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.
Best Answer
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 $\mathbf{x}=\mathbf{0}$.