Proving consequences of vector space axioms I have a couple of proofs that I have no idea are right or not. Here are my attempts:
Claim 1: Given a scalar $c$ in a field, $F$, and vector $v$ in a vector space, $V$, $-(cv) = c(-v)$.
Proof.
$$ -(cv) = -1 ⋅ (cv)$$
$$= (-1 ⋅ c)v $$
$$= (c ⋅ -1)v $$
$$= c ⋅ (-1v) $$
$$= c(-v) $$
QED.
Claim 2: Given any scalars $c$ and $d$ in a field, $F$, and a non-zero vector $v$ in a vector space, $V$, if $cv = dv$, then $c = d$.
Proof.
$$cv = dv \implies cv - dv = 0$$
$$\implies v(c - d) = 0$$
$$\implies v = 0$$ $$or$$ $$c - d = 0$$
$$\implies c = d$$ (since $v$ is nonzero)
QED.
Someone please let me know if these proofs are correct and if not, point out where I may have gone wrong. Thanks!
 A: Your answers are entirely correct, morally speaking. However, I would not consider these pure deductions from the vector space axioms (at least the axioms I am familiar with). For instance, in your first proof you say outright that $-(cv) = -1(cv)$. This is something that can be proved from the field and vector space axioms, so I would not consider it "primitive." In your second proof you say that if $(c - d)v = 0$ then $c - d = 0$ or $v = 0$. It all depends on how much detail you are expected to, and want to, put into this. I do think it's worth thinking through why every single step is a consequence of the axioms, at least when you are first learning this material. It's tedious, but rigor is important.
Here's some help with those two "issues" I pointed out. In the first one, you are claiming that for any $v \in V$, $-v = (-1)v$. Let's see how we can do this in more detail (but I am still skipping a bit).
$-v$, by definition, satisfies $v + (-v) = 0$. Let's now consider $v + (-1)v$.
$$
\begin{align*}
v + (-1)v &= 1v + (-1)v\\
&= (1 + -1)v\\
&= 0v\\
&= (0 + 0)v\\
&= 0v + 0v.
\end{align*}
$$
Thus, $0v = 0v + 0v$. Continuing, we get that $0 = 0v - 0v = 0v$, so $v + (-1)v = 0$. Recall that $v + -v = 0$, so $v + -v = v + (-1)v$. Then by subtracting from $v$ on both sides, we get $-v = (-1)v$.
For the latter, you want to show that $cv = 0$ means that $c = 0$ or $v = 0$. Well suppose that $c \neq 0$. Then we'd have an element $c^{-1} \in F$. Hence, we deduce
$$
\begin{align*}
0 &= cv\\
c^{-1} 0 &= c^{-1}(cv)\\
0 &= (c^{-1} c)v\\
0 &= 1v\\
0 &= v.
\end{align*}
$$
There were a few places where I skipped some steps in these deductions. For instance, when I subtract by $v$ from both sides I really need to mention associativity. Regardless, I hope this is a helpful example of how specific you may have to be to deduce statements directly from axioms.
