{m},} \\ {\text { we define } u \sim v \text { if there exists a vector } x \in \mathbb{R}^{n} \text { such that } C x=u-v} \\ {\text { Prove that for all } u, v, w \in \mathbb{R}^{m}:} \\ {\text { i) } u \sim u} \\ {\text { ii) if } u \sim v \text { then } v \sim u} \\ {\text { iii) if } u \sim v \text { and } v \sim w \text { then } u \sim w} \end{array}$
Answer to i)
$u \sim u \implies C\mathbf{x} = \mathbf{u} - \mathbf{u} = \mathbf{0}$ . We know that in all $R^{m} \ \exists \ \mathbf{0}_{n}$ , hence i) is trivially true
This is what I have scribbled down for ii)
We have an implication. Hence, we need to show that if the LHS holds, then the RHS must hold. Let's assume that $u \sim v$ holds, then what does it mean?
It means that if I have a matrix $C_{m\times n}$ and post-multiply it by a vector $\mathbf{x}_{n}$ then I can write the result as a linear combination of two vectors $\mathbf{u}_{m} , \mathbf{v}_{m}$
$\mathbf{v} \sim \mathbf{u} \implies C\mathbf{x} = \mathbf{v} - \mathbf{u}$. But note that $\mathbf{v} - \mathbf{u} = -(\mathbf{u} - \mathbf{v})$ , and then can be written as a linear combination of $\mathbf{u} \sim \mathbf{v}.$ Hence, $\mathbf{u} \sim \mathbf{v} \implies \mathbf{v} \sim \mathbf{u}$
Is this what is being asked for? If so, are there any giant leaps of logic?
I get what you're trying to get at for the second part, but it feels a little ... floaty I guess? Like you yourself aren't sure. The third paragraph in addition feels like you initially assume the conclusion right off - you work backwards to the implication, which is fine sometimes, but a bit of a curveball to read. Whenever possible, try to start with your given premises, and work from them to your desired conclusion, as opposed to the other way around.
Let's say $\vec u \sim \vec v$. Then there exists some vector $\vec x$ for our fixed matrix $C$ such that $C\vec x = \vec u-\vec v$. With the premise as given, we want to show $\vec v \sim \vec u$.
We can augment the $\vec u \sim \vec v$ relation by multiplying through by $-1$ as so:
$$C\vec x = \vec u-\vec v = -(\vec v-\vec u) \implies C(-\vec x) = \vec v-\vec u$$
Thus, there exists some vector, namely $-\vec x$, such that $\vec v-\vec u$ is $C$ times that vector, giving $\vec v \sim \vec u$ whenever $\vec u \sim \vec v$.