Proving that if $a\mathbf{x}=\mathbf{0}$ then  $a=0$ or $\mathbf{x}=\mathbf{0}$ Prove that if x is a vector and a is a scalar, then the following relation holds ?
1) if ax = 0, then either a = 0 or x = 0 ( or both).
This is trivial although i am unsure if my steps are correct.
step 1 
Lets pick $a = 0$ and $x  = (x_1, x_2,\ldots, x_n)$ where for all $x_1,\ldots, x_n$ are not zero.
$ax = ax_1 + ax_2 + \cdots + ax_n$ as multiplication by scalars is distributive.
Now can I just state $ax = ax_1 + ax_2 + \cdots + ax_n = 0$ as a is zero every where there doesn't seem to be any axiom of vector space which i could quote as reasoning there or is there ?
step 2
Lets pick $a\neq 0$ and $x  = (x_1, x2,\ldots, x_n)$ where for all $x_1,\ldots x_n$ are zero.
Again as before $ax = ax_1 + ax_2 + \cdots + ax_n$ as multiplication by scalars is distributive.
I know there is a zero vector in the vector space and according to my assumption $x$ is a zero vector but how do i justify $ax = 0$ ?
Any help or guidance would be highly appreciated.
 A: Your operations are incorrect, even assuming that you are allowed to assume that a vector is a "tuple". Note that for $\alpha$ a scalar and $\mathbf{x}=(x_1,\ldots,x_n)$, the usual scalar multiplication is defined to be
$$\alpha\mathbf{x} = \alpha(x_1,\ldots,x_n) = (\alpha x_1,\ldots,\alpha x_n).$$
You have $\alpha x_1+\cdots \alpha x_n$, which would make it a scalar, not a vector. And distributivity of scalar multiplication has nothing to do with it.
Also: even if correct, your argument would only have established that (i) if $a=0$ and $\mathbf{x}\neq\mathbf{0}$, then $a\mathbf{x}=\mathbf{0}$; and (ii) if $a\neq 0$ and $\mathbf{x}= \mathbf{0}$ then $a\mathbf{x}=\mathbf{0}$. But this is not what you need to prove! What you need to prove is the implication going the other way: if $a\mathbf{x}=\mathbf{0}$, then $a=0$ or $\mathbf{x}=\mathbf{0}$ (or both).
To that end, you would begin with: "assume that $a$ is a scalar, and $\mathbf{x}$ is a vector, and $a\mathbf{x}=\mathbf{0}$..."
Hint. If $a=0$, then we are done. The only other alternative is that $a\neq 0$. If $a\neq 0$, then $\frac{1}{a}$ makes sense, and is a scalar. Now use two of the properties of scalar multiplication to show that if $a\mathbf{x}=\mathbf{0}$ but $a\neq 0$, then $\mathbf{x}=\mathbf{0}$.
A: I'll assume that your vectors are $n$-tuples of the form $(x_1,x_2,\ldots,x_n)$.
As mentioned by Matt in his comment, you are not going about this in the right way.
The hypothesis is that $a\bf x=o$. So, 
what you need to assume is that $a\bf x=0$. 
Then you need to show that the result of the statement is true: you need to demonstrate that either $a=0$ or $\bf x=0$.
To do this: let's assume that $a\ne0$.  Now we need to demonstrate that $\bf x=0$. 
By the definition of scalar multiplication:
$$
a{\bf x}=(ax_1, ax_2,\ldots,ax_n).
$$
But, as we asummed at the start, we  have $(ax_1, ax_2,\ldots,ax_n)=\bf0$. The only way this can happen is for each component to be zero. That is
$$
ax_1=0, ax_2=0,\ldots ax_n=0.
$$
Now, recalling that we have $a\ne0$, what can we say about each $x_i$? Then, what can we say about $\bf x$? 
