# 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.

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You've (almost) said that $0 x = 0$ and that $a 0 = 0$ for all $a$ and $x$. Note that this is not the same as showing that if $ax = 0$, then $a = 0$ or $x = 0$. You need to start with the statement "suppose $ax = 0$", and then work from there. – Matt Jan 28 '12 at 23:40
Are you allowed to assume that your vector space is $\mathbb{R}^n$ with the usual scalar multiplication? Note that $ax$ is not equal to $ax_1+ax_2+\cdots + ax_n$; rather, $$ax = a(x_1,\ldots,x_n) = (ax_1,\ldots,ax_n).$$ – Arturo Magidin Jan 28 '12 at 23:42
Hardy, when you make a question, deal longer to read the answers before making a new one. You made two question in the last 30 minutes. – emiliocba Jan 28 '12 at 23:48
@emiliocba : i do read the replies and ask questions under the same discussion , but i have a half dozen problems here out many more which i could not solve, i am not going to wait hours between each question. This is an open forum, i am not here to spam anyone i have a genuine question and i honestly value what people have to say as i am learning and clearing my doubts, so please just let me create and ask questions that would be much appreciated. – Comic Book Guy Jan 29 '12 at 0:04

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 $\bf x$?

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I guess we can say each of xi is the ith component of the zero vector, hence x must be the zero vector. Quick question why did u have to add the assumption "vectors are n-tuples of the form", are n't they always ? – Comic Book Guy Jan 29 '12 at 0:16
@hardy I assumed they had $n$ components. You actually want to say "each $x_i$ is the scalar 0; so, $\bf x$ is the zero vector". – David Mitra Jan 29 '12 at 0:19

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

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Thanks buddy, i appreciate your help, i guess i did n't even understand the question correctly. I will attempt a proof based on your hint and write back here. – Comic Book Guy Jan 29 '12 at 0:24
I am unsure how i could use the scalar 1/a ? – Comic Book Guy Jan 29 '12 at 0:33
Not much you can do with a scalar and a vector, except multiply them. What happens if you multiply $\frac{1}{a}$ by $a\mathbf{x}$? – Arturo Magidin Jan 29 '12 at 0:39
we get x right as 1/a cancels with a as the multiplication with a scalar is associative, what would we do next ? – Comic Book Guy Jan 29 '12 at 0:51
@Hardy: That's just one side of the equation $a\mathbf{x}=\mathbf{0}$, though. – Arturo Magidin Jan 29 '12 at 0:53