Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 2 down vote accepted

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 $\bf x$?

share|cite|improve this answer
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 $\mathbf{x}=\mathbf{0}$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.