In any vector space, ax=bx implies a=b The above statement is listed as false in my text, and I wanted to be sure I understood why that is. (I guess if it were written "properly" it would be $a\mathbf{x} = b\mathbf{x}$ implies $a = b$). 
Given the axioms we were given, it would seem that the statement should be true, no? 
A related statement -- also listed as false -- is that "in any vector space, $a\mathbf{x} = a\mathbf{y}$ implies that $\mathbf{x} = \mathbf{y}$." Again, given the axioms we have I am not sure why this is the case. 
 A: Directly from the axioms defining a vector space:

Claim: in a $K$-vector space $V$, $\lambda\cdot x=0_V$ if and only if $\lambda =0_K$ or $x=0_V$.

Proof: the facts that $0_K\cdot x=0_V$ and $\lambda\cdot 0_V=0_V$ for all $\lambda \in K$ and all $x\in V$ are two of the axioms. Conversely, assume $\lambda \cdot x=0_V$. If $\lambda \neq 0$, then it is invertible and
$$
0_V=\lambda^{-1}\cdot 0_V=\lambda^{-1}\cdot(\lambda\cdot x)=(\lambda^{-1}\lambda)\cdot x=1_K\cdot x=x.
$$
So $\lambda =0_K$ or $x=0_V$. QED.
First question: this is false as
$$a\cdot x= b\cdot x\iff (a-b)\cdot x=0_V\iff a-b=0_K\mbox{ or } x=0_V\iff a=b\mbox{ or } x=0_V.
$$
Take $a=0_K$, $b=1_K$, and $x=0_V$ for a counterexample. But what you mention becomes true with th assumption $x\neq 0_V$.
Second question: this is false as
$$
a\cdot x=a\cdot y\iff a\cdot(x-y)=0_V\iff a=0_K\mbox{ or } x-y=0_K\iff a=0_K\mbox{ or } x=y.
$$
Take $a=0$ and any $x\neq y$ for a counterexample. This becomes true if you further assume $a\neq 0_K$.
