# Vector space properties: distributivity

Tags:
1. Oct 1, 2016

### Math_QED

1. The problem statement, all variables and given/known data
I want to proof, using the axioms of a vector space, that:

$(\alpha - \beta)\overrightarrow a = \alpha \overrightarrow a - \beta \overrightarrow a$

2. Relevant equations

Definition vector space:

3. The attempt at a solution

$(\alpha - \beta)\overrightarrow a = (\alpha + (-\beta))\overrightarrow a = \alpha \overrightarrow a + (-\beta) \overrightarrow a$

I'm stuck here, I want to show that:

$(-\beta) \overrightarrow a = -(\beta \overrightarrow a)$

I showed that both have the same symmetric element. But I did not show that symmetric elements are unique. Is there an easier way as this is an exercise and I think I'm overseeing something.

EDIT: I cannot use $(-1)\overrightarrow a = - \overrightarrow a$

Last edited: Oct 1, 2016
2. Oct 1, 2016

### micromass

Staff Emeritus
Then prove it, so you can use it.

3. Oct 1, 2016

### Math_QED

That's funny, because I already proved it but it uses this result.

Can I derive, from the definition of vector space that:

$\overrightarrow a + \overrightarrow b = \overrightarrow a+ \overrightarrow c \Rightarrow \overrightarrow b = \overrightarrow c$. If yes, I can show uniqueness of inverse elements and then it's proven. Is there an easier way?

4. Oct 1, 2016

### micromass

Staff Emeritus
Yes, you should be able to derive this easily

This is the easiest way imo.

5. Oct 1, 2016

### Math_QED

I don't see how I can derive it that easily. Can you give me a hint?

6. Oct 1, 2016

### micromass

Staff Emeritus
Add the additive inverse of $a$.

7. Oct 1, 2016

### Math_QED

Yeah, thought about that too. Why am I allowed to do that on both sides?

8. Oct 1, 2016

### micromass

Staff Emeritus
What's wrong with it?

9. Oct 1, 2016

### Math_QED

I would use the property then, which I did not prove:

a + b = a + c => a + c + d = a + c + d

10. Oct 1, 2016

### micromass

Staff Emeritus

11. Oct 1, 2016

### Math_QED

Yes, for vectors:

$+ : V \times V \rightarrow V$ but how can I conclude that then?

12. Oct 1, 2016

### micromass

Staff Emeritus
What does it mean by definition that $+$ is a function?

13. Oct 1, 2016

### Math_QED

It means that:

$+ \subset ((V \times V) \times V)$ such that for every 2 vectors $(\overrightarrow x, \overrightarrow y) \in (V \times V)$ there is a unique vector $\overrightarrow z \in V$ such that $+(\overrightarrow x, \overrightarrow y) = \overrightarrow z$

14. Oct 1, 2016

### micromass

Staff Emeritus
And do you see that it means that if $x=x'$ and $y = y'$ then $+(x,y) = +(x',y')$? To prove this, let $z = +(x,y) = +(x',y')$ and use uniqueness of $z$.

15. Oct 1, 2016

### Math_QED

Hm let me try:

Proof: Let $(x,y) \in V \times V$

Then, there is a unique $z \in V$ such that $z = +(x,y)$. But, since x = x' and y = y', we have that $+(x',y') = z$ too. Since z is unique, we conclude that $+(x',y') = +(x,y)$

16. Oct 1, 2016

### Ray Vickson

How you can or should do it depends on exactly what vector-space axioms you start with. I have seen different books using different systems of axioms, so being very precise at this point is important.

BTW: you cannot "proof" something, but you might be able to prove it, or to give a proof.

17. Oct 1, 2016

### Math_QED

V is a commutative group for addition and for the scalar multiplication we have:

$1\overrightarrow a = \overrightarrow a$
$(\alpha \beta)\overrightarrow a = \alpha (\beta \overrightarrow a)$
$(\alpha + \beta)\overrightarrow a = \alpha \overrightarrow a + \beta \overrightarrow a$
$\alpha (\overrightarrow a + \overrightarrow b) = \alpha \overrightarrow a + \alpha \overrightarrow b$

where $\alpha, \beta \in \mathbb{K}, \overrightarrow a, \overrightarrow b \in V$